Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'd like to compute the following integral:

$$I = \int_{\mathbb{R}^n} {\rm d}^n x \; \frac{e^{i \vec x \cdot \vec k}}{\vec x^2}$$

My first step is to use generalized spherical coordinates and then I rewrite $I$ as (I ommit integration limits):

$$I = \int {\rm d}\Omega \int {\rm d}r \, r^{n-3} e^{i r |\vec k| \cos \phi_1}$$ The problem is that this does not separete nicely into integrals over angles and integral over $r$. Hence I have to somehow evaluate the following integral:

$$\int_0^\pi {\rm d}\phi_1 \, \sin^{n-2} \phi_1 e^{i r |\vec k| \cos \phi_1}$$ How should I proceed with this one? Maybe there is a better way to tackle $I$?

share|improve this question

1 Answer 1

You may be interesting in the following, which evaluates this integral for $n=3$, from my honours thesis (though I don't claim the result, obviously). Consider $$ G(r) = \frac{1}{(2\pi)^3}\iiint\frac{e^{i\boldsymbol{\kappa}\cdot\mathbf{r}}}{\kappa^2}\ \mbox{d}^3\kappa. $$ This integral can be evaluated using spherical coordinates in $\kappa$-space, with the vertical axis aligned with $\mathbf{r}$, the position vector, so that $\boldsymbol{\kappa}\cdot\mathbf{r} = r\kappa_3$, where $\kappa_3$ denotes the vertical axis. Using the transform coordinates $\rho$, $\theta$ and $\phi$ as normal, $$ G(r) = \frac{1}{(2\pi)^3} \int^{2\pi}_0 \!\int^{\pi}_0 \!\int^{\infty}_0 \frac{e^{i\kappa_3r}}{\rho^2}\rho^2\sin{\theta} \, \mbox{d}\rho \, \mbox{d}\theta \, \mbox{d}\phi, $$ now, since $\kappa_3=\rho\cos{\theta}$ in spherical coordinates, $$ G(r) = \frac{1}{(2\pi)^3} \int^{2\pi}_0 \!\int^{\pi}_0 \!\int^{\infty}_0 e^{i\rho r\cos{\theta}}\sin{\theta} \, \mbox{d}\rho \, \mbox{d}\theta \, \mbox{d}\phi, $$ integrating trivially with respect to $\phi$, $$ G(r) = \frac{1}{(2\pi)^2} \int^{\pi}_0 \!\int^{\infty}_0 e^{i\rho r\cos{\theta}}\sin{\theta} \, \mbox{d}\rho \, \mbox{d}\theta. $$ A simple substitution, $t=\cos{\theta}$, $\mbox{d}t = -\sin{\theta} \ \mbox{d}\theta$, transforms the integral to $$ G(r) = \frac{1}{(2\pi)^2} \int^{1}_{-1} \!\int^{\infty}_0 e^{i\rho r t} \, \mbox{d}\rho \, \mbox{d}t, $$ and interchanging the integrals gives that $$ G(r) =\frac{1}{2\pi^2r} \int^{\infty}_0 \frac{\sin{\rho r}}{\rho} \, \mbox{d}\rho. $$ This can be further evaluated using contour integration in the following way. Consider the Cauchy principal value integral, $$ I = P\!\!\!\!\!\int^{\infty}_{-\infty} \frac{e^{izr}}{z} \ \mbox{d}z, $$ spanning the entire real axis, connected in a semi-circle, which is deformed with radius $|z|=\epsilon$ clockwise around the pole at $z=0$.

The integral around the closed contour, which contains no poles, is zero, so $$ I = \oint_{\mathcal{C}}\frac{e^{izr}}{z} = \left(\int_{-R}^{-\epsilon}+\int_{\epsilon}^{R}\right)\frac{e^{izr}}{z} \ \mbox{d}z + \int_{\mathcal{C}_{\epsilon}}\frac{e^{izr}}{z} \ \mbox{d}z + \int_{\mathcal{C}_{R}}\frac{e^{izr}}{z} \ \mbox{d}z = 0, $$ where the integral around the semicircle $\mathcal{C}_R$ as $R\rightarrow \infty$ is zero. Since $r>0$, the integrand satisfies Jordan's Lemma, that it goes to zero as $R\rightarrow \infty$. Now, $$ \displaystyle{\lim_{\epsilon\rightarrow 0}}\int_{\mathcal{C}_{\epsilon}} f(z) \ \mbox{d} z = i \theta C_{-1}, $$ where $\theta$ is the angle transversed and $C_{-1}=\mbox{Res}(f(z);z_0)$, where $z_0$ is the position of the pole. So, $\theta=-\pi$, since the rotation is clockwise, and $$ C_{-1}= \mbox{Res}\left(\frac{e^{izr}}{z};z=0\right) = \left[\frac{e^{izr}(z-0)}{z}\right]_{z=0} = 1, $$ so $$ \int_{\mathcal{C}_{\epsilon}}\frac{e^{izr}}{z} \ \mbox{d}z = -i\pi, $$ and, from previously, $$ \int_{-\infty}^{\infty} \frac{e^{izr}}{z} \ \mbox{d}z = i\pi. $$ Taking the imaginary part and halving the domain gives that $$ \int^{\infty}_0 \frac{\sin{\rho r}}{\rho} \ \mbox{d}{\rho} = \frac{\pi}{2}, $$ giving the final result that $$ G(r) = \frac{1}{2\pi^2 r}\cdot\frac{\pi}{2} = \frac{1}{4\pi r}. $$

share|improve this answer
    
Thanks, I've already got an answer on physics.SE: physics.stackexchange.com/questions/83103/… –  qoqosz Nov 2 '13 at 9:04

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.