<s>[INST] 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,
$$$$
\begin{aligned}
\int_0^\infty \frac{\ln t}{\left(1+t^2\right)^4} d t=& \frac{\pi}{24} \cdot \frac{15}{8}\left(-\gamma -\ln 4-\frac{46}{15}+\gamma +\ln 4\right) 
=-\frac{23 \pi}{96}
\end{aligned}
$$
$$
My Question: How to find a closed form for $\frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right) $?
 [/INST] The generalized Leibniz rule gives
$$\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.
</s>