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

The angular part is easy to do as the integrand is spherically symmetric so the result is $V_{d-1}\times A$ where $V_{d-1}$ is the volume of the Euclidean n-ball of unit radius. It remains to compute the radial part but I don't know how it's done. This integral arises in the calculation of loop corrections in the propagators of scalar fields

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Are $a$ and $b$ integers? –  joriki Jan 18 '13 at 15:41
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It's very confusing to use $d$ for a variable in an integral... –  Thomas Andrews Jan 18 '13 at 15:45
    
Yes , these are integers. –  nabil Jan 18 '13 at 17:13
    
d is the dimension of Euclidean space , q is the norm of vectors in d-dimensional Euclidean space . –  nabil Jan 18 '13 at 17:15
    
@nabil Could you replace the dimension $d$ with $n$? And what are the limits of the integral? –  user17762 Jan 18 '13 at 17:39

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With the spherical coordinate change followed by the substitution $r = \sqrt{D}\tan\theta$, \begin{align*} &\frac{1}{(2\pi)^{n}} \int_{\Bbb{R}^{n}} \frac{r^{2a}}{(r^2 + D)^{b}} \, dx \\ &=\frac{|S^{n-1}|}{(2\pi)^{n}} \int_{0}^{\infty} \frac{r^{2a}}{(r^2 + D)^{b}} \, r^{n-1}dr \\ &=\frac{2\pi^{n/2}}{(2\pi)^{n}\Gamma(n/2)} D^{a-b+n/2} \int_{0}^{\frac{\pi}{2}} \frac{\tan^{2a+n-1}\theta}{\sec^{2b}\theta} \sec^{2}\theta \, d\theta \\ &=\frac{1}{(4\pi)^{n/2}\Gamma(n/2)} D^{a-b+n/2} \cdot 2 \int_{0}^{\frac{\pi}{2}} \sin^{2a+n-1}\theta \cos^{2b-2a-n-1}\theta \, d\theta \\ &=\frac{1}{(4\pi)^{n/2}\Gamma(n/2)} D^{a-b+n/2} \beta\left(a+\frac{n}{2},b-a-\frac{n}{2}\right) \\ &=\frac{1}{(4\pi)^{n/2}\Gamma(n/2)} D^{a-b+n/2} \cdot \frac{\Gamma\left(a+\frac{n}{2}\right)\Gamma\left(b-a-\frac{n}{2}\right)}{\Gamma(b)}. \end{align*} Here we used the beta function identity $$ 2\int_{0}^{\frac{\pi}{2}} \sin^{2z-1}\theta \cos^{2w-1}\theta \, d\theta = \beta(z, w) = \frac{\Gamma(z)\Gamma(w)}{\Gamma(z+w)}. $$

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Thanks alot , I forgot about the beta function . It's actually straightforward –  nabil Jan 19 '13 at 14:00

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