3
$\begingroup$

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 $I$?

$\endgroup$

1 Answer 1

0
$\begingroup$

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 $$ G(r) = \frac{1}{2\pi^2 r}\cdot\frac{\pi}{2} = \frac{1}{4\pi r}. $$

$\endgroup$
1

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .