I Convolution integral

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1. Dec 15, 2016

kaniello

Dear "General Math" Community,
my goal is to calculate the following integral $$\mathcal{I} = \int_{-\infty }^{+\infty }\frac{f\left ( \mathbf{\vec{x}} \right )}{\left | \mathbf{\vec{c}}- \mathbf{\vec{x}} \right |}d^{3}x$$ in the particular case in which $f\left ( \mathbf{\vec{x}} \right )=f\left ( x \right )$ where $x=\left | \mathbf{\vec{x}} \right |$.
I switched to spherical coordinates and wrote $\left | \mathbf{\vec{c}- \mathbf{\vec{x}}} \right |= \sqrt{c^{2}+x^{2}-2cx\cos \vartheta }$ and $d^{3}x=x^{2}\cos \vartheta d\varphi d\vartheta dx$. After the integration in $\varphi$ and $\vartheta$ it just remains $$\mathcal{I} = \frac{1}{c}\int_{0}^{c}fx^{2}dx+\int_{c}^{+\infty }fxdx.$$
The integral can also be seen as the convoultion of the function $f$ with the function $\frac{1}{\left | \mathbf{\vec{x}} \right |}$ so I expect to find the same result if I evaluate $$\mathscr{F}^{-1}\left \{ \hat{f} \cdot \frac{1}{k^{2}}\right \}$$ where $\hat{f}$ and $\frac{1}{k^{2}}$ are the Fourier transforms of the function $f$ and the Coulomb potential up to some coefficients respectively.
So now I can write $$\int_{0}^{2\pi }\int_{-\frac{\pi }{2}}^{\frac{\pi}{2}}\int_{0}^{\infty }\frac{\hat{f}}{k^{2}}k^{2}e^{i\vec{\mathbf{k}}\cdot \vec{\mathbf{x}}}\cos \vartheta d\vartheta dkd\varphi = 2\pi\int_{-\frac{\pi }{2}}^{\frac{\pi}{2}}\int_{0}^{\infty }\frac{\hat{f}}{k^{2}}k^{2}e^{ikx\sin \vartheta } \cos \vartheta d\vartheta dk=$$$$=2\pi \int_{0}^{\infty }\hat{f}dk\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}e^{ikx\sin \vartheta } \cos \vartheta d\vartheta.$$
Performing the integration in $\vartheta$ yelds $\frac{1}{kx}\sin \left ( kx \right )$ which finally brings to $$\frac{1}{x}\int_{0}^{\infty }\frac{f}{k}\sin \left ( kw \right )dk.$$
I am stuck at this point and I do not see how to recover the first solution.
Can anybody help me out?
Thank you very much in advance

2. Dec 16, 2016

BvU

But these $\theta$ are not one and the same !

($d^3 x$ has a $\sin\theta$!)

(Anyone know why $\TeX$ comes with a different font for the x in the d3x above ?)

Last edited: Dec 18, 2016
3. Dec 16, 2016

ShayanJ

In general you're correct. But you can choose your z axis to be parallel with $\vec c$ or $\vec x$. Then the angle between them is the same as the $\theta$ of the spherical coordinates.

4. Dec 16, 2016

blue_leaf77

Don't you think you need to change $\cos \theta$ in $|\vec c - \vec x|$ to $\sin \theta$?

5. Dec 17, 2016

BvU

kaniello ?

6. Dec 18, 2016

kaniello

Hello and sorry for not being online yesterday.

Can you please explain me better what do you mean by :

7. Dec 18, 2016

blue_leaf77

Can you tell us which angle $\theta$ is in your formula for volume element $d^{3}x=x^{2}\cos \vartheta d\varphi d\vartheta dx$? Please describe in terms of angle subtended by which vectors.

8. Dec 18, 2016

kaniello

I did my best to draw it

9. Dec 18, 2016

blue_leaf77

I see, now what about vector $\vec c$? Since it's fixed in the integration, you need to pick certain direction for it. To which direction did you align $\vec c$?

10. Dec 18, 2016

kaniello

This is maybe more clear

11. Dec 18, 2016

blue_leaf77

Yes I understand what $\theta$ is, it's the angle between $\vec x$ and its projection in the $(x_1,x_2)$ plane. But we also need to know where $\vec c$ is in your diagram.

12. Dec 18, 2016

kaniello

What if pick the red line?

13. Dec 18, 2016

blue_leaf77

You can but that's not a wise choice. If you do that, the expression for $|\vec c - \vec x|$ will get very messy, for instance the angle between $\vec x$ and $\vec c$ is not always $\theta$ as you defined in the picture. Only for $\vec x$ that lies in the same plane as $\vec c$ and $x_3$ axis can be described by your $\theta$. I suggest that you pick $x_3$ to align your vector $\vec c$.

14. Dec 18, 2016

My intention was in fact to place the red lines in the plane $x_1 x_2$ so that it forms the angle $\theta$ with $\vec x$. As you suggested $\vec x$ should be in the plane $\vec c$ x_3$axis. 15. Dec 18, 2016 blue_leaf77 No, since you integrate over the entire space,$\vec x$can have arbitrary direction. There is actually no difficulty in changing$\vec c$to$x_3$axis while expressing$|\vec c -\vec x|$in terms of$\theta$. You just need to use sine instead of cosine which was the reason I asked in post #4. 16. Dec 18, 2016 kaniello Thanks a lot for the explanation. With that hint I repeated the calculations and found:$\int_{0}^{\infty }\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{f\left ( x \right )}{\sqrt{c^{2}+x^{2}-2cx\sin \vartheta }}x^{2}\cos \vartheta d\vartheta dx=\int_{0}^{\infty }f\left ( x \right )x^{2}\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1}{\sqrt{c^{2}+x^{2}-2cx\sin \vartheta }}\cos \vartheta d\vartheta dx\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1}{\sqrt{c^{2}+x^{2}-2cx\sin \vartheta }}\cos \vartheta d\vartheta = \int_{-1}^{1}\frac{1}{\sqrt{c^{2}+x^{2}-2cxt }}dt$(where$t = \sin \vartheta \cos \vartheta d\vartheta =dt$)$=...=-\frac{2}{cx}\left ( \sqrt{c^{2}+x^{2}-2cx}-\sqrt{c^{2}+x^{2}+2cx} \right )$. Reinserting this in the integral in x I get :$ \frac{1}{c}\int_{0}^{c}fx^{2}dx+\int_{c}^{+\infty }fxdx$How can I link this result with$\frac{1}{x}\int_{0}^{\infty }\frac{f}{k}\sin \left ( kw \right )dk$coming from the inverse Fourier transform? 17. Dec 18, 2016 blue_leaf77 How do you get that result? That doesn't seem correct to me. For instance try setting$\vec c = 0$, using your calculation you get  \frac{1}{c} \int_0^{\infty} xf(x) dx  while using the original integral you get  \int_0^{\infty} \frac{f(x)}{x} dx  18. Dec 18, 2016 kaniello The integral in$\vartheta$returns \sqrt{c^{2}+x^{2}-2cx}-\sqrt{c^{2}+x^{2}+2cx} which must be inserted in the integral in$x$taking care to break the integral into$0\leqslant x < c$and$c< x < \infty $due to the first radicand. If$c=0$from the original integral one gets$\mathcal{I}=\int_{-\infty }^{+\infty }\frac{f}{\left | \vec{x} \right |}dx^{3}=2\pi *2\int_{0}^{\infty }\frac{f}{x}x^{2}dx \sim\int_{0}^{\infty }fxdx$and from the calcuation$\lim_{c \to 0} \left (\frac{1}{c}\int_{0}^{c}fx^{2}dx+\int_{c}^{\infty }fxdx \right )=\int_{0}^{\infty }fxdx$. So actually it looks correct to me. 19. Dec 18, 2016 blue_leaf77 Ah my bad I forgot that there is another$x^2$coming from the volume element. 20. Dec 23, 2016 kaniello Hi blue_leaf77, so, up to now we have proven that the result$\mathcal{I}=\int_{-\infty }^{+\infty }\frac{f\left ( \left | \vec{x} \right | \right )}{\left | \vec{c}-\vec{x} \right |}d^{3}x = \frac{1}{c}\int_{0}^{c}fx^{2}dx+\int_{c}^{\infty }fxdx## is correct.

Still my question is : how can I recover this result from the convolution integral?
$$\mathcal{I}=\int_{-\infty }^{+\infty }\frac{f\left ( \left | \vec{x} \right | \right )}{\left | \vec{c}-\vec{x} \right |}d^{3}x = \mathcal{F}^{-1}\left \{ \frac{\hat{f}}{k^{2}} \right \}=...=\frac{1}{x}\int_{0}^{\infty }\frac{f}{k}\sin \left ( kx \right )$$