text:
Le muscle fléchisseur commun des orteils ou muscle long fléchisseur des orteils (Musculus Flexor Digitorum Longus) est un muscle du plan profond de la face postérieure de la jambe qui s'étend de la face postérieure du tibia aux phalanges distales des quatre derniers orteils

Anatomie
Origine : il s'insère sur la face postérieure du tibia depuis en dessous de la ligne du muscle soléaire jusqu'à 7 à  au-dessus de l'extrémité inférieure du tibia et en dedans de l'insertion du muscle tibial postérieur ainsi que sur l'aponévrose recouvrant le muscle tibial postérieur.

Trajet : les fibres musculaires se jettent sur un fin tendon, qui parcourt pratiquement toute la longueur de la face postérieure du muscle. Ce tendon passe derrière la malléole interne, dans une rainure, commune avec celle du muscle tibial postérieur, mais séparé de celui-ci par un septum fibreux, chaque tendon étant contenu dans sa propre gaine synoviale au niveau de la cheville.

Après être passé superficiellement par rapport au ligament latéral interne de la cheville, il passe sous la plante du pied où il passe en dessous du tendon du muscle long fléchisseur de l'hallux.

Terminaison : il se termine par quatre tendons qui reçoivent chacun un muscle lombrical du pied sur leur bord externe avant d'aller se fixer sur la base de la face inférieure des quatre derniers orteils (pas sur le gros orteil) après être passé dans le tendon du muscle court fléchisseur des orteils. Le tendon du muscle long flechisseur des orteils passe au-dessus du sustentaculum tali (situé à la face médiale du calcaneum) et va sous croiser le tendon du muscle long flechisseur de l'hallux en se dirigeant latéralement sous la voûte plantaire.

Innervation

Il est innervé par le nerf tibial

Vascularisation

Il est vascularisé par une branche de l'artère tibiale postérieure.

Action

Il est essentiellement fléchisseur des orteils, fléchissant la  sur la , la  sur la 1.

Muscles antagonistes

Le muscle extenseur commun des orteils et le muscle court extenseur des orteils.

Galerie

fléchisseur commun des orteils

redpajama_set_name: RedPajamaWikipedia


There are some files that can be opened in the Showbie iOS app but you don't have all of the editing capabilities when it's opened in Showbie. In this case, opening the file in another app might give you the features you're looking for.
If you're working with a different file, it's the same process! Just make sure you choose the right app for your file.


Kaylee was born on a beautiful summer day, born healthy and normal with an Apgar score of 9. Never once did the word "autism" cross my mind as a possibility for her. No one in my family had ever been diagnosed with autism before.
Kaylee hit her developmental milestones on time, she was a happy baby who was social, loved music, and enjoyed reading books a few times a day with me. She used to love to share her food with me, and she loved it when I would say "How big is Kaylee?" and she would happily respond with her hands over her head "so big!" At twelve months of age, Kaylee pulled my face to hers and gave me a big kiss.
A few weeks later, Kaylee became an empty shell of herself. She no longer kissed me, or said "so big" anymore.
She didn't respond to her name, and she lost her speech altogether.
I sloughed it off as just a little setback; I never dreamed that there was something very wrong with my baby. At 16 months she still had not redeemed her lost speech, so one day I decided to do an internet search about "child not talking at 16 months." The dreaded word, Autism, appeared. And although I knew in my heart that it was true, I didn't let myself believe it.
One year later, my daughter was officially diagnosed with what was called a "moderate" case of Autism Spectrum Disorder by one of the top specialists in the state of Illinois. I broke down right there in the doctor's office. I had still held out hope that maybe I was wrong, maybe she didn't have Autism.
She was very thin, pale, had dark circles under her eyes. She couldn't sleep through the night; she had terrible bouts of diarrhea, a constantly runny nose, and transient low grade fevers, all of this besides her autistic behaviors.
One late night, I was awakened by the sound of Kaylee laughing and stimming in her room. I fell to my knees and cried out to God to help me figure out what was wrong with her. I wrote in my journal that night through tears, "I want to know her, I want to know who Kaylee Joy is!" That night I felt that He told me He would lead me to the right place and that she would get better.
I went to bed hurting but with some peace.
One month later, a friend told me about a clinic she went to in New York that did biomedical treatment for Autism by addressing immune system dysfunction. I went to New York with Kaylee one month after that and saw Dr. Bruce Russell and we started treatment right away. Kaylee went on antiviral and antifungal medication and she started to open up and come into our world.
Three weeks after starting Valtrex (an antiviral medication) Kaylee looked into my eyes and called me "Mom". My heart swelled with joy because she had never addressed me as a person before. Kaylee started sleeping through the night and answering simple questions, everyday slowly getting better and better.
Then we addressed her allergies with a special treatment called P/N (otherwise known as Provocation/Neutralization), and her tantrums, hyperactivity, allergies, and food intolerances improved immediately.
Over one year ago, Kaylee was retested by her school district and the result was that she no longer fit the diagnosis of Autism Spectrum Disorder. Today, Kaylee does not receive speech therapy, no longer has "sensory" needs, and operates in a mainstream 1st grade classroom without an aide.
Don't believe the widespread lie that Autism is genetic and that children are born this way. Don't believe the therapists when they tell you everything your child won't do. Don't believe that there is nothing you can do for your child. Don't give up. Don't be afraid to hope.


This task-specific rocker patch identifies how your service dog assists you. Our "Guide" patch is 3.5" wide and fits nicely on the top or side of a vest, shirt, or sweater.
This patch may be used individually or in combination with other patches, although most handlers use one on each side of the garment.
Patches are sold individually, not in pairs. This is not an iron-on patch. Although we recommend sewing patches onto a vest, shirt, sweater or jacket, an appropriate adhesive may be also be used.
Our Guide Patch is a 3.5" wide crescent-shaped embroidered patch, lettering is White/Red with a black border, with the word "Guide".
This patch can be used individually or in combination with other patches. These are not iron on patches.
Patches are sold individually, not in pairs. There are no laws that dictate where or how many patches should be on a vest. Most people use two of each patch, one for each side.
For every purchase made, a portion of the proceeds goes to helping care for homeless pets across the United States.


Tilgate Community Centre is in Shackleton Road, close to Loppets Road and a short distance behind the shopping parade in Ashdown Drive.
Click on the link to find Tilgate Community Centre on Google Maps.
Follow the link below for a video tour of Tilgate Community Centre. (Please note the link will take you to YouTube).
Here is some information about Tilgate Community Centre. If you have questions about this centre, please contact us. Our details are on the right.
DisabledGo has produced an access guide for Tilgate Community Centre .


We're pleased to announce the launch of the new Vibratech TVD website. Vibratech TVD is the largest U.S. manufacturer of viscous dampers. Don't worry...I didn't know what a viscous damper was prior to working with Vibratech TVD, but it's actually a pretty cool product. One of the best parts of helping manufacturers succeed online is learning about all of this cool stuff!
In a nutshell, a viscous damper helps reduce engine ware by absorbing vibration and shock. Obviously there is more to it than that, but I am happy to say that Vibratech TVD does a great job educating people through their website content about viscous dampers, so I'll leave that explanation to the pros!
You can check out their website at www.VibratechTVD.com. We'd love to hear what you think in the comments below!


text:
Q: Keyspaces in same cassandra cluster with different passwords I am connecting to two different keyspaces with different credentials. These keyspaces are in the same cluster.
Currently, I am needing to create two different cluster beans in my spring boot app to achieve this as the credentials are set in the cluster. Credentials are not set in the session object.
For this scenario is it right to have two separate cluster beans? Can I avoid making two different cluster beans?

A: In Cassandra there is no such things as a password on the keyspace. Password is set for a user that has some roles, and then this role is given particular access to the keyspace - modify, read, etc. So to have different access rights to different keyspaces, you need to have different users, and to connect as different user from the same application, you need to have different Cluster objects for every user (except the case if you're using DSE with DSE Java driver, where you can have so-called proxy user).

A: You can have a single spring Cassandra Cluster bean and create a two separate session beans by using the single cluster bean and by setting the corresponding key space while creating the session beans. A sample implementation from spring data docs

redpajama_set_name: RedPajamaStackExchange


text:
Q: How can I have ruby logger log for each line of execution? I am very new to ruby, just started learning for doing a automation with cucumber and Gherkin.
I wanted to capture the each and every action into the logs!!!
Is there a way to capture!!! Thanks for your advice in advance!!!
Thanks
Ashok Natarajan

A: If your question is about cucumber logging, check out the man page for available options. Try:
cucumber -v -b -x

-b, --backtrace                  Show full backtrace for all errors.
-v, --verbose                    Show the files and features loaded.
-x, --expand                     Expand Scenario Outline Tables in output.

A: The easiest way to log things in cucumber is using simple puts statements. if you don't want it to show up in the reports you could use kernel.puts
I have also used pry for real time debugging like this:
http://www.alanmacdougall.com/blog/2012/06/08/interactive-debugging-with-pry/
You must also remember to gather any output from gems that you are using. For example if you use the sequel gem you will probably want to log the database statements. you could use something like
require 'logger'
DB.loggers << Logger.new($stdout)


text:
List of athletics events
Title: List of athletics events
Subject: 2014 World Junior Championships in Athletics, 2012 World Junior Championships in Athletics, 2010 World Junior Championships in Athletics, 55 metres hurdles, 55 metres
Athletics comprises a variety of running, jumping, throwing and walking events.
The sport of athletics is defined by the many events which make up its competition programmes. All events with the sport are forms of running, walking, jumping or throwing. These events are divided into the sub-sports of track and field, road running, racewalking and cross country running.
The Olympic athletics programme has played a significant role in shaping the most common events in the sport. The IAAF World Championships in Athletics is the foremost World Championship event, holding the vast majority of World Championship-level events within the competition. A small number of events, such as the 60 metres, are exclusive to the IAAF World Indoor Championships.
Two further, separate World Championship events are held for their specific events: the IAAF World Half Marathon Championships and the IAAF World Cross Country Championships. Cross country is one of many events which have appeared at Olympics but are now not part of the Olympic athletics schedule. Some events, such as the mile run, remain very popular at competitions, despite having neither Olympic, nor World Championship status.
1 Olympic and World Championship events
1.1 Current events
1.2 Former championship events
2 Other common events
Olympic and World Championship events
Key: Present in current competition programme
Olympic status
World Championship status
World Indoor Championship status
60 metres 1900 – 1904
1985 – Present 1985 – Present
100 metres 1896 – Present 1928 – Present 1983 – Present 1983 – Present
200 metres 1900 – Present 1948 – Present 1983 – Present 1983 – Present 1985 – 2004 1985 – 2004
400 metres 1896 – Present 1964 – Present 1983 – Present 1983 – Present 1985 – Present 1985 – Present
800 metres 1896 – Present 1928, 1960 – Present 1983 – Present 1983 – Present 1985 – Present 1985 – Present
1500 metres 1896 – Present 1972 – Present 1983 – Present 1983 – Present 1985 – Present 1985 – Present
1983 – 1993 1985 – Present 1985 – Present
5000 metres 1912 – Present 1996 – Present 1983 – Present 1995 – Present
10,000 metres 1912 – Present 1988 – Present 1983 – Present 1987 – Present
60 metres hurdles
100 metres hurdles
110 metres hurdles 1896 – Present
(except 1912) 1984 – Present 1983 – Present 1983 – Present
3000 metres steeplechase 1920 – Present 2008 – Present 1983 – Present 2005 – Present
Half marathon (road)
(except 2006) 1992 – Present
(except 2006)
Marathon (road) 1896 – Present 1984 – Present 1983 – Present 1983 – Present
20 kilometres race walk (road) 1956 – Present 2000 – Present 1983 – Present 1999 – Present
50 kilometres race walk (road) 1932 – Present
Cross country running 1912 – 1924
4×100 metres relay 1912 – Present 1928 – Present 1983 – Present 1983 – Present
4×400 metres relay 1912 – Present 1972 – Present 1983 – Present 1983 – Present 1991 – Present 1991 – Present
Pole vault 1896 – Present 2000 – Present 1983 – Present 1999 – Present 1985 – Present 1997 – Present
High jump 1896 – Present 1928 – Present 1983 – Present 1983 – Present 1985 – Present 1985 – Present
Long jump 1896 – Present 1948 – Present 1983 – Present 1983 – Present 1985 – Present 1985 – Present
Triple jump 1896 – Present 1996 – Present 1983 – Present 1993 – Present 1985 – Present 1993 – Present
Shot put 1896 – Present 1948 – Present 1983 – Present 1983 – Present 1985 – Present 1985 – Present
Discus throw 1896 – Present 1928 – Present 1983 – Present 1983 – Present
Hammer throw 1900 – Present 2000 – Present 1983 – Present 1999 – Present
Javelin throw 1908 – Present 1932 – Present 1983 – Present 1983 – Present
Pentathlon 1912 – 1924 1964 – 1980
Decathlon 1904 – Present
Former championship events
200 metres hurdles 1900 – 1904
3000 metres team race 1912 – 1924
Short course cross country
1998 – 2006 1998 – 2006
3 kilometres race walk (track)
10 kilometres race walk (track) 1912 – 1924, 1948 – 1952 1992 – 1996
10 kilometres (road)
Medley relay 1908
Marathon road relay
Standing high jump 1900 – 1912
Standing long jump 1900 – 1912
Standing triple jump 1900 – 1904
56 pound weight throw 1904, 1920
NB: This list excludes Summer Olympic events which featured only once on the Olympic athletics programme during the first six editions
Other common events
100-yard dash
200 metres straight
Middle-distance
880-yard run
Mile run
Two miles
4×800 metres relay
4×1500 metres relay
Distance medley relay
Sprint medley relay
Swedish relay
Ekiden
Road events
5 kilometres road race
5-mile road race
10-mile road race
Quarter marathon
25 kilometres road race
One hour run
12-hour run
Multiday race
Octathlon, combined event featured at the IAAF World Youth Championships in Athletics
Throws Pentathlon a championship combined event at the World Masters Athletics Championships
Fierljeppen, Frisian sport of pole vaulting for distance
Softball throw, athletics event throwing for distance using a softball
List of world records in athletics
Disabled athletics
Common track and field events. IAAF. Retrieved on 2010-05-21.
Athletics events
Racewalking
Backward running
One mile
2000 m steeplechase
4 × 100 m
4 × 1500 m
Weight throw
Softball throw
High jump (standing)
Long jump (standing)
Triple jump (standing)
Women's pentathlon
Throws pentathlon
Current Olympic events shown in italics
Athletics (track and field)-related lists
Events in athletics (track and field)
England, Track and field, Road running, International Cross Country Championships, Scotland
Iaaf, Long jump, Shot put, High jump, Pole vault
2014 World Junior Championships in Athletics
United States, Kenya, United Kingdom, Germany, Russia
United States, Kenya, Russia, United Kingdom, Germany
United States, Kenya, United Kingdom, Russia, Canada
Hurdling, Ekiden, 60 Metres Hurdles, Sport of athletics, Track and field
Sprint (running), Track and field, 100 Metres, 60 Metres, United States

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Azure - Basic App Comunication c# vs 2012 I have to write an application with azure in c# that:
has a worker role comparing gps information
stores user profile information
and can get and set both of these from user device.
for example what i want is for my device to periodically upload its gps co-ordinates to the worker by opening a connection with the worker role, authenticate the user, allow saving the gps co -ordibates over any existing co-ordinates for this user. Also or as a separate worker i don't know, i need the worker role to be checking and performing calculations on the gps' of all devices. Which at certain points would then push the some information down to a particular device.
What i want doesnt seem difficult but finding just a simple example of a tcp message from  a device to an azure service ive spent hours looking its the comunications between devices that im not understanding

A: TCP programming with Azure is possible, but not practical if you need the TCP endpoint available at all times. A TCP endpoint cannot remain inactive for long periods of time in Azure.
You can certainly use Web Services as Ondrej suggests. Or you can also use Azure Queues.  I wrote a blog about scaling with Azure Queues. It's a message based solution that runs on top of REST calls.
What I am not clear about is your need to reach out to all the devices from Azure directly. Unless I misunderstood your scenario, you want an Azure role to "connect" to all existing devices, get data from them, perform some calculation in Azure, then push back the data to a device? This seems rather hard to do and error prone if your devices are not accessible - if your GPS devices register with the worker roles, and can be accessed by public DNS I guess you could do that. But personally I would lean towards a solution that only pushes data from the device into the cloud; it should be much easier that way.  What kind of devices are you referring to? And what are you trying to do?


I had the pleasure of photographing Amy and Matt's wedding at The Blazing Donkey in Ham, near Sandwich (makes me chuckle every time!) It was another incredibly sunny day but, fortunately, some shady spots were forthcoming. Their wedding gallery will be appearing shortly.


Barker men's shoes aren't just about keeping you looking smart – they're also great at casual styles too, as the Barker Axel trainers prove. This modern take on a sneaker is already proving to be popular with our customers, and that's partly due to the incredible range of colours these trainers are available in. They also benefit from stylish punch and stitch detailing, a rubber cup sole for all day comfort and a full leather lining. Choose your favourite colour from calf leather or suede, and take your casual style to a new level of sophistication.


Gordon Ripley is a photographer who specializes in nature and landscape photography. When he is not out taking pictures he may be found doing photo restorations or giving lessons in using Photoshop to process digital photographs or basic aspects of photography for beginners. While interested in photography throughout most of his life, he did not begin to take his photography seriously until 1999 after retiring from a long career in computers. At that time he joined a local camera club and made good use of the club competitions to learn the rules of �competition photography�.
Over the past few years Mr. Ripley has been the president of two major camera clubs and is currently past president of the New England Shutterbugs Camera Club located in Leominster, Massachusetts. Since 1999, Mr. Ripley has planned and led members of his camera club on extensive trips to Yellowstone National Park, The Grand Tetons National Park, Western Colorado (for wild flowers and for fall foliage), The Everglades, and most recently the Eastern Sierras of California. He also has visited Oregon to photograph the coast as well as waterfall photography while hiking into distant locations in the Columbia River Gorge. In the New England area he has led an extensive number of photography trips with extensive knowledge of the best Photography locations in Maine, New Hampshire and Vermont.
Gordon is a member of the photographic Society of America, The New England Camera Club Council, and the Monadnock Area Artists Association. He resides in southwestern New Hampshire.


The Butterfly ® Mini Table Tennis Table is perfect for youngsters learning to play the game, and is suitable for play at home, in the hallway, lounge, garden, garage, playroom . . . in fact anywhere.
12mm thick wooden playing surface.
4 easy to assemble legs.
Comes in its own carrying case with convenient handle.
Comes complete with 2 mini sponge bats & 2 40mm balls.
Size (in use): 58 x 29cm.
Size (folded): 30 x 29cm.


Well, whereas I felt the problem against Quinnipiac was effort, or lack thereof during the beginning of the game, I thought the effort was there last night against Boston College. Execution was the issue however. Good idea passes ended up in skates or a foot behind players, shots went just wide, and checks in the corners were missed. That pretty much summed up the night for the Minutemen. They certainly had their chances and a strong effort by Paul Dainton helped keep them in the game but in the end they just were not able to capitalize.
I'm worried with how conservative the offense has looked suddenly. They suddenly look much crisper when they have to abandon the offensive scheme and go all out in order to come from behind. Why not play that type of hockey from the start? I thought with BC's inexperience it would be important to forecheck and do everything possible to test their young defensemen. Yet at the beginning of the game UMass was trying to generate all their shots from the point instead of cycling it down low and the forecheck was hardly used at all. UMass simply must use their speed to generate more quality shots. Of the Minutemen's 24 shots 13 came from the forwards and 11 from the defense. This doesn't even count the many shots taken from the point that never got through. I like our defensemen's offensive play but the skilled forwards must be worked into the offense better, otherwise UMass will continue to make inexperienced defenses look spectacular like last night.
Tonight's game should be interesting as you have two once-hot teams that are now struggling with UMass dropping a couple and Lowell losing their third straight. Is this a must-win for the teams? Probably not, it's only December after all. But I would hope to see some signs of determination from the team tonight.
Could I have picked James Marcou for his nifty wraparound backhand goal in the first? Sure. Could I have picked Paul Dainton for his 29 save effort that kept UMass in the game when they weren't generating any offense? Definitely. But I really liked the game that Hobbs played last night. He was in the corners digging pucks and played a great two way game in my opinion. I thought he looked good nearly every shift he was out there.
Here's the recap from the official athletics site as well as one from USCHO.
Toot says BC deserved the fate they got in the Republican's story.
Here's Matty Vautour's take in the Gazette.
As mentioned Lowell lost their third in the row as they were beaten by UNH. Elsewhere Merrimack continues to play well with their win over Northeastern, Maine beat Providence up north, and UVM and BU tied.
Coach Cahoon will be speaking at the Garrett Regan Hockey Summit, a combination of high school hockey games, seminars, and an expo featuring local colleges and hockey vendors. The summit is named in honor of the late Hingham hockey coach.


text:
Public Service Recognition Week
This week we proudly celebrate the more than 600,000 members and 400,000 retirees of the New York State and Local Retirement System (NYSLRS) for their service to the people of New York State.
A Brief History of Public Service Recognition Week
Public Service Recognition Week was created in 1985 to honor the men and women who serve our nation as federal, state, county and local government employees. They dedicate their careers — and sometimes their lives — to keep others safe and provide for the common good. Their work makes life in our communities better.
Congress officially designated the first week of May as Public Service Recognition Week. This year, it is being celebrated May 6 through May 12.
The Public Servants of NYSLRS
NYSLRS is full of stories about State workers and municipal employees finding value and meaning in the work they do, especially when they help other New Yorkers. The NYSLRS members that work here in the Office of the State Comptroller (OSC) are no exception.
For example, there's Dan Acquilano, who works in the Local Government and School Accountability division of the Comptroller's Office. He travels across the state teaching municipalities how to manage their financial operations. His efforts have helped many municipalities and school districts manage their way out of financial distress. Or, there's Derrick Senior, who began his career as a mental health counselor. But, for the past 13 years, he's helped keep OSC's technology up and running at the CIO-Service Delivery Department. And, there's Stacy Marano. She leads a team of attorneys, investigators and forensic auditors to ensure entities receiving state money – vendors, politicians, municipalities and state agencies – don't misuse public funds. Her work helps law enforcement develop cases and root out fraud.
These are stories you may not hear about, but their work and the efforts of thousands of other public servants help make New York State a better place to live.
View more NYSLRS proud public servants
Whether they are protecting our communities, fighting fires, clearing our roads after snowstorms or simply helping government function better, NYSLRS members deliver the critical resources and services many New Yorkers depend on. Even outside of work, many NYSLRS members and retirees give back to our state by serving their communities as volunteers and supporters of charitable causes.
Comptroller DiNapoli's Faith in Public Service
New York State Comptroller Thomas P. DiNapoli is the administrator of NYSLRS and trustee of the Common Retirement Fund. His public service career began when he was elected as a trustee to the Mineola Board of Education at the age of 18, making him the first 18-year-old in New York State to hold public office. Comptroller DiNapoli is understandably proud about the career path he has chosen, and he often speaks about the contributions that New York's public employees make, not just as engaged citizens, but as individuals who bring value to the communities where they live.
Content last revised June 14, 2021.
This entry was posted in General News and tagged Comptroller Thomas P. DiNapoli, New York State & Local Retirement System, New Yorkers, NYSLRS, Office of the State Comptroller, Proud Public Servants, Public servants, Public Service, Public Service Recognition Week on May 7, 2018 by NYSLRS.
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text:
\section{Introduction}

\subsection{Prime number races}
Let $\pi(x;q,a)$ denote the number of primes in the progression $a\!\! \mod q$.
For fixed $q$, the functions $\pi(x;q,a)$
(for $a\in A_q$, the set of residues coprime to $q$)
all satisfy
\begin{equation}\label{PNT}  \pi(x,q,a) \sim \frac{x}{\varphi(q)\log x},
\end{equation}
where $\varphi$ is Euler's totient function \cite{Da}.
There are, however, curious inequities.  For example
$\pi(x;4,3) \ge \pi(x;4,1)$ seems to hold for most $x$,
an observation of Chebyshev from 1853 \cite{Ch}.
In fact, $\pi(x;4,3) < \pi(x;4,1)$ for the first time
at $x = 26,861$ \cite{Le}.
More generally, one can ask various questions about the behavior of
\begin{equation}\label{Delta}
\Delta(x;q,a,b) := \pi(x;q,a)-\pi(x;q,b)
\end{equation}
for distinct $a,b\in A_q$.  Does $\Delta(x;q,a,b)$ change sign
infinitely often?  Where is the first sign change?  How many sign
changes with $x\le X$ ?  What are the extreme values of
$\Delta(x;q,a,b)$?  Such questions are colloquially known
as \emph{prime race problems}, and were studied extensively by
Knapowski and Tur\'an in a series of papers beginning with \cite{KT}.
 See the survey articles
\cite{FK} and \cite{GM} and references therein for an introduction
to the subject and summary of major findings.  Properties
of Dirichlet $L$-functions lie at the heart of such investigations.

Despite the tendency for the function  $\Delta(x;4,3,1)$ to be negative,
Littlewood \cite{Li} showed that it changes sign
infinitely often.   Similar results have been proved for other
$q,a,b$ (see \cite{JS} and references therein).
Still, in light of Chebyshev's observation, we can ask how frequently
$\Delta(x;q,a,b)$ is positive and how often it is negative.
These questions are best addressed in the context of \emph{logarithmic
density}.  A set $S$ of positive integers has logarithmic density
$$
\ensuremath{\delta}(S) =  \lim_{x\to\infty} \frac{1}{\log x}
\sum_{\substack{n\le x \\ n\in S}} \frac{1}{n}
$$
provided the limit exists.  Let $\ensuremath{\delta}(q,a,b) = \ensuremath{\delta}(P(q,a,b))$, where
$P(q,a,b)$ is the set of integers $n$ with $\Delta(n;q,a,b)>0$.
In 1994, Rubinstein and Sarnak \cite{RS}  showed that
$\delta(q;a,b)$ exists, assuming two
hypotheses (i) the Extended Riemann Hypothesis for Dirichlet
$L$-functions modulo $q$ (ERH$_q$), and (ii) the imaginary parts of
zeros of each Dirichlet $L$-function are linearly independent over the
rationals (GSH$_q$ - Grand Simplicity Hypothesis).
The authors also gave methods to accurately estimate the ``bias'',
for example showing that
$\delta(4;3,1)\approx 0.996$ in Chebyshev's case.
More generally, $\delta(q;a,b)=\frac12$ when $a$ and $b$ are either
both quadratic
residues modulo $q$ or both quadratic nonresidues (unbiased prime races), but
$\delta(q;a,b)>\frac12$ whenever $a$ is a quadratic
non-residue and $b$ is a quadratic residue.  A bit later we will
discuss the reasons behind these phenomena.  Sharp asymptotics for
$\delta(q;a,b)$ have recently been given by Fiorilli and Martin
\cite{FM}, which explain other properties of these densities.

\subsection{Quasi-prime races}

In this paper we develop a parallel theory for comparison of functions
$\pi_2(x;q,a)$, the number of integers $\le x$ which are in the
progression $a\!\! \mod q$ and which are the product
of two primes $p_1p_2$ ($p_1=p_2$ allowed).
Put
$$
\Delta_2(x;q,a,b) := \pi_2(x;q,a)-\pi_2(x;q,b),
$$
let $P_2(q,a,b)$ be the set of integers $n$ with $\Delta_2(n;q,a,b)>0$, and set $\ensuremath{\delta}_2(q,a,b)=\ensuremath{\delta}(P_2(q,a,b))$.
The table below shows all such
quasi-primes up to 100 grouped in residue classes modulo 4.

\vspace{.1in}
\begin{center}
\begin{tabular}{|c|c|}
\hline $pq \equiv 1 \pmod 4$ & $pq \equiv 3 \pmod 4$ \\
\hline 9 & 15\\
\hline 21 & 35\\
\hline 25 & 39\\
\hline 33 & 51\\
\hline 49 & 55\\
\hline 57 & 87\\
\hline 65 & 91\\
\hline 69 & 95\\
\hline 77 & \\
\hline 85 & \\
\hline 93 & \\
\hline
\end{tabular}
\end{center}
\vspace{.1in}

Observe that $\Delta_2(x;4,3,1) \le 0$ for $x\le 100$, and in fact
the smallest $x$ with $\Delta_2(x;4,3,1) > 0$ is $x=26747$ (amazingly
close to the first sign change of $\Delta(x;4,3,1)$).
Some years ago Richard Hudson conjectured that the bias for products
of two primes is always reversed from that of primes; i.e.,
$\delta_2(q;a,b)<\frac12$ when $a$ is a quadratic non-residue modulo
$q$ and $b$ is a quadratic residue.
Under the same assumptions as
\cite{RS}, namely ERH$_q$ and GSH$_q$, we confirm Hudson's conjecture
and also show that the bias is less pronounced.

\begin{thm}\label{thm1}
Let $a,b$ be distinct elements of $A_q$.  Assuming ERH$_q$ and GSH$_q$,
$\delta_2(q;a,b)$ exists.  Moreover, if $a$ and $b$ are both quadratic
residues modulo $q$ or both quadratic non-residues, then
$\delta_2(q;a,b)=\frac12$.  Otherwise, if $a$ is a quadratic
nonresidue and $b$ is a quadratic residue, then
$$
1 - \delta(q;a,b) < \delta_2(q;a,b) < \frac12.
$$
\end{thm}

We can accurately estimate $\delta_2(q;a,b)$ borrowing methods from
\cite[\S 4]{RS}.  In particular we have
$$
\delta_2(4;3,1) \approx 0.10572.
$$
We deduce Theorem \ref{thm1} by connecting the distribution of
$\Delta_2(x;q,a,b)$ with the distribution of $\Delta(x;q,a,b)$.
Although the relationship is ``simple'', there is no elementary way
to derive it, say by writing
$$
\pi_2(x;q,a) = \frac12 \sum_{p\le x} \pi\(\frac{x}{p};q,ap^{-1}\!\! \mod q\)
+ \frac12 \sum_{\substack{p\le \sqrt{x} \\ p^2\equiv a\!\!\! \pmod{q}}} 1.
$$
In particular, our result depends strongly on the assumption that the
zeros of the $L$-functions modulo $q$ have only simple zeros.
Let $N(q,a)$ be the number of $x\in A_q$ with $x^2\equiv a\pmod{q}$,
and let $C(q)$ be the set of nonprincipal Dirichlet characters modulo $q$.

\begin{thm}\label{DeltaDelta2}
Assume ERH$_q$ and for each $\chi\in C(q)$, $L(\frac12,\chi)\ne 0$ and
the zeros of $L(s,\chi)$ are simple.  Then
$$
\frac{\Delta_2(x;q,a,b) \log x}{\sqrt{x} \log\log x} = \frac{N(q,b)-N(q,a)}{2\phi(q)} -
\frac{\log x}{\sqrt{x}} \Delta(x;q,a,b) + \Sigma(x;q,a,b),
$$
where $\frac{1}{Y} \int_1^Y |\Sigma(e^y;q,a,b)|^2\, dy = o(1)$ as $Y\to \infty$.
\end{thm}

The expression for $\Delta_2$ given in Theorem \ref{DeltaDelta2} must be modified if some $L(s,\chi)$ has multiple zeros;
see \S \ref{sec:outline} for details.

Figures 1,2 and 3 show
 graphs corresponding to $(q,a,b)=(4,3,1)$, plotted on a
logarithmic scale from $x=10^3$ to $x=10^9$.
While $\Sigma(x;4,3,1)$ appears to be oscillating around $-0.2$, this is
caused by some terms in $\Sigma(x;4,3,1)$ of order $1/\log\log x$, and
$\log\log 10^9 \approx 3.03$.  By Theorem \ref{DeltaDelta2},
$\Sigma(x;4,3,1)$ will (assuming ERH$_4$ and GSH$_4$)
 eventually settle down to oscillating about 0.

\begin{figure}
\input{delta.tex}
\caption{ $\frac{\log x}{\sqrt{x}} \Delta(x;4,3,1)$ }
\end{figure}

\begin{figure}
\input{delta2.tex}
\caption{ $\frac{\log x}{\sqrt{x}\log\log x} \Delta_2(x;4,3,1)$ }
\end{figure}

\begin{figure}
\input{Sigma.tex}
\caption{$\Sigma(x;4,3,1)$}
\end{figure}

It is not immediate that Theorem \ref{thm1} follows from Theorem
\ref{DeltaDelta2}.  One first needs more precise information about
the distribution of $\Delta(x;q,a,b)$ from \cite{RS}.

\newtheorem*{theoremRS}{Theorem RS}\label{Theorem RS}
\begin{theoremRS}{\cite[\S 1]{RS}}\label{RSthm}
Assume ERH$_q$ and  GSH$_q$.   For any distinct $a,b\in A_q$, the
function
\begin{equation}\label{Deltanorm}
\frac{u \Delta(e^u;q,a,b)}{e^{u/2}}
\end{equation}
has a probabilistic distribution.  This distribution
(i) has mean $(N(q,b)-N(q,a))/\phi(q)$, (ii) is
symmetric with respect to its mean, and (iii) has
 a continuous density function.
\end{theoremRS}

Assume $a$ is a quadratic nonresidue modulo $q$ and $b$ is a quadratic
residue.  Then $N(q,b)-N(q,a)>0$.
Let $f$ be the density function for the distribution of
\eqref{Deltanorm}, that is,
$$
f(t) = \frac{d}{dt} \lim_{U\to\infty} \frac{1}{U} \text{meas} \{ 0\le u\le U :
ue^{-u/2} \Delta(e^u;q,a,b) \le t \}.
$$
We see from Theorem RS that
$$
\delta(q,a,b) = \int_0^\infty f(t) \, dt > \frac12
$$
and from Theorem \ref{DeltaDelta2} that
$$
\delta_2(q,a,b) = \int_{-\infty}^{\frac{N(q,b)-N(q,a)}{2\phi(q)}} f(t) \, dt,
$$
from which Theorem \ref{thm1} follows.

Theorem \ref{DeltaDelta2} also determines the joint distribution
of any vector function
\begin{equation}\label{vector2}
\frac{u}{e^{u/2}\log u} \( \Delta_2(e^u;q,a_1,b_1), \ldots,
\Delta_2(e^u;q,a_r,b_r) \).
\end{equation}

\begin{thm}\label{thmjoint}
If $f(x_1,\ldots,x_r)$ is the density function of
$$
\frac{u}{e^{u/2}} \( \Delta(e^u;q,a_1,b_1), \ldots,
\Delta(e^u;q,a_r,b_r) \),
$$
then the joint density function of \eqref{vector2} is
$$
f\( \frac{N(q,b_1)-N(q,a_1)}{2\phi(q)} - x_1, \ldots,
\frac{N(q,b_r)-N(q,a_r)}{2\phi(q)} - x_r \).
$$
\end{thm}

\subsection{Origin of Chebyshev's bias}

From an analytic point of view ($L$-functions), the
weighted sum
\begin{equation}\label{DeltaLambda}
\Delta^*(x;q,a,b)=\sum_{\substack{n\le x \\ n\equiv a \bmod{q} }}
\Lambda(n) - \sum_{\substack{n\le x \\ n\equiv b \bmod{q} }}
\Lambda(n),
\end{equation}
where $\Lambda$ is the von Mangoldt function, is more natural than
\eqref{Delta}.  Expressing $\Delta^*(x;q,a,b)$ in terms of sums over
zeros of $L$-functions in the standard way (\S 19 of \cite{Da}), we
obtain, on ERH$_q$,
$$
e^{-u/2} \phi(q) \Delta^*(e^u;q,a,b) = - \sum_{\chi\in C(q)} \(
\overline{\chi}(a)-\overline{\chi}(b) \) \sum_{\gamma}
\frac{e^{i\gamma u}}{1/2+i\gamma} + O(u^2 e^{-u/2}),
$$
where  $\ensuremath{\gamma}$ runs over imaginary parts of nontrivial zeros of
$L(s,\chi)$ (counted with multiplicity).
 Hypothesis GSH$_q$ implies, in particular, that $L(1/2,\chi)\ne 0$.
Each summand $e^{i\gamma u}/(1/2+i\gamma)$ is thus a harmonic
with mean zero as $u\to\infty$, and GSH$_q$ implies that
the harmonics behave independently.  Hence, we expect that $e^{-u/2}
\phi(q) \Delta^*(e^u;q,a,b)$ will behave like a mean zero random variable.
On the other hand, the right side of \eqref{DeltaLambda} contains not
only terms corresponding to prime $n$
but terms corresponding to powers of primes.
Applying the prime number theorem for arithmetic progressions \eqref{PNT} to the
terms $n=p^2$ in \eqref{DeltaLambda} gives
$$
\Delta^*(x;q,a,b) = \sum_{\substack{p\le x \\ p\equiv a\bmod{q} }}
\log p -  \sum_{\substack{p\le x \\ p\equiv b\bmod{q} }}
\log p + \frac{x^{1/2}}{\phi(q)} \( N(q,a)-N(q,b) \) + O(x^{1/3}).
$$
Hence, on ERH$_q$ and GSH$_q$, we expect the expression
\begin{equation}\label{sump}
\frac{1}{\sqrt{x}} \Biggl( \sum_{\substack{p\le x \\ p\equiv a\bmod{q} }}
\log p -  \sum_{\substack{p\le x \\ p\equiv b\bmod{q} }}
\log p \Biggr)
\end{equation}
to behave like a random variable with mean $(N(q,b)-N(q,a))/\phi(q)$.  Finally,
the distribution of $\Delta(x;q,a,b)$ is obtained from the distribution
of \eqref{sump} and
partial summation.

\subsection{Analyzing $\Delta_2(x;q,a,b)$}

A natural analog of $\Delta^*(x;q,a,b)$ is
\begin{equation}\label{DeltaLambda2}
\sum_{\substack{mn\le x \\ mn\equiv a\bmod{q} }}
\Lambda(m)\Lambda(n)
- \sum_{\substack{mn\le x \\ mn\equiv b\bmod{q} }}
\Lambda(m)\Lambda(n).
\end{equation}
As with $\Delta^*(x;q,a,b)$,
the expression in \eqref{DeltaLambda2} can be easily written as a
sum over zeros of $L$-functions plus a small error.
The main problem now is that the principal summands, namely $\log p_1
\log p_2$ for primes $p_1,p_2$,
are very irregular as a function of $p_1 p_2$, and thus
estimates for $\Delta_2(x;q,a,b)$ cannot be recovered by partial summation.
We get around this problem using a double integration, a method which
goes back to Landau \cite[\S 88]{La}.   We have
\begin{equation}\label{DeltaG}
\begin{split}
\Delta_2(x;q,a,b) &= \frac{1}{\phi(q)} \sum_{\chi \in C(q)} \(
\overline{\chi}(a) -  \overline{\chi}(b) \) \sum_{\substack{n=p_1
    p_2\le x \\ p_1 \le p_2}} \chi(n) \\
&= \frac{1}{2\phi(q)}  \sum_{\chi \in C(q)} \(
\overline{\chi}(a) -  \overline{\chi}(b) \)
\int_0^\infty \int_0^\infty G(x,u,v;\chi) \, du\, dv +
O\pfrac{\sqrt{x}}{\log x},
\end{split}
\end{equation}
where
\begin{equation} \label{Gsum}
G(x,u,v;\chi) = \sum_{p_1 p_2 \leq x} \frac{\chi(p_1 p_2) \log p_1
  \log p_2}{p_1^u p_2^v}.
\end{equation}
The related functions
$$
G^*(x,u,v;\chi) = \sum_{mn \leq x} \frac{\chi(mn) \Lambda(m)
  \Lambda(n)}{m^u n^v}
$$
are more ``natural'' from an analytic point of view, being easily
expressed in terms of zeros of Dirichlet $L$-functions.  By the
reasoning of the previous subsection, each $G^*(x,u,v;\chi)$ is
expected to be unbiased, the bias in $\Delta_2(x;q,a,b)$ originating
from the summands in  $G^*(x,u,v;\chi)$ where $m$ is not prime or $n$
is not prime.

\subsection{A heuristic argument for the bias in $\Delta_2(x;q,a,b)$}

We conclude this introduction with a heuristic
evaluation of the bias in $\Delta_2(x;q,a,b)$, which originates from
the difference between functions $G(x;u,v;\chi)$ and $G^*(x,u,v;\chi)$.
For simplicity of exposition, we'll concentrate on the special case
$(q,a,b)=(4,3,1)$.  In this case, the
bias arises from terms $p_1 p_2^2$ and $p_1^2 p_2^2$
which appear in  $G^*(x;u,v;\chi)$ but not in $G(x,u,v;\chi)$.
Let $\chi$ be the non-principal character modulo 4, so that
$$
\frac12 \int_0^\infty \int_0^\infty
 \bigl( G^*(x,u,v;\chi) - G(x,u,v;\chi) \bigr)\, du\, dv =
\frac12 \sum_{\substack{p_1^a p_2^b \le x \\ \max(a,b)\ge 2}}
\frac{\chi(p_1^a p_2^b)}{ab}.
$$
There are $O(x^{1/2}/\log x)$ terms with $\min(a,b)\ge 2$ and
$\max(a,b)\ge 3$.  By the prime
number theorem and partial summation,
$$
\frac12 \sum_{p_1^2 p_2^2 \le x} \frac14 = \frac18 \sum_{p\le
  \sqrt{x}} \pi \( \sqrt{x/p^2} \)
\sim \frac{x^{1/2} \log\log x}{2\log x}.
$$
Thus,
\begin{align*}
\Delta_2(x;4,3,1) &= - \frac{1}{2} \sum_{mn \leq x} \frac{\chi(mn) \Lambda(m)
  \Lambda(n)}{\log m \log n} - \Bigg( \sum_{k=2}^{\infty} \frac{1}{k}
\sum_{p_1^k \leq x} \chi(p_1^k) \Delta(x/p_1^k;4,3,1) \Bigg) \\
&\qquad +\( \frac12+o(1) \)\frac{x^{1/2} \log\log x}{\log x}.
\end{align*}
By Theorem RS, $\Delta(y;4,3,1)=y^{1/2}/\log y + E(y)$,
where $E(y)$ oscillates with mean 0.  Thus,
$$
\sum_{k=2}^{\infty} \frac{1}{k}
\sum_{p_1^k \leq x} \chi(p_1^k) \Delta(x/p_1^k;4,3,1) =
\sum_{k=2}^{\infty} \frac{2}{k}
\sum_{p_1^k \leq x} \chi(p_1^k) \frac{\sqrt{x/p_1^k}}{\log(x/p_1^k)} +
E'(x),
$$
where $E'(x)$ is expected to oscillate with mean zero.  The $k=2$ terms are
$$
\sum_{p_1^2 \leq x} \frac{\sqrt{x/p_1^2}}{\log(x/p_1^2)} \sim
\frac{\sqrt{x} \log\log x}{\log x},
$$
while the terms corresponding to $k\ge 3$ contribute
$$
\ll \sum_{k=3}^{\infty} \frac{1}{k}
\sum_{p_1^k \leq x} \frac{\sqrt{x/p_1^k}}{\log(x/p_1^k)} \ll
\frac{\sqrt{x}}{\log x}.
$$
Thus, we find that
\begin{align*}
\Delta_2(x;4,3,1) &= - \frac{1}{2} \sum_{mn \leq x} \frac{\chi(mn) \Lambda(m)
  \Lambda(n)}{\log m \log n} -
\( \frac12+o(1) \)\frac{x^{1/2} \log\log x}{\log x} + E'(x).
\end{align*}

\subsection{Further problems}
It is natural to consider the distribution, in arithmetic
progressions, of numbers
composed of exactly $k$ prime factors, where $k\ge 3$ is fixed.  As with the
cases $k=1$ and  $k=2$, we expect there to be no bias if we count all
numbers $p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ with weight $(a_1
\cdots a_k)^{-1}$.  If, however, we count terms which are the product
of precisely $k$ primes (that is, numbers $p_1^{a_1} \cdots p_j^{a_j}$
with $a_1+\cdots+a_j=k$), then there will be a bias.  Hudson has
conjectured that the bias will be in the same direction as for primes
when $k$ is odd, and in the opposite direction for even $k$.  We
conjecture that, in addition, the bias becomes less pronounced as $k$
increases.

\section{Preliminaries}

With $\chi$ fixed, the letter $\gamma$, with or without subscripts, denotes the
imaginary part of a zero of $L(s,\chi)$ inside the critical strip.  In
sums over $\ensuremath{\gamma}$, each term appears with its multiplicity $m(\ensuremath{\gamma})$ unless we
specify that we sum over distinct $\ensuremath{\gamma}$.  Constants implied by $O-$ and
$\ll-$symbols depend only on $\chi$ (and hence, on $q$)
unless additional dependence is indicated with a subscript.
Let
$$
A(\chi) = \begin{cases} 1 & \chi^2=\chi_0 \\ 0 & \text{else} \end{cases},
$$
where $\chi_0$ is the principal character modulo $q$.  That is,
$A(\chi)=1$ if and only if $\chi$ is a real character.
For $\chi\in C(q)$, define
$$
F(s,\chi)= \sum_{p} \frac{\chi(p) \log p}{p^s}.
$$
The following estimates are standard; see e.g. \cite[\S 15,16]{Da}.

\begin{lem} \label{FsHs} Let $\chi\in C(q)$,
assume ERH$_q$ and fix $c>\frac13$.
Then $F(s,\chi)= -\frac{L'}{L}(s,\chi) +
A(\chi) \frac{\zeta'}{\zeta}(2s) + H(s,\chi)$, where $H(s,\chi)$ is
analytic and uniformly  bounded in the half-plane $\Re s \ge c$.
\end{lem}

\begin{lem} \label{NTchi}
Let $\chi$ be a Dirichlet character modulo $q$.
Let $N(T,\chi)$ denote the
number of zeros of $L(s, \chi)$ with $0<\Re s<1$ and
$|\Im s|<T$.   Then
\begin{enumerate}
\item  $N(T,\chi)=O(T\log(qT))$ for $T\ge 1$.
\item  $N(T,\chi) - N(T-1,\chi) = O(\log(qT))$ for $T\ge 1$.
\item  Uniformly for $s=\sigma+it$ and $\sigma\ge -1$,
$$
\frac{L'(s,\chi)}{L(s,\chi)} =
  \sum_{|\gamma-t|<1} \frac{1}{s-\rho} + O(\log q(|t|+2)).
$$
\item $-\frac{\zeta'}{\zeta}(\sigma)  = \frac{1}{\sigma -1}+O(1) $ uniformly
  for $\sigma\ge \frac12$, $\sigma\ne 1$.
\item $\big|\frac{\zeta'}{\zeta}(\sigma+iT) \big| \le
  -\frac{\zeta'}{\zeta}(\sigma)$ for $\sigma > 1$.
\end{enumerate}
\end{lem}

For a suitably small, fixed $\delta>0$, we say that a number $T\ge 2$ is
\emph{admissible} if for all $\chi\in C(q)\cup \{\chi_0\}$ and all zeros $\frac12+i\ensuremath{\gamma}$ of $L(s,\chi)$,
$|\ensuremath{\gamma}-T| \ge \delta (\log T)^{-1}$.
By Lemma \ref{NTchi}, we can choose $\ensuremath{\delta}$ small enough, depending
on $q$, so that
there is an admissible $T$ in $[U,U+1]$ for all $U\ge 2$.
From Lemma \ref{NTchi} we obtain

\begin{lem}\label{zeta}
 Uniformly for $\sigma \ge \frac25$ and admissible $T\ge 2$,
$$
|F(\sigma+iT,\chi)| = O(\log^2 T).
$$
\end{lem}

\begin{lem} \label{sumg1g2}
Fix $\chi\in C(q)$ and assume $L(\frac12,\chi)\ne 0$.
For $A\ge 0$ and real $k\ge 0$,
$$
\sum_{\substack{ |\ensuremath{\gamma}_1|, |\ensuremath{\gamma}_2| \ge A \\ |\ensuremath{\gamma}_1-\ensuremath{\gamma}_2| \geq
     1}} \frac{{\log}^k (|\ensuremath{\gamma}_1|+3) {\log}^k (|\ensuremath{\gamma}_2|+3)}{{|\ensuremath{\gamma}_1|}
   {|\ensuremath{\gamma}_2|} |\ensuremath{\gamma}_1-\ensuremath{\gamma}_2|} \ll_{k} \frac{(\log (A+3))^{2k+3}}{A+1}.
$$
\end{lem}

\begin{proof}
The sum in question is at most twice the sum
of terms with $|\ensuremath{\gamma}_2| \ge |\ensuremath{\gamma}_1|$, which is
$$
\ll \sum_{|\ensuremath{\gamma}_2|\ge A}
\frac{{\log}^{2k} (|\ensuremath{\gamma}_2|+3)}{{|\ensuremath{\gamma}_2|}} \bigg(\frac{1}{|\ensuremath{\gamma}_2|}
\sum_{|\ensuremath{\gamma}_1| < \frac{|\ensuremath{\gamma}_2|}{2}} \frac{1}{{|\ensuremath{\gamma}_1|}} +
\frac{1}{{|\ensuremath{\gamma}_2|}} \sum_{\substack{\frac{|\ensuremath{\gamma}_2|}{2} \leq |\ensuremath{\gamma}_1| \leq
  |\ensuremath{\gamma}_2| \\ |\ensuremath{\gamma}_2-\ensuremath{\gamma}_1| \ge 1}} \frac{1}{|\ensuremath{\gamma}_2-\ensuremath{\gamma}_1|} \bigg).
$$
By Lemma \ref{NTchi} (1), the two sums over $\ensuremath{\gamma}_1$ are $O(\log^2 (|\ensuremath{\gamma}_2|+3))$.  A further
application of Lemma \ref{NTchi} (1) completes the proof.
\end{proof}

We conclude this section with
a truncated version of the Perron formula for $G(x,u,v;\chi)$.

\begin{lem}\label{Perron} Uniformly for $x \leq T \leq 2x^2$,
  $x \ge 2$, $u\ge 0$ and $v\ge 0$, we have
\begin{equation} \label{Ginterror}
G(x,u,v;\chi) = \frac{1}{2\pi i} \int_{c-iT}^{c+iT} F(s+u,\chi)F(s+v,\chi)
\frac{x^s}{s} ds + O(\log^3 x),
\end{equation}
where $c=1+\frac{1}{\log x}$.
\end{lem}

\begin{proof}
For $\Re s > 1$, we have
$$
F(s+u,\chi)F(s+v,\chi) = \sum_{n=1}^\infty f(n) n^{-s}, \qquad
f(n)=\sum_{p_1p_2=n} \frac{\chi(p_1 p_2) \log p_1 \log p_2}{p_1^{u} p_2^v}.
$$
Using the trivial estimate
$|f(n)| \le \log^2 n$
and a standard argument \cite[\S 17, (3) and (5)]{Da}, we
obtain the desired bounds.
\end{proof}

\section{Outline of the proof of Theorem \ref{DeltaDelta2}}\label{sec:outline}

Throughout the remainder of this paper, fix $q$, assume ERH$_q$
and that $L(\frac12,\chi) \ne 0$ for each $\chi\in C(q)$.  Let
$$
\ensuremath{\varepsilon} = \frac{1}{100}.
$$
We next define a function $T(x)$ as follows.  For each positive
integer $n$, let $T_n$ be an admissible value of $T$
satisfying $\exp (2^{n+1}) \le T_n \le
\exp(2^{n+1})+1$ and set $T(x)=T_n$ for $\exp(2^n) < x \le \exp
(2^{n+1})$.  In particular, we have
$$
x \le T(x) \le 2x^2 \qquad (x\ge e^2).
$$

Our first task is to express the double integrals in \eqref{DeltaG}
in terms of sums over zeros of $L(s,\chi)$.  This is proved in
Section \ref{sec:analytic}.

\begin{lem}\label{analytic}
Let $\chi\in C(q)$ and let $T=T(x)$.  Then
\begin{multline*}
 x^{-1/2} \int_0^\infty \int_0^\infty G(x,u,v;\chi)\, du\, dv \\
= 2\int_{0}^{2\ensuremath{\varepsilon}} \!\!\! \int_{0}^{2\ensuremath{\varepsilon}} \sum_{|\ensuremath{\gamma}|\le T}
\frac{F(\frac{1}{2}+u-v+i\ensuremath{\gamma},\chi)x^{-v+i\ensuremath{\gamma}}}{\frac{1}{2}-v+i\ensuremath{\gamma}}
du\, dv + \frac{A(\chi) \log\log x + \Sigma_1(x;\chi)+O(1)}{\log x},
\end{multline*}
where
$\int_{1}^{Y} |\Sigma_1 (e^y;\chi)|^2 dy = O(Y)$.
\end{lem}

The aggregate of terms $A(\chi)\log\log x/\log x$ account for the
bias for products of two primes.  As with the Chebyshev bias for primes,
these terms arise from poles of $F(s)$ at $s=\frac12$ when $A(\chi)=1$
(see Lemma \ref{FsHs}) and correspond to the contribution to $F(s)$
from squares of primes.
The double integral on the right side in Lemma \ref{analytic} is
complicated to analyze.  In Section \ref{sec:doubleint} we prove the following.

\begin{lem}\label{doubleint}  Let $\chi\in C(q)$.
Let $n$ be a positive integer, $2^n < \log x \le 2^{n+1}$ and  $T=T(x)$.  Then
\begin{align*}
2 \int_0^{2 \ensuremath{\varepsilon}} \int_0^{2 \ensuremath{\varepsilon}} &\sum_{|\ensuremath{\gamma}| \leq T}
  \frac{ F(\frac{1}{2}+u-v+i \ensuremath{\gamma},\chi) x^{-v+i \ensuremath{\gamma}}}{\frac{1}{2}-v+i \ensuremath{\gamma}}
  du\ dv
= \frac{\Sigma_2(x;\chi)}{\log x}\\
& + 2\sum_{\substack{|\ensuremath{\gamma}| \leq T \\
\ensuremath{\gamma} \text{ distinct}}} m^2(\ensuremath{\gamma}) x^{i\ensuremath{\gamma}} (\frac{1}{2}+i\ensuremath{\gamma})
\int_0^{2 \ensuremath{\varepsilon}-2^{-n}} \frac{ x^{-v}}{\frac{1}{2}-v+i \ensuremath{\gamma}}
\int_{v+2^{-n}}^{2 \ensuremath{\varepsilon}} \frac{du}{(u-v)(\frac{1}{2}-u+i \ensuremath{\gamma})} dv,
\end{align*}
where $\int_{1}^{Y} |\Sigma_2(e^y;\chi)|^2 dy = o(Y\log^2 Y)$.
\end{lem}

The terms on the right in Lemma \ref{doubleint} with small $|\ensuremath{\gamma}|$
will give the main term, and terms with larger $|\ensuremath{\gamma}|$ are considered
as error terms.
The next lemma is proved in Section \ref{sec:TT0}.

\begin{lem}\label{TT0}
Let $\chi\in C(q)$.
Let $n$  be a positive integer, $2^n < \log x \le 2^{n+1}$,  $T=T(x)$
and $2\le T_0 \le T$.  Then
\begin{align*}
&2\sum_{\substack{|\ensuremath{\gamma}| \leq T \\ \ensuremath{\gamma} \text{ distinct}}} m^2(\ensuremath{\gamma}) x^{i\ensuremath{\gamma}}
(\frac{1}{2}+i\ensuremath{\gamma}) \int_0^{2 \ensuremath{\varepsilon}-2^{-n}} \frac{ x^{-v}}{\frac{1}{2}-v+i \ensuremath{\gamma}}
 \int_{v+2^{-n}}^{2 \ensuremath{\varepsilon}} \frac{du}{(u-v)(\frac{1}{2}-u+i \ensuremath{\gamma})} dv \\
&\qquad = \frac{2\log\log x}{\log x} \sum_{\substack{|\gamma|\le T_0 \\
\ensuremath{\gamma} \text{ distinct}}}
 \frac{m^2(\gamma) x^{i\gamma}}{1/2+i\ensuremath{\gamma}} + O\pfrac{\log^3 T_0}{\log x} +
\frac{\Sigma_3(x,T_0;\chi)}{\log x},
\end{align*}
where
$$
\frac{1}{Y}\int_1^Y |\Sigma_3(e^y,T_0;\chi)|^2\, dy
\ll \frac{\log^5 T_0}{T_0} \log^2 Y.
$$
\end{lem}

Combining Lemmas \ref{analytic}, \ref{doubleint} and \ref{TT0}
with \eqref{DeltaG} yields (for fixed, large $T_0$)
\begin{multline*}
\Delta_2(x;q,a,b) = \frac{\sqrt{x}}{2\phi(q)} \sum_{\chi\in C(q)}
\( \overline{\chi}(a)-\overline{\chi}(b) \) \Bigg[
\frac{\log\log x}{\log x} \Bigg( A(\chi)+2 \sum_{\substack{|\ensuremath{\gamma}|\le T_0
\\ \ensuremath{\gamma} \text{ distinct}}} \frac{m^2(\ensuremath{\gamma}) x^{i\ensuremath{\gamma}}}{1/2+i\ensuremath{\gamma}}\Bigg) \\
+ \frac{\Sigma_1(x;\chi)+\Sigma_2(x;\chi)+\Sigma_3(x,T_0;\chi)
+O(\log^3 T_0)}{\log x}
\Bigg],
\end{multline*}
where
$$
\lim_{T_0\to \infty} \(\limsup_{Y\to \infty}
\frac{1}{Y\log^2 Y} \sum_{\chi\in C(q)}  \int_{1}^Y |\Sigma_1(e^y;\chi)+
\Sigma_2(e^y;\chi)+\Sigma_3(e^y;T_0;\chi)|^2\, dy
\) =0.
$$
On the other hand (cf. \cite{RS}),
$$
\Delta(x;q,a,b) = \frac{\sqrt{x}}{\log x} \( \frac{N(q,b)-N(q,a)}{\phi(q)}
- \sum_{\chi\in C(q)} \( \overline{\chi}(a)-\overline{\chi}(b) \)
\sum_{|\ensuremath{\gamma}| \le T_0}\frac{x^{i\ensuremath{\gamma}}}{1/2+i\ensuremath{\gamma}} + \Sigma_4(x;T_0) \),
$$
where
$$
\lim_{T_0\to \infty} \(\limsup_{Y\to\infty} Y^{-1} \int_{1}^Y |\Sigma_4(e^y;T_0)|^2\, dy
\) =0.
$$
Now assume $m(\ensuremath{\gamma})=1$ for all $\ensuremath{\gamma}$, and note that
$$
 \sum_{\chi\in C(q)}
\( \overline{\chi}(a)-\overline{\chi}(b) \)
 A(\chi) = N(q,a)-N(q,b).
$$
Letting $T_0\to \infty$ finishes the proof of Theorem \ref{DeltaDelta2}.

\section{proof of Lemma \ref{analytic}}\label{sec:analytic}

Assume ERH$_q$ throughout.
We first estimate $G(x,u,v;\chi)$ for different ranges of $u,v$.

\begin{lem}\label{Gbounds}
Let $\chi\in C(q)$, $\chi\ne \chi_0$.
For $x\ge 4$, the following hold:
\begin{enumerate}

\item For $u,v \geq \ensuremath{\varepsilon}$,
$G(x,u,v;\chi) \ll x^{\frac{1}{2} - \frac {\ensuremath{\varepsilon}}{2}} \log^5 x$.

\item For $u \geq 2\ensuremath{\varepsilon}$, $v \leq \ensuremath{\varepsilon}$ and $T=T(x)$,
\begin{align*}
x^{-1/2} G(x,u,v;\chi) &= \sum_{|\ensuremath{\gamma}|\le T} \frac{F(\frac{1}{2}+u-v+i\ensuremath{\gamma}
  ,\chi)x^{-v+i\ensuremath{\gamma}}}{\frac{1}{2}-v+i\ensuremath{\gamma}}
-A(\chi) \frac{F(\frac{1}{2}+u-v,\chi)x^{-v}}{1-2v} \\
&\qquad\qquad + O(x^{-\frac{3\ensuremath{\varepsilon}}{2}} \log^5 x).
\end{align*}

\item For $u\le 2\ensuremath{\varepsilon}$, $v\le 2\ensuremath{\varepsilon}$, $u\ne v$
and $T=T(x)$,
\begin{multline*}
x^{-1/2} G(x,u,v;\chi) = \sum_{|\ensuremath{\gamma}| \le T}
 \frac{F(\frac{1}{2}+u-v+i\ensuremath{\gamma}
   ,\chi)x^{-v+i\ensuremath{\gamma}}}{\frac{1}{2}-v+i\ensuremath{\gamma}} +
 \frac{F(\frac{1}{2}-u+v+i\ensuremath{\gamma}
   ,\chi)x^{-u+i\ensuremath{\gamma}}}{\frac{1}{2}-u+i\ensuremath{\gamma}} \\  \qquad -
  A(\chi) \( \frac{F(\frac{1}{2}+u-v,\chi)x^{-v}}{1-2v}+
  \frac{F(\frac{1}{2}-u+v,\chi)x^{-u}}{1-2u} \)
  + O (x^{-3\ensuremath{\varepsilon}} \log^5 x).
\end{multline*}
\end{enumerate}
\end{lem}

\begin{proof}
Assume $u\ge \ensuremath{\varepsilon}$ and $v\ge \ensuremath{\varepsilon}$.  Start with the approximation of
$G(x,u,v;\chi)$ given by Lemma \ref{Perron}, then deform the segment
of integration to the contour consisting of three straight segments connecting
$c-iT$, $b-iT$, $b+iT$ and $c+iT$, where $b=\frac12 - \frac{\ensuremath{\varepsilon}}{2}$
and $T=T(x)$.  The rectangle formed by the new and old contours
does not contain any poles of $F(s+u,\chi)F(s+v,\chi)s^{-1}$.
On the three new segments, by Lemmas \ref{FsHs}, \ref{NTchi} and \ref{zeta}, we
have $|F(s+u,\chi)F(s+v,\chi)| \ll \log^4 T$.  Hence the integral of
$F(s+u,\chi)F(s+v,\chi) x^s s^{-1}$ over the three segments is
$$
\ll (\log^4 x) \Bigl( \int_b^c \frac{x^\sigma}{|\sigma+iT|}\, d\sigma
+ \int_{-T}^T \frac{x^b}{|b+it|}\, dt \Bigr) \ll x^b \log^5 x.
$$
This proves (1).

We now consider the case $v \leq \ensuremath{\varepsilon}$ and $u \geq 2\ensuremath{\varepsilon}$.
We set $b = \frac{1}{2}-\frac{3\ensuremath{\varepsilon}}{2}$ and deform the
contour of integration as in the previous case.
Since $u+b \ge \frac12 + \frac{\ensuremath{\varepsilon}}{2}$ and $v+b \le \frac12 -
\frac{\ensuremath{\varepsilon}}{2}$, we have by Lemma \ref{zeta} that
$|F(s+u,\chi)F(s+v,\chi)| \ll \log^4 T \ll \log^4 x$
on all three new segments.  As in the proof of (1), the integral over
the new contour is $\ll x^b \log^5 x$.
We pick up residue terms from poles of
$F(s+v,\chi)$ inside the rectangle coming from the nontrivial zeros of
$L(s,\chi)$, plus a pole at $s=\frac12-v$ from
the $\frac{\zeta'}{\zeta}(2s+2v)$ term if $\chi^2=\chi_0$.  The sum of
the residues is
$$
\sum_{|\ensuremath{\gamma}|\le T}
\frac{F(\frac{1}{2}+u-v+i\ensuremath{\gamma}
  ,\chi)x^{\frac{1}{2}-v+i\ensuremath{\gamma}}}{\frac{1}{2}-v+i\ensuremath{\gamma}}
-A(\chi) \frac{F(\frac{1}{2}+u-v,\chi)x^{\frac{1}{2}-v}}{1-2v},
$$
and (2) follows.

Finally, consider the case $0\le u,v \le 2\ensuremath{\varepsilon}$.
Let $b = \frac{1}{2} - 3\ensuremath{\varepsilon}$ and deform the contour as in the
previous cases.  As before, the integral over the new contour is
$O(x^b \log^5 x)$.  This time, we pick up
residues from poles of both $F(s+u,\chi)$ and $F(s+v,\chi)$.  The sum of the
residues is
\begin{align*}
\sum_{|\ensuremath{\gamma}|\le T}  \Bigl( \frac{F(\frac{1}{2}+u-v+i\ensuremath{\gamma},\chi)
x^{\frac{1}{2}-v+i\ensuremath{\gamma}}}{\frac{1}{2}-v+i\ensuremath{\gamma}}+ \frac{F(\frac{1}{2}-u+v+i\ensuremath{\gamma}
   ,\chi)x^{\frac{1}{2}-u+i\ensuremath{\gamma}}}{\frac{1}{2}-u+i\ensuremath{\gamma}} \Bigr) \\
- A(\chi) \( \frac{F(\frac{1}{2}+u-v,\chi)x^{\frac{1}{2}-v}}{1-2v}+
\frac{F(\frac{1}{2}-u+v,\chi)x^{\frac{1}{2}-u}}{1-2u} \),
\end{align*}
and (3) follows.
\end{proof}

\begin{proof}[Proof of Lemma \ref{analytic}]
Begin with
$$
 \int_0^\infty \int_0^\infty G(x,u,v;\chi)\, du\, dv  = I_1 + I_2 +
 2I_3 + I_4,
$$
where $I_1$ is the integral over  $\max(u,v)\ge \log x$,
$I_2$ is the integral over $2\ensuremath{\varepsilon} \le \max(u,v) \le \log
x$ and $\min(u,v)\ge \ensuremath{\varepsilon}$, $I_3$ is the integral over $0\le v\le
\ensuremath{\varepsilon}$, $2\ensuremath{\varepsilon} \le u \le \log x$, and $I_4$ is the integral over $0\le
u,v \le 2\ensuremath{\varepsilon}$.  For $\max(u,v)\ge \log x$,
$$
|G(x,u,v;\chi)| \le \sum_{p\le x} \frac{\log p}{p^u} \sum_{p\le x}
\frac{\log q}{q^v} \ll \frac{x}{2^{\max(u,v)}},
$$
whence $I_1 \ll x^{1-\log 2}$.  By Lemma \ref{Gbounds} (1), $I_2 \ll
x^{1/2-\ensuremath{\varepsilon}/2} \log^7 x$.

By Lemma \ref{Gbounds} (2),
\begin{equation}\label{I3}
\begin{split}
I_3 &= x^{1/2} \int_0^\ensuremath{\varepsilon} \int_{2\ensuremath{\varepsilon}}^{\log x}
\sum_{|\ensuremath{\gamma}|\le T} \frac{F(\frac{1}{2}+u-v+i\ensuremath{\gamma}
  ,\chi)x^{-v+i\ensuremath{\gamma}}}{\frac{1}{2}-v+i\ensuremath{\gamma}}
-A(\chi) \frac{F(\frac{1}{2}+u-v,\chi)x^{-v}}{1-2v}\, du\, dv \\
&\qquad\qquad + O(x^{1/2-\frac{3\ensuremath{\varepsilon}}{2}} \log^6 x).
\end{split}
\end{equation}
By Lemmas \ref{NTchi} and \ref{zeta},
\begin{equation}\label{I31}
\int_{0}^{\ensuremath{\varepsilon}} \int_{2\ensuremath{\varepsilon}}^{\log x} \frac{F(\frac{1}{2}+u-v
  ,\chi)x^{-v}}{1-2v} du\ dv \ll \int_{0}^{\ensuremath{\varepsilon}} x^{-v}\, dv
\ll \frac{1}{\log x}.
\end{equation}
Let
$$
\Sigma_1(x) = (\log x) \int_0^\ensuremath{\varepsilon} \int_{2\ensuremath{\varepsilon}}^{\log x}
\sum_{0<|\ensuremath{\gamma}|<T} \frac{F(\frac{1}{2}+u-v+i\ensuremath{\gamma}
  ,\chi)x^{-v+i\ensuremath{\gamma}}}{\frac{1}{2}-v+i\ensuremath{\gamma}}\, du\, dv.
$$
Since $\frac{1}{2}+u-v \ge \frac{1}{2}+\ensuremath{\varepsilon}$ for $0 \le v  \le \ensuremath{\varepsilon}$ and $2
\ensuremath{\varepsilon} \le u \le \log x$, by Lemmas \ref{FsHs}, \ref{NTchi}, and
\ref{zeta},
$$
F(\frac{1}{2}+u-v+i\ensuremath{\gamma},\chi) = -\frac{L'}{L}(\frac{1}{2}+u-v+i\ensuremath{\gamma},
\chi) + O(1) \ll \log(|\ensuremath{\gamma}|+3).
$$
We also have $F(1/2+u-v+i\ensuremath{\gamma},\chi) \ll 2^{-u}$ for $u\ge 2$.
Thus, for positive integers $n$,
\begin{align*}
\int_{2^n}^{2^{n+1}} |\Sigma_1 (e^y)|^2 dy &\ll 2^{2n}
\sum_{|\ensuremath{\gamma}_1|,|\ensuremath{\gamma}_2|\le T} \frac{\log (|\ensuremath{\gamma}_1|+3)
  \log(|\ensuremath{\gamma}_2|+3)}{|\ensuremath{\gamma}_1 \ensuremath{\gamma}_2|} \\
& \qquad\qquad \times \int_{0}^{\ensuremath{\varepsilon}}
\int_{0}^{\ensuremath{\varepsilon}}  \Bigg| \int_{2^n}^{2^{n+1}}
e^{y(-v_1+i\ensuremath{\gamma}_1-v_2-i\ensuremath{\gamma}_2)} dy \Bigg| dv_1 dv_2.
\end{align*}
The summands with $|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2| <  1$ contribute, by Lemma
\ref{NTchi},
\begin{align*}
&\ll 2^{2n} \sum_{\substack{|\ensuremath{\gamma}_1|, |\ensuremath{\gamma}_2| \le T
      \\ |\ensuremath{\gamma}_1-\ensuremath{\gamma}_2| < 1}}   \frac{\log (|\ensuremath{\gamma}_1|+3) \log
    (|\ensuremath{\gamma}_2|+3)}{|\ensuremath{\gamma}_1||\ensuremath{\gamma}_2|} \int_{2^n}^{2^{n+1}}
  \Big(\int_{0}^{\ensuremath{\varepsilon}} e^{-vy} dv\Big)^2 dy \\
&\ll 2^n \sum_{|\ensuremath{\gamma}|\le  T }  \frac{{\log}^3
  (|\ensuremath{\gamma}|+3)}{|\ensuremath{\gamma}|^2} \ll 2^n.
\end{align*}
The summands with $|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2| \ge 1$ contribute, by Lemma \ref{sumg1g2},
\[
\ll \sum_{\substack{|\ensuremath{\gamma}_1|, |\ensuremath{\gamma}_2| < T \\ |\ensuremath{\gamma}_1-\ensuremath{\gamma}_2|
    \geq 1}} \frac{2^{2n} \log (|\ensuremath{\gamma}_1|+3) \log
  (|\ensuremath{\gamma}_2|+3)}{|\ensuremath{\gamma}_1||\ensuremath{\gamma}_2||\ensuremath{\gamma}_1-\ensuremath{\gamma}_2|}
\Big(\int_{0}^{\ensuremath{\varepsilon}} e^{-v2^n} dv \Big)^2
 \ll 1.
\]
Thus, $\int_{2^n}^{2^{n+1}} |\Sigma_{1} (e^y)|^2 dy = O (2^n)$.
Summing over $n\le \frac{\log Y}{\log 2}+1$ yields
$\int_1^Y |\Sigma_1(e^y)|^2\, dy =  O(Y)$.

Finally, using Lemma \ref{Gbounds} (3) gives
\begin{equation}\label{I4}
\begin{split}
I_4 &= x^{1/2} \int_0^{2\ensuremath{\varepsilon}} \int_0^{2\ensuremath{\varepsilon}}
 \sum_{|\ensuremath{\gamma}|\le T}
 \frac{F(\frac{1}{2}+u-v+i\ensuremath{\gamma}
   ,\chi)x^{-v+i\ensuremath{\gamma}}}{\frac{1}{2}-v+i\ensuremath{\gamma}} +
 \frac{F(\frac{1}{2}-u+v+i\ensuremath{\gamma}
   ,\chi)x^{-u+i\ensuremath{\gamma}}}{\frac{1}{2}-u+i\ensuremath{\gamma}} \\
& \qquad -
  A(\chi) \( \frac{F(\frac{1}{2}+u-v,\chi)x^{-v}}{1-2v}+
  \frac{F(\frac{1}{2}-u+v,\chi)x^{-u}}{1-2u} \) \, du\, dv
  + O (x^{\frac12-3\ensuremath{\varepsilon}} \log^3 x).
\end{split}
\end{equation}
Now assume $\chi^2=\chi_0$.  We will show that
\begin{equation}\label{I41}
- \int_0^{2\ensuremath{\varepsilon}} \int_0^{2\ensuremath{\varepsilon}}
 \frac{F(\frac{1}{2}+u-v,\chi)x^{-v}}{1-2v}+
  \frac{F(\frac{1}{2}-u+v,\chi)x^{-u}}{1-2u} \, du\, dv
= \frac{\log\log x  + O(1)}{\log x}.
\end{equation}
Together with \eqref{I3}, \eqref{I31} and \eqref{I4}, this completes
the proof of Lemma \ref{analytic}.

Note that $F(\frac{1}{2}+w)=-\frac{1}{2w}+O(1)$ by Lemmas \ref{FsHs}
and \ref{zeta}.
Replacing $x$ with $e^y$, the left side of \eqref{I41} is
$$
=\frac12 \int_{0}^{2\ensuremath{\varepsilon}} \int_{0}^{2\ensuremath{\varepsilon}}
  \frac{e^{-yv}}{(u-v)(1-2v)} +
  \frac{e^{-yu}}{(v-u)(1-2u)}du\ dv + O\Big(\int_{0}^{2\ensuremath{\varepsilon}}
  \int_{0}^{2\ensuremath{\varepsilon}} e^{-yv} du\ dv\Big).
$$
The error term above is $O(1/y)$.
In the main term, when $|u-v|<1/y$, the integrand is $O(y e^{-vy})$
and the corresponding part of the double integral is $O(1/y)$.
When $u \ge v + 1/y$, the integrand is
$$
\frac{e^{-vy}}{u-v} + O\pfrac{v e^{-vy}+e^{-uy}}{u-v}
$$
and the corresponding part of the double integral is
$$
\int_0^{2\ensuremath{\varepsilon}} e^{-vy} \log\pfrac{y}{2\ensuremath{\varepsilon}-v}\, dv + O\pfrac{1}{y} =
\frac{\log y + O(1)}{y}.
$$
The contribution from $u\le v-1/y$ is, by symmetry, also $\frac{\log y+O(1)}{y}$.
The asymptotic \eqref{I41} follows.
\end{proof}

\section{Proof of Lemma \ref{doubleint}}\label{sec:doubleint}

\begin{lem}\label{Qint}
 Uniformly for $y\ge 1$, $0 < |\xi| \leq 1$, $|w| \geq \frac{1}{2}$
 and $a\ge 0$ we have
$$
\Biggl| \int_0^{2 \ensuremath{\varepsilon}} \int_0^{2 \ensuremath{\varepsilon}} \frac{v^a e^{-vy}}{(u-v+i\xi)(w-v)}
 du\ dv \Biggr|   \ll \frac{(4\ensuremath{\varepsilon})^{a}\log \min(2y, \frac{2}{|\xi|})}{y|w|}.
$$
\end{lem}

\begin{proof}  Let $I$ denote the double integral in the
Lemma.  If $|\xi| \geq \frac{1}{y}$, then
\begin{align*}
I&\ll \frac{1}{|w|} \int_0^{2 \ensuremath{\varepsilon}} v^a e^{-vy} \int_0^{2\ensuremath{\varepsilon}}
\min\( \frac{1}{|u-v|}, \frac{1}{|\xi|} \)\, du\, dv \\
&\ll \frac{(2\ensuremath{\varepsilon})^a}{|w|} \(1 + \log \frac{2}{|\xi|}\)
 \int_0^{2 \ensuremath{\varepsilon}} e^{-vy}\, dv
\ll \frac{(2\ensuremath{\varepsilon})^a \log({\frac{2}{|\xi|}})}{y|w|}.
\end{align*}
If $|\xi| < \frac{1}{y}$, let $I=I_1+I_2+I_3$, where $I_1$ is the part of
 $I$ coming from $|u-v| \leq |\xi|$, $I_2$ is the part of $I$ coming from
 $|\xi| < |u-v| \leq \frac{1}{y}$, and $I_3$ is the part of $I$ coming
 from $|u-v|>\frac{1}{y}$.  We have
$$
I_1 \ll \frac{1}{|w\xi|} \; \;\;  \iint
\limits_{\substack{0\le u,v\le 2\ensuremath{\varepsilon} \\ |u-v|\leq |\xi|}}
v^a e^{-vy} du\ dv \ll \frac{(2\ensuremath{\varepsilon})^a}{y|w|}.
$$
and
$$
I_3 \ll \frac{(2\ensuremath{\varepsilon})^a}{|w|}\;\;  \iint
\limits_{\substack{0\le u,v \le 2\ensuremath{\varepsilon} \\ |u-v|\geq \frac{1}{y}}}
\frac{e^{-vy}}{|u-v|} du\ dv
\ll \frac{(2\ensuremath{\varepsilon})^a}{|w|} \int_0^{2 \ensuremath{\varepsilon}} e^{-vy}(\log y + 1) dv
\ll \frac{(2\ensuremath{\varepsilon})^a \log(2y)}{y|w|}.
$$
By symmetry,
$$
I_2 = \frac12 \iint\limits_{|\xi| < |u-v| \le 1/y} \frac{v^a
  e^{-vy}}{(u-v+i\xi)(w-v)} + \frac{u^a e^{-uy}}{(v-u+i\xi)(w-u)}\,
  du\, dv.
$$
Since, $|u^a-v^a|\le a|u-v| (2\ensuremath{\varepsilon})^{a-1}$,
\begin{equation}\label{euy}\begin{split}
u^a e^{-uy} - v^a e^{-vy} &= e^{-vy} v^a \( e^{(v-u)y}-1 \) +
e^{-vy} (u^a-v^a) e^{(v-u)y} \\&\ll e^{-vy} y |u-v| (4\ensuremath{\varepsilon})^a.
\end{split}\end{equation}
We deduce that
\begin{align*}
I_2&= \!\!\!\iint\limits_{\substack{0\le u,v\le 2\ensuremath{\varepsilon} \\ |\xi| < |u-v| \le 1/y}}
 \!\! \frac{(w-u)(u-v)(u^ae^{-uy}-v^ae^{-vy})+u^ae^{-uy}(u-v)^2+O(|\xi w| (2\ensuremath{\varepsilon})^a
  e^{-vy})}{2(u-v+i\xi)(v-u+i\xi)(w-u)(w-v)} du\ dv \\
&\ll \frac{(4\ensuremath{\varepsilon})^a}{|w|}\;\;\;
 \iint\limits_{\substack{0\le u,v\le 2\ensuremath{\varepsilon} \\ |\xi|<|u-v| \le 1/y}}
y e^{-vy} + \frac{|\xi| e^{-vy}}{|u-v|^2} \, du\, dv
\ll \frac{(4\ensuremath{\varepsilon})^a}{y|w|}.
\end{align*}
\end{proof}

\begin{proof}[Proof of Lemma \ref{doubleint}]
Let $y=\log x$.
We first note by Lemmas \ref{FsHs} and \ref{NTchi},
$$
F(\frac{1}{2}+u-v +i \ensuremath{\gamma},\chi) = \frac{m(\ensuremath{\gamma})}{u-v}+R(\ensuremath{\gamma},u-v)+R'(\ensuremath{\gamma},u-v),
$$
where
$$
R(\ensuremath{\gamma},w) = \sum_{0< |\ensuremath{\gamma}'-\ensuremath{\gamma}| \leq 1} \frac{1}{w+i(\ensuremath{\gamma} - \ensuremath{\gamma}')}, \qquad
R'(\ensuremath{\gamma},u-v) =  O(\log (|\ensuremath{\gamma}|+3)).
$$
Then, the double integral in Lemma \ref{doubleint} is
$$
=\sum_{i=1}^4 \Sigma_{2,i}(y)+2 \sum_{\substack{|\ensuremath{\gamma}| \leq T \\
\ensuremath{\gamma} \text{ distinct}}} m^2(\ensuremath{\gamma}) e^{iy\ensuremath{\gamma}}
(\frac{1}{2}+i\ensuremath{\gamma}) \int_0^{2 \ensuremath{\varepsilon}-2^{-n}} \frac{ e^{-yv}}{\frac{1}{2}-v+i \ensuremath{\gamma}}
\int_{v+2^{-n}}^{2 \ensuremath{\varepsilon}} \frac{du}{(u-v)(\frac{1}{2}-u+i \ensuremath{\gamma})} dv,
$$
where
\begin{align*}
\Sigma_{2,1}(y) &= 2 \int_0^{2 \ensuremath{\varepsilon}} \int_0^{2 \ensuremath{\varepsilon}} \sum_{|\ensuremath{\gamma}| \leq T}
\frac{ R(\ensuremath{\gamma},u-v) e^{y(-v+i \ensuremath{\gamma})}}{\frac{1}{2}-v+i \ensuremath{\gamma}} du\ dv, \\
\Sigma_{2,2}(y) &= 2  \int_0^{2 \ensuremath{\varepsilon}} \int_0^{2 \ensuremath{\varepsilon}}
\frac{ R'(\ensuremath{\gamma},u-v) e^{y(-v+i \ensuremath{\gamma})}}{\frac{1}{2}-v+i \ensuremath{\gamma}}\, du\, dv, \\
\Sigma_{2,3}(y)& =  \sum_{\substack{|\ensuremath{\gamma}| \leq T \\
\ensuremath{\gamma} \text{ distinct}}} m^2(\ensuremath{\gamma}) e^{iy \ensuremath{\gamma}} (\frac{1}{2} + i\ensuremath{\gamma})
 \iint \limits_{\substack{0 \leq u,v \leq 2 \ensuremath{\varepsilon} \\ |u-v| \leq 2^{-n}}}
\frac{ e^{-yv}-e^{-uy}}{(u-v)(\frac{1}{2}-v+i \ensuremath{\gamma})(\frac{1}{2}-u+i \ensuremath{\gamma})} dv\ du, \\
\Sigma_{2,4}(y) &= 2 \sum_{\substack{|\ensuremath{\gamma}| \leq T \\
\ensuremath{\gamma} \text{ distinct}}} m^2(\ensuremath{\gamma}) e^{iy \ensuremath{\gamma}} (\frac{1}{2} + i\ensuremath{\gamma})
 \int_{2^{-n}}^{2 \ensuremath{\varepsilon}} \int_{0}^{v-2^{-n}}
\frac{ e^{-yv}}{(u-v)(\frac{1}{2}-v+i \ensuremath{\gamma})(\frac{1}{2}-u+i \ensuremath{\gamma})} du\ dv.
\end{align*}
We show that $\sum_{j=1}^4 \Sigma_{2,j}(y)$ is small in mean square.
Note that for  $2^n < y \le 2^{n+1}$, $T=T(e^y)$ is constant.
Also, by Lemma \ref{NTchi}, we have
\begin{equation}\label{mgam}
m(\ensuremath{\gamma}) \ll \log (|\ensuremath{\gamma}|+3).
\end{equation}

First, by Lemmas \ref{NTchi} and \ref{sumg1g2},
\begin{equation}\label{Sigma22}
\begin{split}
\int_{2^n}^{2^{n+1}}  \!\! &|\Sigma_{2,2}(y)|^2\, dy = 4
\iiiint\limits_{[0,2\ensuremath{\varepsilon}]^4} \!\!\! \sum_{\substack{|\ensuremath{\gamma}_1|\le T \\ |\ensuremath{\gamma}_2| \le T}}
\frac{R'(\ensuremath{\gamma}_1,u_1-v_1)\overline{R'(\ensuremath{\gamma}_2,u_2-v_2)}}
{(\frac12-v_1+i\ensuremath{\gamma}_1)(\frac12-v_2-i\ensuremath{\gamma}_2)} \\
&\qquad\qquad \times
\int_{2^n}^{2^{n+1}} \!\!\!\! e^{y(-v_1-v_2+i\ensuremath{\gamma}_1-i\ensuremath{\gamma}_2)}\, dy \, du_j dv_j \\
&\ll \sum_{|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2|>1} \frac{\log (|\ensuremath{\gamma}_1|+3) \log (|\ensuremath{\gamma}_2|+3)}
{|\ensuremath{\gamma}_1\ensuremath{\gamma}_2|\cdot|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2|} \iiiint\limits_{[0,2\ensuremath{\varepsilon}]^4}
 e^{-2^n(v_1+v_2)}\, du_j dv_j \\
&\qquad +  \sum_{|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2|\le 1} \frac{\log (|\ensuremath{\gamma}_1|+3) \log (|\ensuremath{\gamma}_2|+3)}
{|\ensuremath{\gamma}_1\ensuremath{\gamma}_2|} \int_{2^n}^{2^{n+1}}\iiiint\limits_{[0,2\ensuremath{\varepsilon}]^4} e^{-y(v_1+v_2)}\,
du_j dv_j dy \\
&\ll 2^{-n}.
\end{split}
\end{equation}

For the remaining sums, for brevity we define
$$
\rho_1 = \frac12 + i\ensuremath{\gamma}_1, \qquad \rho_2 = \frac12 - i\ensuremath{\gamma}_2.
$$
Next,
\begin{multline*}
\int_{2^n}^{2^{n+1}} |\Sigma_{2,3}(y)|^2 dy = \int_{2^{n}}^{2^{n+1}}
 \sum_{|\ensuremath{\gamma}_1|, |\ensuremath{\gamma}_2| \leq T} m(\ensuremath{\gamma}_1) m(\ensuremath{\gamma}_2) e^{i y (\ensuremath{\gamma}_1 - \ensuremath{\gamma}_2)} \rho_1 \rho_2 \\ \times
 \iiiint \limits_{\substack{[0,2\ensuremath{\varepsilon}]^4 \\ |u_j-v_j| \leq 2^{-n}}}\!\!\!
 \frac{(e^{-v_1y}-e^{-u_1y})(e^{-v_2 y}-e^{-u_2 y})}
{\prod_{j=1}^2 (u_j-v_j)(\rho_j-v_j)(\rho_j-u_j)}
 dv_j dv_j\, dy.
\end{multline*}
By \eqref{euy}, the integrand in the quadruple integral is
$\ll y^2 e^{-uy-u_1 y}|\rho_1 \rho_2|^{-2}.$
By Lemma \ref{NTchi}, for a given $\ensuremath{\gamma}_1$, there are $\ll \log (|\ensuremath{\gamma}_1|+3)$ zeros $\ensuremath{\gamma}_2$ with $|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2| < 1$.  Hence, the contribution from terms with $|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2| < 1$ is
$$
\ll 2^{-n} \sum_{|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2| < 1} \frac{m(\ensuremath{\gamma}_1)m(\ensuremath{\gamma}_2)}{|\rho_1 \rho_2|}
\ll 2^{-n} \sum_{\ensuremath{\gamma}_1} \frac{\log^3 (|\ensuremath{\gamma}_1|+3)}{|\ensuremath{\gamma}_1|^2} \ll 2^{-n}.
$$
Using integration by parts,  we have
$$
\int_{2^n}^{2^{n+1}} e^{iy(\ensuremath{\gamma}_1-\ensuremath{\gamma}_2)} (e^{-v_1y}-e^{-u_2y})(e^{-v_1y}-e^{-u_2y})\, dy
\ll \frac{2^{3n} |u_1-v_1|\, |u_2-v_2| e^{-2^n(u_1+u_2)}}{|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2|}
$$
uniformly in $u_1,v_1,u_2,v_2$.  Thus, by \eqref{mgam} and Lemma \ref{sumg1g2},
the contribution from terms with  $|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2| \ge 1$ is
$$
\ll 2^{-n} \sum_{|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2| \ge 1} \frac{m(\ensuremath{\gamma}_1)m(\ensuremath{\gamma}_2)}
{|\rho_1 \rho_2| \cdot |\ensuremath{\gamma}_1-\ensuremath{\gamma}_2|} \ll 2^{-n}.
$$
Combining these estimates, we have
\begin{equation} \label{Sigma23}
\int_{2^n}^{2^{n+1}} |\Sigma_{2,3}(y)|^2 dy \ll 2^{-n}.
\end{equation}

In the same manner, we have
$$
\int_{2^n}^{2^{n+1}} \!\!\! |\Sigma_{2,4}(y)|^2 dy\ =  \!\!
\sum_{\substack{|\ensuremath{\gamma}_1|\le T \\ |\ensuremath{\gamma}_2| \leq T}} \!\! m(\ensuremath{\gamma}_1)m(\ensuremath{\gamma}_2)
\rho_1 \rho_2 \int_{2^{n}}^{2^{n+1}} \!\!\!\!
\iiiint\limits_{\substack{[0,2\ensuremath{\varepsilon}]^4 \\ u_j \le v_j - 2^{-n}}}
\frac{ e^{y(-v_1-v_2+i(\ensuremath{\gamma}_1-\ensuremath{\gamma}_2))}du_j dv_j}{\prod_{j=1}^2
(u_j-v_j)(\rho_j-v_j)(\rho_j-u_j)}  dy.
$$
The contribution to the right side from terms with
 $|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2|<1$ is
\begin{align*}
&\ll \sum_{|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2| < 1}
\frac{m(\ensuremath{\gamma}_1)m(\ensuremath{\gamma}_2)}{|\ensuremath{\gamma}_1\ensuremath{\gamma}_2|} \int_{2^{n}}^{2^{n+1}}
\Biggl( \int_{2^{-n}}^{2 \ensuremath{\varepsilon}} \int_{0}^{v-2^{-n}} \frac{e^{-yv}}{(v-u)}\, du\, dv
\Biggr)^2  \\
&\ll \sum_{\ensuremath{\gamma}_1} \frac{\log^3 (|\ensuremath{\gamma}_1|+3)} {|\ensuremath{\gamma}_1|^2}
\int_{2^{n}}^{2^{n+1}} \Biggl( \int_{1/y}^{\infty} e^{-yv} \log(y v)\, dv \Biggr)^2
\ll 2^{-n}.
\end{align*}
The terms with $|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2|>1$ contribute
\begin{align*}
&\ll \sum_{\substack{|\ensuremath{\gamma}_1|, |\ensuremath{\gamma}_2| < T \\ |\ensuremath{\gamma}_1-\ensuremath{\gamma}_2| > 1}}  \frac{m(\ensuremath{\gamma}_1)m(\ensuremath{\gamma}_2)}
{|\ensuremath{\gamma}_1\ensuremath{\gamma}_2| \cdot |\ensuremath{\gamma}_1-\ensuremath{\gamma}_2|} \Biggl( \int_{2^{-n}}^{2 \ensuremath{\varepsilon}} \int_{0}^{v-2^{-n}}
\frac{ e^{-2^nv}}{v-u}\, du\ dv \Biggr)^2 \\
&\ll \sum_{|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2| > 1}  \frac{\log (|\ensuremath{\gamma}_1|+3) \log (|\ensuremath{\gamma}_2|+3)}
{|\ensuremath{\gamma}_1\ensuremath{\gamma}_2| \cdot |\ensuremath{\gamma}_1-\ensuremath{\gamma}_2| } \pfrac{1}{2^n}^2 \ll \frac{1}{2^{2n}}.
\end{align*}
Therefore,
\begin{equation} \label{Sigma24}
\int_{2^n}^{2^{n+1}} |\Sigma_{2,4}(y)|^2 dy \ll 2^{-n}.
\end{equation}

Estimating an average of $\Sigma_{2,1}(y)$ is more complicated, since
$R(\ensuremath{\gamma},w)$ could be very large if $|w|$ is small and there is another
 $\ensuremath{\gamma}'$ very close to $\ensuremath{\gamma}$.
We get around the problem by noticing that $R(\ensuremath{\gamma},w)+R(\ensuremath{\gamma},-w)$ is always small.
We first have, by \eqref{euy} and Lemma \ref{NTchi},
\begin{multline}\label{Sig211}
 \int_{2^n}^{2^{n+1}} |\Sigma_{2,1}(y)|^2 dy \ll
\sum_{\ensuremath{\gamma}_1,\ensuremath{\gamma}_2} \log^2(|\ensuremath{\gamma}_1|+3) \log^2(|\ensuremath{\gamma}_2|+3)
\max_{\substack{0<|\ensuremath{\gamma}_1-\ensuremath{\gamma}_1'|\le 1  \\
0<|\ensuremath{\gamma}_2-\ensuremath{\gamma}_2'|\le 1}} \int_{2^n}^{2^{n+1}} e^{iy(\ensuremath{\gamma}_1-\ensuremath{\gamma}_2)}  \\
\times
 \iiiint\limits_{[0,2\ensuremath{\varepsilon}]^4}
\frac{ e^{y(-v_1-v_2)}}
{(u_1-v_1+i\xi_1)(\rho_1-v_1)(u_2-v_2+i\xi_2)(\rho_2-v_2)}
du_j dv_j\, dy,
\end{multline}
where $\xi_1=\ensuremath{\gamma}_1-\ensuremath{\gamma}_1'$ and $\xi_2=-(\ensuremath{\gamma}_2-\ensuremath{\gamma}_2')$.
Let
$$
M(\ensuremath{\gamma}) = \max_{\substack{|\ensuremath{\gamma}-\ensuremath{\gamma}_1|\leq 1 \\ 0 < |\ensuremath{\gamma}_1-\ensuremath{\gamma}_1'| < 1}} \frac{2}{|\ensuremath{\gamma}_1-\ensuremath{\gamma}_1'|}.
$$
By Lemmas \ref{zeta} and \ref{Qint},
the terms with $|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2|<1$ contribute
\begin{align*}
&\ll  \sum_{|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2| < 1} \frac{
\log^2(|\ensuremath{\gamma}_1|+3)\log^2(|\ensuremath{\gamma}_2|+3)}
{|\ensuremath{\gamma}_1\ensuremath{\gamma}_2|} \int_{2^n}^{2^{n+1}} \frac{1}{y^2}
  \prod_{j=1}^2  \log
  \(\min\(2y,\frac{2}{|\ensuremath{\gamma}_j - \ensuremath{\gamma}_j'|}\)\)  \, dy \\
&\ll \frac{1}{2^{n}} \sum_{\ensuremath{\gamma}_1} \frac{\log^5 (|\ensuremath{\gamma}_1|+3)}{|\ensuremath{\gamma}_1|^2}
\log^2 \( \min(2^{n+2},M(\ensuremath{\gamma})) \) \, = o\pfrac{n^2}{2^n} \qquad (n\to\infty).
\end{align*}

Now suppose $|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2|>1$.  With $\ensuremath{\gamma}_1,\ensuremath{\gamma}_2,\ensuremath{\gamma}_1',\ensuremath{\gamma}_2'$ all fixed, let
$\Delta=\ensuremath{\gamma}_1-\ensuremath{\gamma}_2$.
Fixing $u_1,v_1,u_2,v_2$, we integrate over $y$ first.  The quintuple
integral in \eqref{Sig211} is $J(2^{n+1})-J(2^n)$, where
$$
J(y)=e^{iy\Delta} \iiiint\limits_{[0,2\ensuremath{\varepsilon}]^4} \frac{e^{-y(v_1+v_2)}}
{(i\Delta-v_1-v_2)\prod_{j=1}^2 (u_j-v_j+i\xi_j)(\rho_j-v_j)} \, du_j dv_j.
$$
Using
$$
\frac{1}{i\Delta-v_1-v_2}=\frac{1}{i\Delta}\sum_{k=0}^\infty
\pfrac{v_1+v_2}{i\Delta}^k=\sum_{a,b\ge 0} \binom{a+b}{a} \frac{v_1^a v_2^b}
{(i\Delta)^{a+b}},
$$
together with Lemma \ref{Qint}, yields
$$
|J(y)| \ll \frac{\log^2 y}{|\rho_1 \rho_2 \Delta| y^2} \sum_{a,b\ge 0}
\binom{a+b}{a} \pfrac{4\ensuremath{\varepsilon}}{|\Delta|}^{a+b} \ll
 \frac{\log^2 y}{|\rho_1 \rho_2 \Delta| y^2}.
$$
Therefore, by Lemma \ref{sumg1g2},
$$
\sum_{\ensuremath{\gamma}_1,\ensuremath{\gamma}_2} \log^2(|\ensuremath{\gamma}_1|+3) \log^2(|\ensuremath{\gamma}_2|+3)
\max_{\substack{0<|\ensuremath{\gamma}_1-\ensuremath{\gamma}_1'|\le 1 \\
0<|\ensuremath{\gamma}_2-\ensuremath{\gamma}_2'|\le 1}} |J(2^{n+1})-J(2^n)| \ll \frac{n^2}{2^{2n}},
$$
and hence
\begin{equation}\label{Sigma21}
\int_{2^n}^{2^{n+1}} \left| \Sigma_{2,1}(y) \right|^2 = o(n^2 2^{-n}).
\end{equation}
Define
$$
\Sigma_2(x;\chi) = (\log x) \sum_{j=1}^4 \Sigma_{2,j}(\log x).
$$
By \eqref{Sigma22}, \eqref{Sigma23}, \eqref{Sigma24}
and \eqref{Sigma21},
\[
\int_2^Y |\Sigma_2(e^y;\chi)|^2\, dy \le 4 \sum_{j=1}^4
\sum_{n\le \frac{\log Y}{\log 2} + 1} 2^{2n}
 \int_{2^n}^{2^{n+1}}  |\Sigma_{2,j}(y)|^2\, dy
= o (Y \log^2 Y) \qquad (Y\to\infty).
\]
This completes the proof of Theorem \ref{doubleint}.
\end{proof}

\section{Proof of Lemma \ref{TT0}}\label{sec:TT0}

Put $y=\log x$.  For any $\ensuremath{\gamma}$ we have
\begin{align*}
\int_0^{2 \ensuremath{\varepsilon}-2^{-n}} &\frac{ e^{-yv}}{\frac{1}{2}-v+i \ensuremath{\gamma}} \int_{v+2^{-n}}^{2 \ensuremath{\varepsilon}} \frac{du}{(u-v)(\frac{1}{2}-u+i \ensuremath{\gamma})} dv \\
&= \int_0^{2 \ensuremath{\varepsilon}-2^{-n}}  e^{-yv} \Big(\frac{1}{\frac{1}{2}+i\ensuremath{\gamma}} +
 O(\frac{v}{\frac{1}{4}+{\ensuremath{\gamma}}^2}) \Big)
\int_{v+2^{-n}}^{2 \ensuremath{\varepsilon}} \Big(\frac{1}{\frac{1}{2}+i\ensuremath{\gamma}} + O(\frac{u}{\frac{1}{4}+{\ensuremath{\gamma}}^2}) \Big) \frac{du}{u-v} dv \\
&= \frac{M+E}{(1/2+i\ensuremath{\gamma})^2},
\end{align*}
where
$$
M =
\int_0^{2 \ensuremath{\varepsilon}-2^{-n}}  e^{-yv} \( \log(2\ensuremath{\varepsilon}-v)+\log 2^n \)\, dv =
\frac{\log y + O(1)}{y}
$$
and
\begin{align*}
E &\ll \int_0^{2\ensuremath{\varepsilon}-2^{-n}} e^{-yv} \int_{v+2^{-n}}^{2\ensuremath{\varepsilon}}
\frac{u}{u-v}\, du\, dv \\
&\ll \int_0^{2\ensuremath{\varepsilon}-2^{-n}} e^{-yv} \( 1 + v\log 2^n + v\log (2\ensuremath{\varepsilon}-v) \)
\, dv \ll \frac{1}{y}.
\end{align*}
Hence, the zeros with $|\ensuremath{\gamma}| \le T_0$ contribute
$$
 \frac{2\log\log x}{\log x} \sum_{\substack{|\ensuremath{\gamma}|\le T_0 \\ \ensuremath{\gamma} \text{ distinct}}}
 \frac{m^2(\ensuremath{\gamma})x^{i\ensuremath{\gamma}}}{1/2+i\ensuremath{\gamma}} + O\pfrac{\log^3 T_0}{\log x}.
$$

Next, let $\Sigma_3(x;T_0)$ be the sum over zeros with $T_0 < |\ensuremath{\gamma}| \le T$.
We have
\begin{multline}\label{lem3.7}
\int_{2^n}^{2^{n+1}} |\Sigma_3(e^y,T_0)|^2 dy \le
 \sum_{T_0 \leq |\ensuremath{\gamma}_1|, |\ensuremath{\gamma}_2| \leq T}  2^{2n+2} m(\ensuremath{\gamma}_1) m(\ensuremath{\gamma}_2)
\(\frac{1}{2} + i\ensuremath{\gamma}_1\)
 \(\frac{1}{2} - i\ensuremath{\gamma}_2\)
\\ \int_{2^n}^{2^{n+1}} e^{y i (\ensuremath{\gamma}_1 - \ensuremath{\gamma}_2)}
\iiiint\limits_{u_j\ge v_j+2^{-n}}
\frac{e^{-yv_1-yv_2}}{\prod_{j=1}^2 (u_j-v_j)(\frac{1}{2}-v_j+i \ensuremath{\gamma}_j)
(\frac{1}{2}-u_j+i \ensuremath{\gamma}_j)} du_j dv_j\ dy.
\end{multline}

The sum over $|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2|<1$ on the right side of \eqref{lem3.7} is
\begin{align*}
&\ll \sum_{\substack{T_0 \leq |\ensuremath{\gamma}_1|, |\ensuremath{\gamma}_2| \leq T \\ |\ensuremath{\gamma}_1-\ensuremath{\gamma}_2| < 1}}
 \frac{2^{2n}m(\ensuremath{\gamma}_1)m(\ensuremath{\gamma}_2)}{|\ensuremath{\gamma}_1||\ensuremath{\gamma}_2|} \int_{2^n}^{2^{n+1}}
 \iiiint\limits_{u_j\ge v_j+2^{-n}}
 \frac{ e^{-yv_1-yv_2}}{(u_1-v_1)(u_2-v_2)\, }du_j dv_j\,  dy \\
&\ll \sum_{\substack{T_0 \leq |\ensuremath{\gamma}_1|, |\ensuremath{\gamma}_2| \leq T \\ |\ensuremath{\gamma}_1-\ensuremath{\gamma}_2| < 1}}
  \frac{n^22^nm(\ensuremath{\gamma}_1)m(\ensuremath{\gamma}_2)}{|\ensuremath{\gamma}_1||\ensuremath{\gamma}_2|}
\ll n^2 2^n \sum_{|\ensuremath{\gamma}| \ge T_0} \frac{\log^3(|\ensuremath{\gamma}|+3)}
{|\ensuremath{\gamma}|} \ll \frac{n^2 2^n \log^5 T_0}{{T_0}},
\end{align*}
applying Lemma~\ref{NTchi}.
The terms where $|\ensuremath{\gamma}_1-\ensuremath{\gamma}_2|>1$ on the right hand side of \eqref{lem3.7} total
\begin{align*}
&\ll \sum_{\substack{T_0 \leq |\ensuremath{\gamma}_1|, |\ensuremath{\gamma}_2| \leq T \\ |\ensuremath{\gamma}_1-\ensuremath{\gamma}_2| > 1}}
  \frac{2^{2n}m(\ensuremath{\gamma}_1)m(\ensuremath{\gamma}_2)}{|\ensuremath{\gamma}_1||\ensuremath{\gamma}_2||\ensuremath{\gamma}_1-\ensuremath{\gamma}_2|}
\iiiint\limits_{u_j\ge v_j+2^{-n}}
  \frac{ e^{-2^nv_1-2^nv_2}}{(u_1-v_1)(u_2-v_2)}du_j d v_j \\
&\ll \sum_{\substack{T_0 \leq |\ensuremath{\gamma}_1|, |\ensuremath{\gamma}_2| \\ |\ensuremath{\gamma}_1-\ensuremath{\gamma}_2| > 1}}
\frac{n^2\log(|\ensuremath{\gamma}_1|+3)\log(|\ensuremath{\gamma}_2|+3)}{|\ensuremath{\gamma}_1||\ensuremath{\gamma}_2||\ensuremath{\gamma}_1-\ensuremath{\gamma}_2|} \ll n^2
\frac{\log^5 T_0}{T_0}.
\end{align*}
by Lemma \ref{sumg1g2}.  Summing over $n$ proves the lemma.

redpajama_set_name: RedPajamaArXiv


text:
BNHC reports profitable year
Bahrain's National Hotels Company (NHC), which owns the Diplomat Radisson SAS Hotel, made a profit of BD1.4 million last year.
The BD4 million hotel refurbishment project will be completed in July, NHC chairman Ahmed Abdul Rahman Al Zayani told the company's annual general meeting at the hotel.
He also said tenders for the new BD6.5 million furnished apartment project at the hotel complex would be issued by the end of the year.
The company's profit was BD1,403,417 last year, against BD1,413,714 in 2001.
"The Diplomat managed to achieve very good results in comparison to the previous years despite the fact that the refurbishment project for the entire hotel's rooms and restaurants started in July last year," said Al Zayani.
"In fact, the number of rooms available for sale had significantly reduced by three floors and sometimes four.
"To date, we have managed to complete eight out of the 12 floors earmarked for refurbishment. It is expected that the whole project will be completed in July."
Work on the Mediterranean restaurant has also been completed, and work has started on the Italian and the Polynesian restaurants.
"We have practically completed the design stage of the furnished apartment project," said Al Zayani.
"Despite the threat of war we are going ahead with the expansion project. Once the war clouds clears it will be business as usual, we hope.
"We expect to put the apartment project for general tender during this year and we have signed with Radisson SAS for taking up the responsibility of managing the project."
Another dream project of the company, said Al Zayani, is a multi-million dinar 20-storey tower and a shopping arcade adjacent to the hotel.
"Plans are under way to purchase the land for this project in the near future," he said.
NHC general manager Abdul Rahman Morshed said all the 246 rooms in the hotel were being renovated to high international standard. The health club will also undergo a total renovation.
"The new furnished apartment project will be a 14-storey tower, north of the hotel," he noted.
"The project is being designed by Mohammed Salahuddin Consulting Engineering Bureau. There will be 114 apartments of different sizes. Work is expected to start early next year.


text:
using Google.Picasa;
using NUnit.Framework;

namespace Google.GData.Client.UnitTests.Picasa
{

    ///
    ///This is a test class for PhotoAccessorTest and is intended
    ///to contain all PhotoAccessorTest Unit Tests
    ///
    [TestFixture]
    [Category("Picasa")]
    public class PhotoAccessorTest
    {

        ///
        ///Gets or sets the test context which provides
        ///information about and functionality for the current test run.
        ///
        public TestContext TestContext { get; set; }

        ///
        ///A test for AlbumId
        ///
        [Test]
        public void AlbumIdTest()
        {
            Photo target = new Photo();
            string expected = "TestValue";
            string actual;
            target.AlbumId = expected;
            actual = target.AlbumId;
            Assert.AreEqual(expected, actual);
        }

        ///
        ///A test for Checksum
        ///
        [Test]
        public void ChecksumTest()
        {
            Photo target = new Photo();
            string expected = "TestValue";
            string actual;
            target.Checksum = expected;
            actual = target.Checksum;
            Assert.AreEqual(expected, actual);
        }

        ///
        ///A test for CommentCount
        ///
        [Test]
        public void CommentCountTest()
        {
            Photo target = new Photo();
            uint expected = 12; // TODO: Initialize to an appropriate value
            uint actual;
            target.CommentCount = expected;
            actual = target.CommentCount;
            Assert.AreEqual(expected, actual);
        }

        ///
        ///A test for CommentingEnabled
        ///
        [Test]
        public void CommentingEnabledTest()
        {
            Photo target = new Photo();
            bool expected = true;
            bool actual;
            target.CommentingEnabled = expected;
            actual = target.CommentingEnabled;
            Assert.AreEqual(expected, actual);
        }

        ///
        ///A test for Height
        ///
        [Test]
        public void HeightTest()
        {
            Photo target = new Photo();
            int expected = 12; // TODO: Initialize to an appropriate value
            int actual;
            target.Height = expected;
            actual = target.Height;
            Assert.AreEqual(expected, actual);
        }

        ///
        ///A test for Id
        ///
        [Test]
        public void IdTest()
        {
            Photo target = new Photo();
            string expected = "TestValue";
            string actual;
            target.Id = expected;
            actual = target.Id;
            Assert.AreEqual(expected, actual);
        }

        ///
        ///A test for Latitude
        ///
        [Test]
        public void LatitudeTest()
        {
            Photo target = new Photo();
            double expected = 12.5F; // TODO: Initialize to an appropriate value
            double actual;
            target.Latitude = expected;
            actual = target.Latitude;
            Assert.AreEqual(expected, actual);
        }

        ///
        ///A test for Longitude
        ///
        [Test]
        public void LongitudeTest()
        {
            Photo target = new Photo();
            double expected = 12.5F; // TODO: Initialize to an appropriate value
            double actual;
            target.Longitude = expected;
            actual = target.Longitude;
            Assert.AreEqual(expected, actual);
        }

        ///
        ///A test for PhotoSummary
        ///
        [Test]
        public void PhotoSummaryTest()
        {
            Photo target = new Photo();
            string expected = "TestValue";
            string actual;
            target.Summary = expected;
            actual = target.Summary;
            Assert.AreEqual(expected, actual);
        }

        ///
        ///A test for PhotoTitle
        ///
        [Test]
        public void PhotoTitleTest()
        {
            Photo target = new Photo();
            string expected = "TestValue";
            string actual;
            target.Title = expected;
            actual = target.Title;
            Assert.AreEqual(expected, actual);
        }

        ///
        ///A test for Rotation
        ///
        [Test]
        public void RotationTest()
        {
            Photo target = new Photo();
            int expected = 45; // TODO: Initialize to an appropriate value
            int actual;
            target.Rotation = expected;
            actual = target.Rotation;
            Assert.AreEqual(expected, actual);
        }

        ///
        ///A test for Size
        ///
        [Test]
        public void SizeTest()
        {
            Photo target = new Photo();
            long expected = 12; // TODO: Initialize to an appropriate value
            long actual;
            target.Size = expected;
            actual = target.Size;
            Assert.AreEqual(expected, actual);
        }

        ///
        ///A test for Timestamp
        ///
        [Test]
        public void TimestampTest()
        {
            Photo target = new Photo();
            ulong expected = 122; // TODO: Initialize to an appropriate value
            ulong actual;
            target.Timestamp = expected;
            actual = target.Timestamp;
            Assert.AreEqual(expected, actual);
        }

        ///
        ///A test for Width
        ///
        [Test]
        public void WidthTest()
        {
            Photo target = new Photo();
            int expected = 5; // TODO: Initialize to an appropriate value
            int actual;
            target.Width = expected;
            actual = target.Width;
            Assert.AreEqual(expected, actual);
        }

        //You can use the following additional attributes as you write your tests:
        //
        //
        //Use ClassInitialize to run code before running the first test in the class
        //[ClassInitialize()]
        //public static void MyClassInitialize(TestContext testContext)
        //{
        //}
        //
        //Use ClassCleanup to run code after all tests in a class have run
        //[ClassCleanup()]
        //public static void MyClassCleanup()
        //{
        //}
        //
        //Use TestInitialize to run code before running each test
        //[TestInitialize()]
        //public void MyTestInitialize()
        //{
        //}
        //
        //Use TestCleanup to run code after each test has run
        //[TestCleanup()]
        //public void MyTestCleanup()
        //{
        //}
        //
    }
}

redpajama_set_name: RedPajamaGithub


You will feel right at home at our Italian Market location. It is the site of the very first Sabrina's and will bring a sense of family and comfort to your private event that is incredibly unique to this cheerful and cozy environment. This location is the home base of our catering services and has the most availability for private events. It is the only location that can host events during weekday brunches, and can host private events on Saturdays and Sundays after 5:30pm.


You have to go on their facebook page (www.facebook.com/sheppedcom) and use the contest app.
4. Get your friends and family to visit Shepped Facebook page & VOTE for your photo, as the photo with the most votes will win dog food ($275 value)"