Proving $\int_0^\infty \frac{x^k}{(x^2 + b^2)^l}\,dx=\frac1{2b^{2l-k-1}}\Gamma(\frac{k+1}2)\Gamma(\frac{2l-k-1}{2})/\Gamma(l)$ .

Question

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?

Answer 1

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. $\square$

Answer 2

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 $$ I=\frac{1}{2 b^{2 l-k-1}} \frac{\Gamma\left(\frac{k+1}{2}\right)\Gamma\left(\frac{2 l-k-1}{2}\right)}{\Gamma(l)} $$