How to solve $\int_0^{\infty}\frac{\log^n(x)}{1+x^2}dx$? As an exercise for myself I constructed the Integral 
$$
\int_0^{\infty}\frac{\log^n(x)}{1+x^2}dx
$$
with $n\in \mathbb{N}$. With the help of Mathematica I found the analytical result 
$$
\int_0^{\infty}\frac{\log^n(x)}{1+x^2}dx=\frac{1+(-1)^n}{4^{n+1}}\Gamma(n+1)\left[\zeta\left(n+1,\frac{1}{4}\right)-\zeta\left(n+1,\frac{3}{4}\right)\right].
$$
For $n=1$ (and probably $n\in \mathbb{N}$) one can employ the methods of complex analysis and find a result. For $n\in \mathbb{N}$ I encountered a nasty recursion relation. I can provide details if needed. Is there another way how to solve the integral at hand? 
 A: Referring to this answer,
$$I_n=\frac{\pi}2\frac{d^n}{dx^n}\sec\left(\frac{\pi x}2\right)\bigg\vert_{x=0}$$
or equivalently,
$$\int^\infty_0\frac{\log^n(x)}{x^2+1}dx=\left(\frac{\pi}2\right)^{n+1}\sec^{(n)}(0)$$
$$\int^\infty_0\frac{\log^{2n}(x)}{x^2+1}dx=(-1)^nE_{2n}\left( \frac{\pi}2\right)^{2n+1} $$
The integral is zero for odd $n$.
A: Let, for $n$, a natural integer,
\begin{align}J_n=\int_0^\infty \frac{\ln^n x}{1+x^2}\,dx\end{align}
First, observe that if $n$ is odd then $J_n=0$ (perform the change of variable $y=\dfrac{1}{x}$ )
Consider, for $n$, a natural integer,
\begin{align}K_n=\int_0^\infty \int_0^\infty\frac{\ln^n(xy)}{(1+x^2)(1+y^2)}\,dx\,dy\end{align}
Observe that,
\begin{align}K_{2n}&=\sum_{k=0}^{2n}\binom{2n}{k} J_kJ_{2n-k}\\
&=\sum_{k=0}^{n}\binom{2n}{2k} J_{2k}J_{2(n-k)}
\end{align}
On the other hand,
Perform the change of variable $u=xy$ ($x$ variable, $y$, a "parameter"),
\begin{align}K_n&=\int_0^\infty \int_0^\infty\frac{y\ln^n u}{(1+y^2)(u^2+y^2)}\,du\,dy\\
&=\frac{1}{2}\int_0^\infty \left[\ln\left(\frac{1+y^2}{u^2+y^2}\right)\right]_{y=0}^{y=\infty}\frac{\ln^{n} u}{u^2-1}\,du\\
&=\int_0^\infty \frac{\ln^{n+1} u}{u^2-1}\,du\\
&=\int_0^1 \frac{\ln^{n+1} u}{u^2-1}\,du+\int_1^\infty \frac{\ln^{n+1} u}{u^2-1}\,du\\
\end{align}
In the second integral perform the change of variable $v=\dfrac{1}{u}$,
\begin{align}K_n&=\int_0^1 \frac{\ln^{n+1} u}{u^2-1}\,du+\int_0^1 \frac{\ln^{n+1} u}{u^2-1}\,du\\
&=\int_0^1 \frac{\left(1+(-1)^n\right)\ln^{n+1} u}{u^2-1}\,du\\
&=\left(1+(-1)^n\right)\int_0^1 \frac{\ln^{n+1} u}{u-1}\,du-\left(1+(-1)^n\right)\int_0^1\frac{u\ln^{n+1} u}{u^2-1}\,du\\
\end{align}
In the latter integral perform the change of variable $y=u^2$,
\begin{align}K_n&=\left(1+(-1)^n\right)\int_0^1 \frac{\ln^{n+1} u}{u-1}\,du-\frac{1+(-1)^n}{2^{n+2}}\int_0^1 \frac{\ln^{n+1} u}{u-1}\,du\\
&=\left(1+(-1)^n\right)\left(1-\frac{1}{2^{n+2}}\right)\int_0^1 \frac{\ln^{n+1} u}{u-1}\,du\\
&=\left(1+(-1)^n\right)\left(1-\frac{1}{2^{n+2}}\right)(-1)^{n+2}(n+1)!\zeta(n+2)
\end{align}
Therefore, one obtains a recursion relation,
\begin{align}
\boxed{\sum_{k=0}^{n}\binom{2n}{2k} J_{2k}J_{2(n-k)}=2\left(1-\frac{1}{2^{2n+2}}\right)(2n+1)!\zeta(2n+2)}
\end{align}
Observe that,
\begin{align}J_0=\frac{\pi}{2}\end{align}
