# A Integration of a polynomial

1. Nov 4, 2016

### spaghetti3451

Consider the following integration:

$$\int \frac{d^{4}k}{(2\pi)^{4}}\ \frac{1}{(k^{2}+m^{2})^{\alpha}}=\frac{1}{(4\pi)^{d/2}} \frac{\Gamma\left(\alpha-\frac{d}{2}\right)}{\Gamma(\alpha)}\frac{1}{(m^{2})^{\alpha-d/2}}.$$

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How does the dependence on $d$ arise in this integral?

Can someone show the intermediate steps in this integration explicitly?

2. Nov 4, 2016

### Fightfish

$d$ is the dimensionality - in this case the LHS of your expression should have $d$s instead of $4$s in the powers.

The hint is to cast the integral in terms of the $d$-dimensional spherical coordinates
$$\int \mathrm{d}^{d} V = \int_0^{R} \mathrm{d} r \int^{2\pi}_0 \mathrm{d} \phi_{d-1} \int^{\pi}_0 \mathrm{d} \phi_{d-2} \cdots \int^{\pi}_0 \mathrm{d} \phi_{1} \sin(\phi_{d-2}) \sin^{2}(\phi_{d-1}) \cdots \sin^{d-2} (\phi_1) \,r^{d-1},$$
noting that the integrand in question is independent of the generalised angles, so that the various angular integrals factorise nicely.

To solve the individual integrals, the following special function (Euler beta function) identity is extremely useful:
$$B(x,y) = 2 \int^{\pi/2}_{0} \mathrm{d}\phi \left(\sin \phi\right)^{2x-1} \left(\cos \phi\right)^{2y-1} = 2 \int^{\infty}_{0} \mathrm{d}t \frac{t^{2x-1}}{(1+t^2)^{x+y}} = \frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}$$

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