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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?

Schnarco
  • 352

3 Answers3

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Referring to this answer and using the definition of $I(a)$ in it, $$\int^\infty_0\frac{x^a\log^n(x)}{x^2+1}dx=I^{(n)}(a)=\left(\frac{\pi}2\right)^{n+1}\sec^{(n)}\left(\frac{\pi a}2\right)$$

Szeto
  • 11,159
1

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

Richard Martin
  • 1,671
1

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

J.G.
  • 115,835