I'm trying to evaluate the following two dimensional integral:

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}e^{i(k_{1}x+k_{2}y)}dxdy$

The paper that i'm following reports the following solution:

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}e^{i(k_{1}x+k_{2}y)}dxdy = \frac{1}{\sqrt{k_{1}^{2}+k_{2}^{2}}}e^{-\sqrt{k_{1}^{2}+k_{2}^{2}}z}$

it might be useful using the following definitions: $r = \sqrt{x^{2}+y^{2}+z^{2}}$ and $k = \sqrt{k_{1}^{2}+k_{2}^{2}}$

I think this should follow from some kind of substitution; anyway i tried to solve it with mathematica but i'm not getting any solution...

Any idea on how to obtain the result shown by the authors? Thanks in advance

  • $\begingroup$ You may want to switch to polar coordinates. $\endgroup$ – hickslebummbumm Oct 14 '14 at 11:09
  • $\begingroup$ do you think that switching to polar coordinates simplifies the integral? $\endgroup$ – SSC Napoli Oct 14 '14 at 11:17
  • $\begingroup$ $\sqrt{x^2 + y^2 + z^2} = r$ for example. $\endgroup$ – hickslebummbumm Oct 14 '14 at 11:19
  • $\begingroup$ yes, however i don't know if the exponential part gets actually simplified adopting polar coordinates... $\endgroup$ – SSC Napoli Oct 14 '14 at 11:21
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    $\begingroup$ I'd have to look up more carefully to be sure, but this smells of something where, if one goes to polar coordinates, an integral representation of a (modified?) Bessel function would be useful. $\endgroup$ – Semiclassical Oct 14 '14 at 11:27

Let me fill in the details of how to perform the angular integration in tired's answer. First, we can make our lives easier by notating that the integrand $e^{i \mathbf{k}\cdot\mathbf{x}}$ with $\mathbf{k}=k_1 \hat{x}+k_2\hat{y}=k\hat{k}$ is invariant under rotations about the $z$-axis. So we're free to orient the coordinates such that $\hat{k}$ as the positive $x$-axis i.e. $\mathbf{k}\mapsto k\hat{x}$. Then, defining $s=\sqrt{x^2+y^2}$, the integral becomes

$$I(k,z)=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}e^{i k x}\,dxdy=\int_{0}^{\infty}\int_{-\pi}^{\pi}\frac{e^{ik s\cos\theta }}{\sqrt{s^2+z^2}}s\,d\theta \,ds.$$

While this integral hardly seems better, the $\theta$-integral is recognizable to Mathematica in terms of a Bessel function. Indeed, if uses the integral representation $J_n(x)=\int\limits_{-\pi}^{\pi} e^{i (n \theta -x\sin \theta)}\,d\theta$ (see eq. 71 on Mathworld's Bessel functions page, or derive by hand using an Jacobi-Anger identity) then the $n=0$ case allows us to write the integration as

$$I(k,z)=2\pi\int_{0}^{\infty}\frac{s\,J_0(ks)}{\sqrt{s^2+z^2}}\,ds=\frac{2\pi}{k}\int_{0}^{\infty}\frac{u\,J_0(u)}{\sqrt{u^2+k^2 z^2}}\,du\quad\text{with }u=ks.$$

What remains is to justify the integral identity cited by tired which allows us to conclude $I(k,z)=2\pi k^{-1} e^{-k |z|}.$ I'll see if I can recall the details.

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  • $\begingroup$ Mathematica automatically solves the integral:$\frac{2\pi}{k}\int_{0}^{\infty}\frac{u\,J_0(u)}{\sqrt{u^2+k^2 z^2}}\,du$ returning $\frac{1}{k} e^{-kz}$; however i don't know why it is not able to solve the entire integral, I need to pass through a manual subsitution everytime and then perform the last integral with mathematica... $\endgroup$ – SSC Napoli Oct 14 '14 at 14:42
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    $\begingroup$ @user3810266: Try doing "\$Assumptions = k > 0 && z > 0" in order to force it to treat $k$ and $z$ as positive values first, and then call "Integrate[ Integrate[ Exp[I k s Sin[\[Theta]]] s/Sqrt[s^2 + z^2], {\[Theta], 0, 2 \[Pi]}], {s, 0, \[Infinity]}]". That should give the integral directly. $\endgroup$ – Semiclassical Oct 14 '14 at 14:50
  • $\begingroup$ I was already using the first assumptions, however your method still goes through a manual substitution of the variables and therefore of the integration boundaries. I was looking for a completely automated method in which i give a generic function $f(x,y,z)$ and the integration is completely automated... as i have to apply this kind of two dimensional fourier transforms to several functions $\endgroup$ – SSC Napoli Oct 14 '14 at 14:58
  • $\begingroup$ @user3810266: Eh, this isn't even so much a substitution as it is the representation in cylindrical coordinates. That's not terribly labour-intensive to implement. And that's unfortunately about the best I can imagine you can do, since Mathematica is really only suited for doing multivariable integrals one integration at a time (which in Cartesian coordinates is a poor direction for an integral like this.) $\endgroup$ – Semiclassical Oct 14 '14 at 15:02
  • $\begingroup$ hi just a quick taught, i guess for the second equation (6.554.1) one should take the series representation of $J_0$ and integrate term by term where convergence allows it. Afterwards one can analytically continue safely... $\endgroup$ – tired Oct 14 '14 at 15:09

Ok let's see:

Denote the Integral by $I(k_x,k_y,z)$

Going over to Polar coordinates we have: $x=r\sin(\phi),r\cos(\phi)$, $|det(J)|=r$

\begin{align} I(k_x,k_y,z)=\int_0^{\infty} dr \frac{r}{\sqrt{r^2+z^2}}\int_0^{2\pi} d\phi e^{-ir[k_x sin(\phi)+k_y cos(\phi)]} \end{align}

It can now be shown that the inner integral equals $2\pi J_0[r\sqrt{k_x^2+k_y^2}]$

So we are left with (1) \begin{align} I(k_x,k_y,z)=2\pi\int_0^{\infty} dr \frac{r}{\sqrt{r^2+z^2}}J_0[r\sqrt{k_x^2+k_y^2}] \end{align}

Using now Formula 6.554.1 from Gradstheyn/Ryzhik which states (2) that $\int_0^{\infty}dq\frac{q}{\sqrt{s^2+q^2}}J_0[qp]=\frac{e^{-sp}}{p}$

We get \begin{align} I(k_x,k_y,z)=\frac{2\pi}{\sqrt{k_x^2+k_y^2}}e^{-z\sqrt{k_x^2+k_y^2}} \end{align} which is the stated result apart from a factor of $2\pi$ which supposedly stems from some on of the definitions of Fouriertransform

Unluckily i'm very busy so i can't fill in the proof which leds us to (1) or try to proof (2) so i would appreciate if someone could fill this gaps

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    $\begingroup$ Working on an answer that has those gaps filled in right now (filling in $(1)$ isn't too bad, still working on $(2)$.) $\endgroup$ – Semiclassical Oct 14 '14 at 13:38

In my first answer, I proceeded by trying to directly compute the momentum integrals. In doing so, I'd forgotten that there is a rather less direct approach which yet has the advantage of not requiring Bessel function identities.

First, note that our task is to take the $2$D Fourier transform of the function $G(\mathbf{r})=(x^2+y^2+z^2)^{-1/2}=r^{-1}$. My choice of $G$ as symbol for this function is not done idly, for this is in fact the free-space Green's function of the $3$D Laplacian i.e. $\nabla^2 G(\mathbb{r})=-4\pi \delta^{3}(\mathbf{r})$. This is a standard Dirac-delta identity which can be verified via the divergence theorem and direct differentiation.

The $3$D Fourier transform of this is readily computed to be $(k_x^2+k_y^2+k_z^2)G(\mathbf{k})=4\pi.$ To get the planar Fourier transform, we can then take the inverse transform with respect to $k_z$. That iss,

\begin{align} G(k_x,k_y,z)=\mathcal{F}^{-1}[G(\mathbf{k})](z) &=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-i k_z z}G(\mathbf{k})\,dk_z\\ &=\int_{-\infty}^\infty \frac{2e^{-i k_z z}}{k_x^2+k_y^2+k_z^2}\,dk_z\\ &=\frac{2}{k}\int_{-\infty}^\infty \frac{e^{-i k z u}}{1+u^2}\,du \end{align} where in the last line I have introduced $k=\sqrt{k_x^2+k_y^2}$ and $k_z=k u$. This last integral can be evaluated via methods of contour integration (see Wikipedia of such) to obtain the desired result

$$\boxed{G(k_x,k_y,z)=2\pi\dfrac{e^{-k |z|}}{k}=2\pi\left(\dfrac{e^{-|z|\sqrt{k_x^2+k_y^2}}}{\sqrt{k_x^2+k_y^2}}\right)}$$

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  • $\begingroup$ @user3810266: I should point out that these two approaches---direct integration via Bessel functions, and indirect integration via Fourier transforms---are complementary. Either or both may be useful in a given context, and so it's best to understand both. $\endgroup$ – Semiclassical Oct 15 '14 at 14:58
  • $\begingroup$ is this true for a generic function of $\boldsymbol{r}$? I mean, can I compute the two dimensional fourier transform of $f(\boldsymbol{r})$ by first computing its three dimensional fourier transform and then anti transform along $z$? $\endgroup$ – SSC Napoli Oct 15 '14 at 14:59
  • $\begingroup$ In principle, yes. (There may be exceptional cases, but they don't spring to mind.) It just may not always be the easiest approach; here we had the advantage that taking $3+1$ Fourier transforms was easier than just $2$ since the first three were immediate. @user3810266 $\endgroup$ – Semiclassical Oct 15 '14 at 15:07

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