1
$\begingroup$

I'm trying to solve the following integral:

$$\int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty}\frac{3xy}{(x^{2}+y^{2}+z^{2})^{5/2})}e^{i(k_{x}x+k_{y}y)} dx dy$$

i'm using the solution posted as the answer to another question: here's the link

The principle behind the solution is to change the coordinates of integration and obtain a separation of the integrals.

In particular following the steps reported in the link. The exponential argument can be written as $e^{\boldsymbol{k}\cdot \boldsymbol{x}}$, where $\boldsymbol{k}= k_{x} \hat{x} +k_{y} \hat{y}$ moreover $\boldsymbol{k} = k \hat{k}$ here we have defined $k=\sqrt{k_{x}^{2}+k_{y}^{2}}$. As reported in the link we can orient $\hat{k}$ such that $\mathbf{k}\mapsto k\hat{x}$.

Considering also $s=\sqrt{x^{2}+y^{2}}$ and substituting the new variables in the integral I obtain:

$\int_{0}^{\infty}\int_{-\pi}^{\pi}\frac{Cos(\theta)Sin(\theta) \; e^{ik s\cos\theta }}{\sqrt{s^2+z^2}}s^3\,d\theta \,ds$

that can be integrated with mathematica for example returning zero because the integral: $\int_{-\pi}^{\pi}Cos(\theta)Sin(\theta) \; e^{ik s\cos\theta }d\theta=0$

however the paper that i'm following reports the following as solution of the integral: $-\frac{k_{x}k_{y}}{k}e^{-kz}$

$\endgroup$
  • $\begingroup$ Apparently you need to use spherical coordinates. $\endgroup$ – Yiorgos S. Smyrlis Oct 15 '14 at 15:46
  • $\begingroup$ @YiorgosS.Smyrlis what would be the advantage over the coordinates that I emplyed? $\endgroup$ – SSC Napoli Oct 15 '14 at 15:57
1
$\begingroup$

Here is an answer in the spirit of my second approach of your linked question. We want to find the planar Fourier transform of $F(\mathbf{x})=\dfrac{3xy}{(x^2+y^2+z^2)^{5/2}}$. Let's start with the 3D Fourier transform. In the previous problem, we were able to take advantage of the fact that the function $G(\mathbf{x})=(x^2+y^2+z^2)^{-1/2}$ was the Green's function of the Laplacian i.e. $\nabla^2 G=-4\pi \delta^{3}(\mathbf{r})$ and thus had the simple Fourier transform $(k_x^2+k_y^2+k_z^2)G(\mathbf{k})=4\pi$.

Happily, we can take advantage of this again: observe that $$\frac{\partial^2 G}{\partial y\partial x} =\frac{\partial}{\partial x}\frac{-y}{(x^2+y^2+z^2)^{3/2}}=\frac{3xy}{(x^2+y^2+z^2)^{5/2}}=F(\mathbf{x}),$$ so the Fourier transform simply gives

$$ F(\mathbf{k})=-k_x k_y G(\mathbf{k})=- \dfrac{4\pi k_x k_y}{k_x^2+k_y^2+k_z^2}.$$

Even better, this only differs from my last answer in the $k_x,k_y$-dependence; the $z$-dependence is unchanged, and so the same result for the inverse Fourier transform in $z$ is applicable. Consequently

$$\boxed{\displaystyle F(k_x,k_y,z)=-2\pi k_x k_y \left(\dfrac{e^{-|z|\sqrt{k_x^2+k_y^2}}}{\sqrt{k_x^2+k_y^2}}\right)},$$ which apart from a factor of $2\pi$ due to definitions is exactly the expected result.

$\endgroup$
  • $\begingroup$ Now that i think about it, all the funtions that i have to transform are derivates of a green function; in particulars are green functions of the stokes equation that actually is a laplace equation! $\endgroup$ – SSC Napoli Oct 15 '14 at 19:44
  • $\begingroup$ @user3810266: That simplifies your task immensely. Something I should point out is that, while the Fourier transform language was convenient, it wasn't strictly speaking necessary for this problem. Instead, you can take the result from your previous question and take various integration by parts; the boundary terms vanish, and the resulting integrals trade derivatives of the Green's function for powers of $k_i$ (just as you'd expect from this problem.) $\endgroup$ – Semiclassical Oct 15 '14 at 19:48
  • $\begingroup$ I have thinked about it, and in reality only a part of the functions that i have to transform are derivatives of $\frac{1}{r}$; another class of functions that I'll have to transform are derivatives of $\frac{\boldsymbol{r}}{r^{3}}$ which is still a green function of the Stokes equation. More importantly the Stokes equation is a vectorial laplace equation, because the laplacian is applied to a vector. Can this simple method be applied easily also to $\frac{\boldsymbol{r}}{r^{3}}$ and its derivatives? $\endgroup$ – SSC Napoli Oct 16 '14 at 6:09
  • $\begingroup$ @user3810266: To avoid cluttering comments, let's continue this in chat. $\endgroup$ – Semiclassical Oct 16 '14 at 12:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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