Gamma functions integral identity

Question

In Srednicki's QFT book, eq. $14.27$ is a result used over and over again for computing loop correction. It is the following integral evaluated in terms of gamma functions:

$$ \int d^dq \frac{(q^2)^a}{(q^2+D)^b} = \frac {\Gamma (b-a-\frac{1}{2}d)\Gamma (a+\frac{1}{2}d)} {(4\pi)^{d/2}\Gamma(b)\Gamma(\frac{1}{2}d)} D^{-(b-a-d/2)} $$

Where $q$ is a $d$-dimensional vector, and $q^2$ denotes the square of its norm. I want to know how to obtain the above result.

Since $q^2$ is the only variable in the integrand, we can use hyper-spherical coordinates, where the integral is non-trivial only for the radial component. The factor from the angular components is the volume of the unit $d$-sphere: $$\frac {2\pi^{d/2}}{\Gamma(d/2)}$$ Then the above expression is then a equation for the radial component, let's call it $r$: $$ \int dr \frac{r^{(2a+d-1)}}{(r^2+D)^b} = \frac {\Gamma (b-a-\frac{1}{2}d)\Gamma (a+\frac{1}{2}d)} {(8\pi)^{d}\Gamma(b)} D^{-(b-a-d/2)} $$ Where the extra $r^{d-1}$ in the numerator comes from the volume element of the $n$-sphere. I would like to know how to obtain this result.

Answer 1

$$ \int_0^\infty \text{d}r\frac{r^{(2a+d-1)}}{(r^2+D)^b}= \frac{1}{2}\int_0^\infty \text{d}x\frac{x^{(a+d/2-1)}}{(x+D)^b}= \frac{1}{2}\int_0^\infty \text{d}y\frac{y^{(a+d/2-1)}}{(y+1)^b}D^{a-b+d/2}, $$ with $x=r^2$ and $y=x/D$. This integral is a beta function (see eq. (22) of Wolfram Mathworld): $$ \int_0^\infty \text{d}y\frac{y^{(a+d/2-1)}}{(y+1)^b} = B(a+d/2,b-a-d/2). $$ I don't know where the factor $(4\pi)^{d/2}$ comes from, you'll have to check that.

Answer 2

We insert the identity $$ 1=\int_0^\infty \frac{dt\; t^{b-1}}{\Gamma(b)}e^{-t} $$ in the integral over r: $$ \int_{[0,\infty)^2} \left(\frac{dt\; t^{b-1}}{(r^2+D)^b}\right)\frac{dr}{\Gamma(b)}r^{2a+d-1}e^{-t}= \int_{[0,\infty)^2} (dx)\;\frac{dr}{\Gamma(b)} x^{b-1} r^{2a+d-1}e^{-x(r^2+D)} $$ where we make the change of variable $t=x(r^2+D)$. We now pass from $r$ to variable $s=r^2 x$: $$ \int_{[0,\infty)^2} dx\frac{2\,r\, x\,dr }{2\,r\, \sqrt{x}\sqrt{x}\,\Gamma(b)} x^{b-1} \frac{(xr^2)^{a+d/2-1/2}}{x^{a+d/2-1/2}}e^{-x(r^2+D)}= \int_{[0,\infty)^2} dx\frac{ds}{2\,\Gamma(b)} x^{b-1-a-d/2} s^{a+d/2-1/2}e^{-s-xD}=\\ =\frac{D^{a+d/2-b}}{2\Gamma(b)}\Gamma(a+d/2+1/2)\Gamma(b-a-d/2) $$