# Evaluate $\int_0^{\infty} dx \, \, \frac{\sin(a \arctan(\frac{x}{k}))}{x^b (k^2 + x^2)^{a/2}}$

I came across this integral $$\int_0^{\infty} dx \, \, \frac{\sin(a \arctan(\frac{x}{k}))}{x^b (k^2 + x^2)^{a/2}},$$ with $a+b > 1>b$. It is stated that this integral is equal to $$\frac{\pi \Gamma(a+b-1)}{2\,k^{a+b-1}\,\sin(\frac{b\pi}{2}) \, \Gamma(a) \Gamma(b)} .$$ I am interested in proving this, but where to start? Using residues?

• Assuming $a$ and $k$ are positive, $\frac{\sin \left(a \arctan \frac{x}{k}\right)}{(k^2+x^2)^{a/2}} = - \text{Im} \, \frac{1}{(k+ix)^{a}}$ – Random Variable Feb 23 '17 at 15:35

If you set $x=kz$ and $dx= k\,dz$ you are left with $$\frac{1}{k^{a+b-1}}\int_{0}^{+\infty}\frac{\sin(a\arctan z)}{z^b (z^2+1)^{a/2}}\,dz=-\frac{1}{k^{a+b-1}}\cdot\text{Im}\int_{0}^{+\infty}\frac{1}{z^b(1+iz)^a}\,dz$$ and the last integral is simple to compute through a substitution and Euler's Beta function,
leading to: $$\frac{1}{k^{a+b-1}}\cdot\text{Im}\left(ie^{-\pi i b/2}\,B(1-b,a+b-1)\right)$$ The claim then follows from the $\Gamma$ reflection formula $\Gamma(b)\,\Gamma(1-b) = \frac{\pi}{\sin(\pi b)}$.