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I've been stuck on this one homework problem for nearly a day now, so I'd be really thankful for any pointers. The problem is to find $\int\int\int_{\mathbb{R}^3} \frac{1}{\mathbf{x}^2+1}e^{-2\pi i (x_1\zeta_1+x_2\zeta_2+x_3\zeta_3)}dx_1dx_2dx_3$ where $\mathbf{x}\in \mathbb{R}^3$. I'm not sure how to go about this; I know that the Fourier transform of $1/(1+x^2)$ in $\mathbb{R}$ is $\pi e^{-2\pi |\zeta|}$, but I don't think that helps at all. I've tried changing it to spherical coordinates as well.

Also, if anyone could briefly explain how this relates to solving the equation $u(x)-\Delta u(x)=f(x)$, $x\in \mathbb{R}^3$ and $\Delta$ being the Laplacian operator, I'd be eternally grateful.

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    So apparently $\mathbf{x}^2$ is the dot product, and after using spherical coordinates I have reduced it to the integral $\int_{-\infty}^\infty \left(\int_0^{2\pi} \int_0^{2\pi} \frac{(r^2\sin \phi) e^{-2\pi i (\zeta_1r\cos \theta \sin \phi+\zeta_2 r\sin \theta \sin \phi+\zeta_3r\cos \theta)}}{r^2+1} d\theta d\phi \right) dr$. Would anybody have any tips on evaluating this, or how it relates to the equation $u(x)-\Delta u(x)=f(x)$? Thanks in advance.2012-05-05

2 Answers 2

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Let's compute the Fourier Transform of the surface measure on a sphere of radius $r$. We will do this by integrating in slices.

The singular measure supported on the sphere is the limit of the volume measure on a thin sphere of thickness $\mathrm{d}r$ divided by $\mathrm{d}r$. When integrating a slice, the angle of intersection of the slice with the sphere must be taken into account. In the diagram below, it is shown that a surface of thickness $\mathrm{d}r$ intersecting a surface of thickness $\mathrm{d}x$ at an angle of $\theta$ has a cross sectional area of $\mathrm{d}x\,\mathrm{d}r\sec(\theta)$. $\hspace{2cm}$singular measures

When integrating along the slice whose intersection with the sphere is a circle of radius $r\cos(\theta)$, the angle of intersection of the slice with the surface of the sphere is $\theta$. Thus, the $\sec(\theta)$ from the angle of the intersection is cancelled by the $\cos(\theta)$ from the radius of the circle. Therefore, $ \begin{align} \int_{rS^2}e^{-2\pi i\,x\cdot\xi}\,\mathrm{d}x &=\int_{-r}^r2\pi re^{-2\pi i\,t|\xi|}\mathrm{d}t\\ &=\frac{r}{-i|\xi|}\left(e^{-2\pi i\,r|\xi|}-e^{2\pi i\,r|\xi|}\right)\\ &=\frac{2r}{|\xi|}\sin(2\pi r|\xi|)\tag{1} \end{align} $ Computing the Fourier Transform

To compute the Fourier transform of $\dfrac{1}{r^2+1}$, we integrate against $(1)$: $ \begin{align} \int_0^\infty\frac{2r}{|\xi|}\frac{\sin(2\pi r|\xi|)}{r^2+1}\mathrm{d}r &=\int_{-\infty}^\infty\frac{r}{|\xi|}\frac{\sin(2\pi r|\xi|)}{r^2+1}\mathrm{d}r\\ &=\frac{1}{|\xi|}\int_{-\infty}^\infty\frac{r\sin(r)\,\mathrm{d}r}{r^2+4\pi^2|\xi|^2}\\ &=\frac{1}{|\xi|}\Im\left(\int_{-\infty}^\infty\frac{re^{ir}\,\mathrm{d}r}{r^2+4\pi^2|\xi|^2}\right)\\ &=\frac{1}{|\xi|}\Im\left(\int_\gamma\frac{re^{ir}\,\mathrm{d}r}{r^2+4\pi^2|\xi|^2}\right)\\ &=\frac{1}{|\xi|}\Im\left(2\pi i\operatorname{Res}\left(\frac{re^{ir}}{r^2+4\pi^2|\xi|^2},2\pi i|\xi|\right)\right)\\ &=\frac{1}{|\xi|}\Im\left(2\pi i\frac{2\pi i|\xi|e^{-2\pi|\xi|}}{4\pi i|\xi|}\right)\\ &=\frac{\pi}{|\xi|}e^{-2\pi|\xi|}\tag{2} \end{align} $ where $\gamma$ is the limit of the path on the real axis from $-M$ to $M$ followed by the semi-circle in the upper half-plane centered at $(0,0)$ from $M$ back to $-M$ as $M\to\infty$.

Therefore, $ \int_{\mathbb{R}^3}\frac{1}{|x|^2+1}e^{-2\pi i\,x\cdot\xi}\;\mathrm{d}x=\frac{\pi}{|\xi|}e^{-2\pi|\xi|}\tag{3} $


Relation to the Laplacian Equation

Taking the Fourier Transform of $ u(x)-\Delta u(x)=f(x)\tag{4} $ yields $ (1+4\pi^2|\xi|^2)\hat{u}(\xi)=\hat{f}(\xi)\tag{5} $ which becomes $ \hat{u}(\xi)=\dfrac{1}{1+4\pi^2|\xi|^2}\hat{f}(\xi)\tag{6} $ and taking the Inverse Fourier Transform yields $ \begin{align} u(x) &=\left(\dfrac{1}{1+4\pi^2|\xi|^2}\right)^\wedge(x)\ast f(x)\\ &=\frac{1}{4\pi|x|}e^{-|x|}\ast f(x)\tag{7} \end{align} $ The convolution in $(7)$ is a Singular Integral.

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    Thanks a bunch, that makes sense now! This method of attack is really nice.2012-05-07
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Let's begin with the second part of the question. Take the Fourier transform on both sides of the equation to obtain $ (1+|\xi|^2)\hat u(\xi)=\hat f(\xi)\implies \hat u(\xi)=\frac{1}{|\xi|^2+1}\,\hat f(\xi). $ If $K(x)$ is the inverse Fourier transform of $(1+|\xi|^2)^{-1}$, then the solution of the equation is $u=K\ast f$.

The integral you want to compute (modulo some factor of $\pi$) is $K(\zeta)$. We know that the Fourier transform commutes with rotations and hence, that the Fourier transform of a radial function is radial. Then $ K(\zeta)=K(\zeta_1,\zeta_2,\zeta_3)=K(0,0,|\zeta|). $ This simplifies the computation of the integral (that I have not carried out).

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    Thanks a bunch-- I'll try going through the motions again. Of course, computing the integral is still a mess...2012-05-05