Htykuut

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.

$$
int_0^{infty}frac{x^{alpha}log^n(x)}{1+x^2}dx=frac{1}{4^{n+1}}Gamma(n+1)left[zetaleft(n+1,frac{1-alpha}{4}right)-zetaleft(n+1,frac{3-alpha}{4}right)+(-1)^nleft(zetaleft(n+1,frac{1+alpha}{4}right)-zetaleft(n+1,frac{3+alpha}{4}right)right)right]
$$

for $ninmathbb{N}$ and $-1<alpha<1$.

$$
int_0^{infty}frac{x^{alpha}log(x)}{(ax^2+b)^n}dx
$$

for $ninmathbb{N}$ and $0<alpha+1<2n$ and $a,b>0$.
How to deal with the integrals at hand?

Referring to this answer and using the definition of $I(a)$ in it,
$$int^infty_0frac{x^alog^n(x)}{x^2+1}dx=I^{(n)}(a)=left(frac{pi}2right)^{n+1}sec^{(n)}left(frac{pi a}2right)$$

By considering
$$ frac{1}{2pi i} int frac{z^a}{1+z^2} , dz $$
around the standard keyhole contour. Then differentiate wrt $a$, $n$ times, and set $a=alpha$.

In the $n=0$ case, substituting $x=tan t$ gives $$int_0^inftytfrac{x^alpha dx}{1+x^2}=int_0^inftytan^alpha dx=tfrac{1}{2}operatorname{B}(tfrac{1+alpha}{2},,tfrac{1-alpha}{2})=tfrac{pi}{2}csc(tfrac{pi}{2}+tfrac{pialpha}{2})=tfrac{pi}{2}sec(tfrac{pialpha}{2}).$$Applying $partial_a^n$,$$int_0^inftytfrac{x^alphaln^nxdx}{1+x^2}=(tfrac{pi}{2})^{n+1}sec^{(n)}(tfrac{pialpha}{2}).$$