Recently I posted a rather similar question in here:
How to solve $\int_0^{\infty}\frac{\log^n(x)}{1+x^2}dx$?
Now I was looking for harder integrals, for which I know how to calculate the case $n=1$. Mathematica provides analtical solutions.
$$
\int_0^{\infty}\frac{x^{\alpha}\log^n(x)}{1+x^2}dx=\frac{1}{4^{n+1}}\Gamma(n+1)\left[\zeta\left(n+1,\frac{1-\alpha}{4}\right)-\zeta\left(n+1,\frac{3-\alpha}{4}\right)+(-1)^n\left(\zeta\left(n+1,\frac{1+\alpha}{4}\right)-\zeta\left(n+1,\frac{3+\alpha}{4}\right)\right)\right]
$$
for $n\in\mathbb{N}$ and $-1<\alpha<1$.
$$
\int_0^{\infty}\frac{x^{\alpha}\log(x)}{(ax^2+b)^n}dx
$$
for $n\in\mathbb{N}$ and $0<\alpha+1<2n$ and $a,b>0$.
How to deal with the integrals at hand?
Best Answer
Referring to this answer and using the definition of $I(a)$ in it, $$\int^\infty_0\frac{x^a\log^n(x)}{x^2+1}dx=I^{(n)}(a)=\left(\frac{\pi}2\right)^{n+1}\sec^{(n)}\left(\frac{\pi a}2\right)$$