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?

After someone suggested the beta function (many thanks!), I managed to figure out a proof by myself. Starting from the key property of the beta function that
$$
B(u,v) = \int_0^1 t^{u-1} (1-t)^{v-1}\mathrm dt = \frac{\Gamma(u) \Gamma(v)}{\Gamma(u + v)}
$$
(the proof of this as given on the Wikipedia page is not too difficult), we substitute a variable $y$ which satisfies
$$
y^2 = \frac{t}{1-t}.
$$
This gives the desired bounds $t=0 \to y = 0$, $t=1 \to y = \infty$. Solving for $t$ and $\mathrm d t$, we get
$$
t = \frac{y^2}{1+y^2}=1- \frac{1}{1+y^2} \implies \mathrm dt = \frac{2y}{(1+y^2)^2}\mathrm dy.
$$
Hence
$$
\begin{align}
B(u,v) &= \int_0^\infty \left(\frac{y^2}{1+y^2}\right)^{u-1}
\left(\frac{1}{1+y^2}\right)^{v-1} \frac{2y}{(1+y^2)^2}\mathrm dy
\\ &= 2 \int_0^\infty \frac{y^{2u - 1}}{(1+y^2)^{u+v}}\
\mathrm d y = \frac{\Gamma(u) \Gamma(v)}{\Gamma(u + v)}.
\end{align}
$$
Letting $k \equiv 2u-1$ and $l \equiv u + v$, we get
$ u = \frac{k+1}{2}$ and $v = \frac{2l-k-1}{2}$. With this,
$$
2 \int_0^\infty \frac{y^{k}}{(1+y^2)^{l}}\
\mathrm d y = \frac{\Gamma(\frac{k+1}{2}) \Gamma(\frac{2l-k-1}{2})}{\Gamma(l)}.
$$
Now, simply substituting $y \to \frac xb$ returns the original identity. $\blacksquare$
