Integral Calculation: Compute l/sqrt(x2+l2)

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SUMMARY

The integral $$\frac{1}{\pi^2} \frac{l}{\sqrt{\vec{x}^2+l^2}}$$ can be computed using spherical coordinates, specifically through the transformation $$\int_0^{2\pi}d\theta\int_0^\pi d\phi \int_0^\infty dp~p^2e^{-lp}e^{ipr\cos (\theta)}$$. The correct volume element includes the factor of $$\sin\theta$$, which is essential for proper integration limits, where $$\theta$$ ranges from 0 to $$\pi$$ and $$\phi$$ from 0 to $$2\pi$$. Attempts to solve the integral without these adjustments resulted in incorrect outcomes.

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Homework Statement
Help please with the integral of figure
Relevant Equations
In the description
I need compute the integral

$$(2\pi)^{-3} \int d^3p e^{-l|p|}e^{i \vec{x} \cdot \vec{p}}$$

The problem does not specified the limits of integration

The result is

$$\frac{1}{\pi^2} \frac{l}{\sqrt{\vec{x}^2+l^2}}$$I saw the references about t-Student and I had not achieved it.
I have tried to split the integral as

$$\int dp_x dp_y dp_z e^{-l\sqrt{p_x^2+p_y^2+p_z^2}} e^{ip_xx+ip_yy+ip_zz}$$
x2 without result.
Also I tried to insert a 2D dirac Delta and after integrate only in dp instead dp^2. Also without result

Could you help me to solve this integral?
 

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Spherical coordinates.
 
Orodruin said:
Spherical coordinates.
Do you say something like that

##\int_0^{2\pi}d\theta\int_0^\pi d\phi \int_0^\infty dp~p^2e^{-lp}e^{ipr\cos (\theta)}##

I also have intented without results
 
Last edited:
You are missing the ##\sin\theta## iof the volume element. (And a } that makes your LaTeX not render)
 
Orodruin said:
(And a } that makes your LaTeX not render)
Fixed the LaTeX...
 
PeteSampras said:
Do you say something like that

##\int_0^{2\pi}d\theta\int_0^\pi d\phi \int_0^\infty dp~p^2e^{-lp}e^{ipr\cos (\theta)}##

I also have intented without results

Orodruin said:
You are missing the ##\sin\theta## iof the volume element. (And a } that makes your LaTeX not render)

In which case the limits of \theta should be [0, \pi] and the limits of \phi should be [0, 2\pi].
 
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