Do we have a closed form for $\int_0^{\infty} \frac{\ln t}{\left(1+t^2\right)^n} d t $? Latest Edit
We are glad to see there are 4 alternative solutions which give the same closed form to the integral:
$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(a^2+t^2\right)^n} d t = \frac{a^{1-2n}\sqrt{\pi}\Gamma\left(n-\frac{1}{2}\right)}{4\Gamma(n)}\left[2\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right)\right] },$$
where $\psi$ denotes the Digamma Function.


In the post, we found that
$$\int_0^{\infty} \frac{\ln x}{a^2+t^2} d x =\frac{\pi \ln a}{2 a }$$
Now I want to generalise the integral as
$$
I_n=\int_0^{+\infty} \frac{\ln t}{\left(a+t^2\right)^n} d t
$$
where $n\in N$.
Replacing a by $\sqrt{a}$ and differentiating both sides w.r.t. $a$ by $n$ times yields
$$
\begin{aligned}
&J(a)=\int_0^{\infty} \frac{\ln t}{a+t^2} d t=\frac{\pi}{4 \sqrt{a}} \ln a\\
&\frac{d^n}{d a^n}(J(a))=\frac{\pi}{4} \frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right) \\ & (-1)^n n ! \int_0^{\infty} \frac{\ln t}{\left(a+t^2\right)} d t= \frac{\pi}{4} \frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right)\\& \boxed{\int_0^{\infty} \frac{\ln t}{\left(a+t^2\right)^{n+1}} d t=\frac{(-1)^n \pi}{4 n !} \frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right)}
\end{aligned}
$$
By Wolfram-alpha, we have
$$
\frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right)=(-1)^n a^{-\frac{1}{2}-n}\left(\frac{1}{2}\right)_n\left(\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(\frac{1}{2}-n\right)\right)\cdots (*)
$$
Hence
$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(a+t^2\right)^{n+1}} d t=\frac{\pi}{4 n !} a^{-\frac{1}{2}-n}\left(\frac{1}{2}\right)_n\left(\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(\frac{1}{2}-n\right)\right)}$$
In particular, when $a=1$, we have
$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(1+t^2\right)^{n+1}} d t=\frac{\pi}{4 n !} \left(\frac{1}{2}\right)_n\left(\psi\left(\frac{1}{2}\right)-\psi\left(\frac{1}{2}-n\right)\right)}$$
For example,
$$$$
\begin{aligned}
\int_0^\infty \frac{\ln t}{\left(1+t^2\right)^4} d t=& \frac{\pi}{24} \cdot \frac{15}{8}\left(-\gamma -\ln 4-\frac{46}{15}+\gamma +\ln 4\right) 
=-\frac{23 \pi}{96}
\end{aligned}
$$
$$
My Question: How to find a closed form for $\frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right) $?

Note
$$ \int_0^\infty\frac{t^a}{(1+t)^n}dt=\int_0^1u^a(1-u)^{n-a-2}du=B(a+1,n-a-1)=\frac{1}{(n-1)!}\Gamma(a+1)\Gamma(n-a-1) $$
and hence
$$ \int_0^\infty\frac{t^a\ln t}{(1+t)^n}dt=\frac{d}{da}\frac{\Gamma(a+1)\Gamma(n-a-1)}{\Gamma(n)}=\frac{1}{(n-1)!}\Gamma (a+1) \Gamma (-a+n-1) (\psi ^{(0)}(a+1)-\psi ^{(0)}(-a+n-1)). $$
So
\begin{eqnarray}
I_n&=&\int_0^{\infty} \frac{\ln t}{\left(1+t^2\right)^n} d t\\
&=&\frac14\int_0^{\infty} \frac{t^{-1/2}\ln t}{\left(1+t\right)^n} d t \\
&=&\frac14\frac{1}{(n-1)!}\Gamma (a+1) \Gamma (-a+n-1) (\psi ^{(0)}(a+1)-\psi ^{(0)}(-a+n-1))\bigg|_{a=-\frac12} \\
&=&\frac{\sqrt{\pi }\Gamma \left(n-\frac{1}{2}\right)}{4(n-1)!}\bigg[ \psi ^{(0)}\left(\frac{1}{2}\right) -\psi ^{(0)}\left(n-\frac{1}{2}\right)\bigg].
\end{eqnarray}
