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) $?

Assuming that $n$ is a positive integer, using hypergeometric functions$$I_n=\int \frac{\log(t)}{\left(1+t^2\right)^n}\, dt=t \log (t) \, _2F_1\left(\frac{1}{2},n;\frac{3}{2};-t^2\right)-t \,
   _3F_2\left(\frac{1}{2},\frac{1}{2},n;\frac{3}{2},\frac{3}{2};-t^2
   \right)$$
$$J_n=\int_0^\infty \frac{\log(t)}{\left(1+t^2\right)^n}\, dt=-\frac{\sqrt{\pi }}{4}\,\frac{\Gamma \left(n-\frac{1}{2}\right)}{\Gamma (n)}\,\left(H_{n-\frac{3}{2}}+2 \log (2)\right)$$
Asymptotically
$$J_n=-\frac{\sqrt{\pi }}{4 \sqrt{n}}(\log (n)+\gamma +2 \log (2))+O\left(\frac{1}{n^{3/2}}\right)$$
Trying to simplify what is given in other answers
$$\frac{d^n}{d a^n}\left(\frac{\log( a)}{\sqrt{a}}\right)=\sqrt \pi \frac{ \left(\log (a)-H_{n-\frac{1}{2}}-2 \log
   (2)\right)}{a^{n+\frac{1}{2}}\,\,\Gamma \left(\frac{1}{2}-n\right)}$$
