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I Convolution integral

  1. Dec 15, 2016 #1
    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 [itex] f\left ( \mathbf{\vec{x}} \right )=f\left ( x \right ) [/itex] where [itex] x=\left | \mathbf{\vec{x}} \right | [/itex].
    I switched to spherical coordinates and wrote [itex]\left | \mathbf{\vec{c}- \mathbf{\vec{x}}} \right |= \sqrt{c^{2}+x^{2}-2cx\cos \vartheta }[/itex] and [itex]d^{3}x=x^{2}\cos \vartheta d\varphi d\vartheta dx[/itex]. After the integration in [itex]\varphi[/itex] and [itex]\vartheta[/itex] 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 [itex]f[/itex] with the function [itex]\frac{1}{\left | \mathbf{\vec{x}} \right |}[/itex] so I expect to find the same result if I evaluate $$\mathscr{F}^{-1}\left \{ \hat{f} \cdot \frac{1}{k^{2}}\right \}$$ where [itex]\hat{f}[/itex] and [itex]\frac{1}{k^{2}}[/itex] are the Fourier transforms of the function [itex]f[/itex] 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 [itex]\vartheta[/itex] yelds [itex]\frac{1}{kx}\sin \left ( kx \right )[/itex] 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. jcsd
  3. Dec 16, 2016 #2

    BvU

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    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 ?)

    answer: my mistake. I did \rm {and} instead of {\rm and} :rolleyes:
     
    Last edited: Dec 18, 2016
  4. Dec 16, 2016 #3

    ShayanJ

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    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.
     
  5. Dec 16, 2016 #4

    blue_leaf77

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    Don't you think you need to change ##\cos \theta## in ##|\vec c - \vec x|## to ##\sin \theta##?
     
  6. Dec 17, 2016 #5

    BvU

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    kaniello ?
     
  7. Dec 18, 2016 #6
    Hello and sorry for not being online yesterday.

    Can you please explain me better what do you mean by :
     
  8. Dec 18, 2016 #7

    blue_leaf77

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    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.
     
  9. Dec 18, 2016 #8
    I did my best to draw it 1482048060189.jpg
     
  10. Dec 18, 2016 #9

    blue_leaf77

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    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##?
     
  11. Dec 18, 2016 #10
    This is maybe more clear 1482049088656.jpg
     
  12. Dec 18, 2016 #11

    blue_leaf77

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    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.
     
  13. Dec 18, 2016 #12
    What if pick the red line? 1482049483683.jpg
     
  14. Dec 18, 2016 #13

    blue_leaf77

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    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##.
     
  15. Dec 18, 2016 #14
    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.
     
  16. Dec 18, 2016 #15

    blue_leaf77

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    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.
     
  17. Dec 18, 2016 #16
    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?
     
  18. Dec 18, 2016 #17

    blue_leaf77

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    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
    $$
    \int_0^{\infty} \frac{f(x)}{x} dx
    $$
     
  19. Dec 18, 2016 #18
    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.
     
  20. Dec 18, 2016 #19

    blue_leaf77

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    Ah my bad I forgot that there is another ##x^2## coming from the volume element.
     
  21. Dec 23, 2016 #20
    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?
    $$\mathcal{I}=\int_{-\infty }^{+\infty }\frac{f\left ( \left | \vec{x} \right | \right )}{\left | \vec{c}-\vec{x} \right |}d^{3}x = \mathcal{F}^{-1}\left \{ \frac{\hat{f}}{k^{2}} \right \}=...=\frac{1}{x}\int_{0}^{\infty }\frac{f}{k}\sin \left ( kx \right )$$
     
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