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?
Best Answer
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 $n$.