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,
$$$$
\begin{aligned}
\int_0^\infty \frac{\ln t}{\left(1+t^2\right)^4} d t=& \frac{\pi}{24} \cdot \frac{15}{8}\left(-\gamma -\ln 4-\frac{46}{15}+\gamma +\ln 4\right) 
=-\frac{23 \pi}{96}
\end{aligned}
$$
$$
My Question: How to find a closed form for $\frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right) $?
 A: The generalized Leibniz rule gives
$$\newcommand{\d}{\mathrm{d}}
\frac{\d^n}{\d x^n} \ln(x) x^{-1/2} \bigg|_{x=1}
= \sum_{k=0}^n \binom n k \left( \frac{\d^{n-k}}{\d x^{n-k}} \ln(x) \bigg|_{x=1} \right)\left( \frac{\d ^{k}}{\d x^{k}} x^{-1/2} \bigg|_{x=1} \right)$$
For the $(n-k)$th derivative of $\ln(x)$, it is easy to show by, e.g., induction that
$$\frac{\d ^{n-k}}{\d x^{n-k}} \ln(x)
= \frac{(n-k-1)! \cdot (-1)^{n-k-1}}{x^{n-k}}$$
Similarly,
$$\begin{align*}
\frac{\d ^k}{\d x^k} x^{-1/2}
&= \frac{-1}{2} \frac{-3}{2} \frac{-5}{2} \cdots \frac{ -(2k-1)}{2} x^{-k-1/2} \\
&= \frac{(-1)^k (2k-1)!!}{2^k} x^{-k-1/2}  
\end{align*}$$
so
$$\begin{align*}
\frac{\d ^n}{\d x^n} \ln(x) x^{-1/2} \bigg|_{x=1}
&= \sum_{k=0}^n \binom n k (n-k-1)! \cdot (-1)^{n-k-1} \cdot \frac{(-1)^k (2k-1)!!}{2^k} \\
&= (-1)^{n-1} \cdot \sum_{k=0}^n \binom n k (n-k-1)! \cdot \frac{(2k-1)!!}{2^k} \\
&= (-1)^{n-1} n! \cdot \sum_{k=0}^n \frac{1}{(n-k) \cdot k!} \cdot \frac{(2k-1)!!}{2^k} \\
\end{align*}$$
I don't know if there's a meaningful simplification beyond this.
A: Assuming that $n$ is a positive integer, using hypergeometric functions$$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)}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{{\displaystyle #1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
{\tt I}_{n} & \equiv\color{#44f}{\int_{0}^{\infty} {\ln\pars{t} \over \pars{a + t^{2}}^{n}}\,\dd t}
\sr{t^{2}\ \mapsto\ t}{=}
\int_{0}^{\infty}{\ln\pars{t}/2 \over
\pars{a + t}^{n}}\pars{{1 \over 2}\,t^{-1/2}}\,\dd t
\\[5mm] & =
\left.{1 \over 4}\partiald{}{\nu}\int_{0}^{\infty}{\,t^{\pars{\nu + 1/2} - 1
\,\,\,} \over
\pars{a + t}^{n}}\,\dd t\right\vert_{\nu\ =\ 0}
\end{align}
Note that
\begin{align}
& \pars{a + t}^{-n} = a^{-n}\sum_{k = 0}^{\infty}{-n \choose k}\pars{t \over a}^{k} =
a^{-n}\sum_{k = 0}^{\infty}{n + k - 1 \choose k}
{\pars{-t}^{k} \over a^{k}}
\\[5mm] = & \
\sum_{k = 0}^{\infty}
{a^{-n - k} \over \pars{n - 1}!}\,\Gamma\pars{n + k}{\pars{-t}^{k} \over k!}
\end{align}
Therefore,
\begin{align}
{\tt I}_{n} & \equiv\color{#44f}{\int_{0}^{\infty} {\ln\pars{t} \over \pars{a + t^{2}}^{n}}\,\dd t}\qquad
\substack{\mbox{With}\\
\ds{Ramanujan's\ Master\ Theorem}}\ :
\\[5mm] & =
{1 \over 4}\partiald{}{\nu}\bracks{\Gamma\pars{\nu + {1 \over 2}}
{a^{-n + \nu + 1/2}\,\,\, \over \pars{n - 1}!}\,\Gamma\pars{n - \nu - {1 \over 2}}}_{\nu\ =\ 0}
\\[5mm] & =
{\root{\pi}a^{-n + 1/2} \over 4\pars{n - 1}!}\Gamma\pars{n - {1 \over 2}}
\bracks{\ln\pars{a} + \Psi\pars{1 \over 2} -
\Psi\pars{n - {1 \over 2}}}
\\[5mm] & =
\color{#44f}{{\root{\pi}a^{-n + 1/2}\,\, \over 4\pars{n - 1}!}\,\Gamma\pars{n - {1 \over 2}}} \times
\\ &
\rule{1cm}{0pt}\color{#44f}{\bracks{\ln\pars{a} - \gamma - 2\ln\pars{2} -
\Psi\pars{n - {1 \over 2}}}}
\\ &
\end{align}
$\ds{\gamma:\ Euler\mbox{-}mascheroni\ Constant.\ \Psi:\ Digamma\ Function}$.
$\ds{\Psi\pars{n - 1/2}}$ can be evaluated with the  $\ds{Gauss\ Digamma\ Theorem}$ for given values of $\ds{n \in \mathbb{N}_{\geq\ 1}}$.
A: As @metamorphy suggested, let $I\left(\lambda\right)=\int_0^{\infty} \frac{t^{2 \lambda-1}}{\left(a^2+t^2\right)^n} d t,$ then
$$I_n=\int_0^{\infty} \frac{\ln t}{\left(a^2 +t^2\right)^n} d t= \frac{1}{2} I^{\prime}\left(\frac{1}{2}\right) .$$
Now we are going to express $I\left(\lambda\right)$ as a beta function by letting $t=a\tan \theta$, then
$$
\begin{aligned}
I(\lambda) &= a^{2(\lambda-n)}\int_0^{\frac{\pi}{2}} \sin ^{2 \lambda-1} \theta \cos ^{2(n-\lambda)-1} \theta d \theta \\
&=\frac{a^{2(\lambda-n)}}{2} B(\lambda, n-\lambda) \\
&=\frac{a^{2(\lambda-n)}}{2\Gamma(n) } \Gamma(\lambda) \Gamma(n-\lambda)
\end{aligned}
$$
By logarithmic differentiation, we get
$$
\begin{aligned}
&\frac{I^{\prime}(\lambda)}{I(\lambda)}=2\ln a+\psi(\lambda)-\psi(n-\lambda) \\
&I^{\prime}(\lambda)=\frac{a^{2(\lambda-n)}}{2 \Gamma(n)} \Gamma(\lambda) \Gamma(n-\lambda)[ 2\ln a+\psi(\lambda)-\psi(n-\lambda)]
\end{aligned}
$$
Putting $\lambda=\frac{1}{2} $ yields
$$
I^{\prime}\left(\frac{1}{2}\right)=\frac{a^{1-2n}\sqrt{\pi}}{2} \cdot \frac{\Gamma\left(n-\frac{1}{2}\right)}{\Gamma(n)}\left[2\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right)\right]
$$
Hence $$\boxed{\int_0^{\infty} \frac{\ln t}{\left(a^2+t^2\right)^n} d t= \frac{1}{2} I^{\prime}\left(\frac{1}{2}\right) = \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] },$$
which is same as the answer provided by @Felix Marin.
Back to our integral,
$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(1+t^2\right)^n} d t= \frac{1}{2} I^{\prime}\left(\frac{1}{2}\right) = \frac{\sqrt{\pi}\Gamma \left(n-\frac{1}{2}\right)}{4\Gamma(n)}\left[\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right)\right] }$$
By Wolfram-Alpha, $\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right) = -H_{n-\frac{3}{2}}-2 \log (2) $ yields
$$I_n= -\frac{\sqrt{\pi}\Gamma\left(n-\frac{1}{2}\right)}{4\Gamma(n)}\left(H_{n-\frac{3}{2}}+2\ln 2\right),$$
which is the same as the answer provided by @Claude Leibovici.
A: Note
$$ \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}
