Calculating the Convolution Integral for General Math Community

In summary: I am confused..If you take ##x_3## in the direction of ##\vec c## then the angle between ##\vec c## and ##\vec x## is always 0. That's why I suggested you take ##x_3## in the plane. But you can take ##x_3## in the direction of ##\vec c## if you change the angle in ##|\vec c - \vec x|## to ##\sin \theta##.I think you forgot to include the ##\sin \theta## factor in the volume element. So the correct volume element should be ##x^2 \sin \theta d\phi d\theta dx##. Then you have ##\cos
  • #1
kaniello
21
0
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
 
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  • #2
kaniello said:
I switched to spherical coordinates and wrote $$|\vec c− \vec x|=\sqrt{c^2+x^2−2cx\cos\theta }\ \ \ \rm { and } \ \ d^3 x= x^2\sin\theta \;d\phi \; d\theta \;dx$$
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:
  • #3
BvU said:
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 ?)
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
kaniello said:
I switched to spherical coordinates and wrote |⃗c−⃗x|=√c2+x2−2cxcosϑ\left | \mathbf{\vec{c}- \mathbf{\vec{x}}} \right |= \sqrt{c^{2}+x^{2}-2cx\cos \vartheta } and d3x=x2cosϑdφdϑdxd^{3}x=x^{2}\cos \vartheta d\varphi d\vartheta dx. After the integration in φ\varphi and ϑ\vartheta it just remains
I=1c∫c0fx2dx+∫+∞cfxdx.​
Don't you think you need to change ##\cos \theta## in ##|\vec c - \vec x|## to ##\sin \theta##?
 
  • #5
kaniello ?
 
  • #6
Hello and sorry for not being online yesterday.

Can you please explain me better what do you mean by :
blue_leaf77 said:
Don't you think you need to change ##\cos \theta## in ##|\vec c - \vec x|## to ##\sin \theta##?
 
  • #7
kaniello said:
Hello and sorry for not being online yesterday.

Can you please explain me better what do you mean by :
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
blue_leaf77 said:
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.
I did my best to draw it [emoji6]
1482048060189.jpg
 
  • #9
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
blue_leaf77 said:
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.
This is maybe more clear
1482049088656.jpg
 
  • #11
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
blue_leaf77 said:
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##?
What if pick the red line?
1482049483683.jpg
 
  • #13
kaniello said:
What if pick the red line?
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
blue_leaf77 said:
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##.

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
kaniello said:
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.
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
blue_leaf77 said:
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.

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
kaniello said:
Reinserting this in the integral in x I get :
1c∫c0fx2dx+∫+∞cfxdx
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
blue_leaf77 said:
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
$$
\int_0^{\infty} \frac{f(x)}{x} dx
$$
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
Ah my bad I forgot that there is another ##x^2## coming from the volume element.
 
  • #20
blue_leaf77 said:
Ah my bad I forgot that there is another ##x^2## coming from the volume element.
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?
$$\mathcal{I}=\int_{-\infty }^{+\infty }\frac{f\left ( \left | \vec{x} \right | \right )}{\left | \vec{c}-\vec{x} \right |}d^{3}x = \mathcal{F}^{-1}\left \{ \frac{\hat{f}}{k^{2}} \right \}=...=\frac{1}{x}\int_{0}^{\infty }\frac{f}{k}\sin \left ( kx \right )$$
 

1. What is the convolution integral?

The convolution integral is a mathematical operation used to combine two functions to produce a third function. It involves multiplying one function by the time-reversed and shifted version of the other function and integrating over the resulting product.

2. How is the convolution integral calculated?

The convolution integral is calculated by multiplying one function by the time-reversed and shifted version of the other function, and then integrating over the resulting product. This is typically done using integration techniques such as substitution or integration by parts.

3. What is the purpose of calculating the convolution integral?

The convolution integral is used to analyze and model systems in various fields such as physics, engineering, and signal processing. It allows us to understand how two functions interact and the resulting output, which is essential in predicting and controlling system behavior.

4. Can the convolution integral be applied to any two functions?

Yes, the convolution integral can be applied to any two functions, as long as they are continuous. However, the resulting integral may be difficult or impossible to solve analytically, and numerical methods may be needed.

5. Are there any real-life applications of the convolution integral?

Yes, there are many real-life applications of the convolution integral. It is commonly used in image and audio processing, circuit analysis, and probability theory. It also has applications in fields such as economics, biology, and medical imaging.

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