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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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Have you tried spherical coordinates? –  Jack May 4 '12 at 23:08
    
I guess my question is more about what the $1/(\mathbb{x}^2+1)$ being in the integrand entails. It's not a map $\mathbb{R}^3\rightarrow \mathbb{R}$, so I don't really know how to evaluate it. I've tried letting $(x_1,x_2,x_3)=(r\cos \phi \sin \theta, r\sin \phi \sin \theta,r\cos \theta)$ but don't see how it helps. –  Felix May 4 '12 at 23:22
    
You may find the answer to question 80308 relevant. –  Sasha May 4 '12 at 23:50
    
I don't really see the relation... –  Felix May 5 '12 at 0:02
    
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. –  Felix May 5 '12 at 2:39

2 Answers 2

up vote 4 down vote accepted

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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What you wrote makes sense, but I'm not sure how it relates to the problem of computing the $\mathbb{R}^3$ Fourier transform? –  Felix May 5 '12 at 17:10
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@Felix: Equation $(1)$ is the Fourier Transform of Euclidean measure on the sphere of radius $r$; it's $4\pi r^2 \mathrm{sinc}(2\pi r|\xi|)$. Any radial function can be built up by integrating these shells of Euclidean measure times the value of the function on that shell. Since the Fourier Transform is linear, we can find the Fourier Transform of any radial function by integrating the value of that function on a given shell against the Fourier Transform of the Euclidean measure on that shell. Does that make sense? –  robjohn May 5 '12 at 17:28
    
Thanks a bunch, that makes sense now! This method of attack is really nice. –  Felix May 7 '12 at 2:34

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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Thomas Klimpel suggested a minor correction and added the following comment to your answer which I repost here: $${}$$ Note Instead of computing the integral, you could also get the answer by solving $k^2u-\Delta u=k^2\delta(x)$ with $k^2=4\pi^2$. Because $u$ is radially symetric, $\Delta u=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 \frac{\partial u}{\partial r})$, so it's easy to verify that $u=\frac{c}{r}\exp(kr)$ is the solution, where the constant $c$ must still be determined. –  t.b. May 5 '12 at 9:38
    
Thanks a bunch-- I'll try going through the motions again. Of course, computing the integral is still a mess... –  Felix May 5 '12 at 17:19

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