Calculate the value of the improper Riemann integral $H(s)=\int_0^{\infty} \frac{t^s}{1+t^2}dt$ for $0

Question

Given $0<s<1$ calculate the value of the improper Riemann integral $H(s)=\int_0^{\infty} \frac{t^s}{1+t^2}dt$.

I thought about using having a semi-circle contour, then use Residue's theorem.

But the problem is that $\frac{t^s}{1+t^2}$ may not be an even function. So we can't just divide it by 2 in the end.

I also know that $\int_0^{\infty}\frac{1}{1+t^2}dt = \frac{\pi}{2}$. I guess the problem here is to deal with $t^s$.

Any help will be appreciated!

Answer 1

Let $t^2=x$

$$H(s)=\frac{1}{2}\int_0^\infty \frac{x^{(s-1)/2}}{1+x}dx=\frac{1}{2}B\left(\frac{s+1}{2},\frac{1-s}2\right)=\frac{\pi}{2\sin(\frac{s+1}2\pi)}=\frac{\pi}{2\cos(\frac{s}2\pi)}$$