Proving $\int_0^\infty \frac{x^k}{(x^2 + b^2)^l}\,dx=\frac1{2b^{2l-k-1}}\Gamma(\frac{k+1}2)\Gamma(\frac{2l-k-1}{2})/\Gamma(l)$ . My quantum mechanics textbook states the following integral without proof:
$$
\int_0^\infty \frac{x^k}{(x^2 + b^2)^l} \mathrm dx = \frac1{2b^{2l-k-1}}\frac{\Gamma(\frac{k+1}2)\Gamma(\frac{2l-k-1}{2})}{\Gamma(l)}
$$
What is this class of integrals called? Why are the gamma functions involved, and how do we prove this identity?

Letting $x=b \tan \theta, \quad$ then $d x=b \sec ^{2} \theta d \theta$ and
$$
I=\int_{0}^{\frac{\pi}{2}} \frac{b^{k} \tan ^{k} \theta}{b^{2 l} \sec ^{2 l} \theta} \cdot b \sec ^{2} \theta d \theta =\frac{1}{b^{2 l-k-1}} \int_{0}^{\frac{\pi}{2}} \sin ^{k} \theta \cos ^{2 l-k-2} \theta d \theta
$$
Using $$
B(x, y)=2 \int_{0}^{\frac{\pi}{2}} \sin ^{2 x-1} \theta \cos^{2y-1} \theta d \theta
$$
yields
$$
I=\frac{1}{2 b^{2 l-k-1}} B\left(\frac{k+1}{2}, l-\frac{k+1}{2}\right)
$$
By the property of Beta function
$$
B(x, y)=\frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)},
$$
we have
$$
I=\frac{1}{2 b^{2 l-k-1}} \frac{\Gamma\left(\frac{k+1}{2}\right)\Gamma\left(\frac{2 l-k-1}{2}\right)}{\Gamma(l)}
$$
