Integrating a Vector Modulus: Dealing with Square Roots in Exponential Form

[go4321]

I am trying to do this integral
[tex]\int_{-\infty}^{\infty}\frac{e^{-a| \vec{r}|}e^{-\vec{b}. \vec{r}}}{|\vec{r}|}[/tex]
But if I write the modulus as
[tex]\vec{r}=\sqrt{x^{2}+y^{2}+z^{2}}[/tex]
I can't figure out how to handle the annoying square root in the exponential.

Would be grateful for hints, thanks in advance.

 

[arkajad]

Use spherical coordinates. Use the appropriate volume element.

 

[poonintoon]

Use spherical coordinates. Use the appropriate volume element.

Doesn't the [tex]\vec{b}.\vec{r}[/tex] term then become a horrible mess though?
Like [tex]b_{1}cos\theta sin \phi +b_{2}sin\theta sin \phi+...[/tex] which is a pain because it is in the exponential so at best you can get it in the denominator of a fraction when you integrate r out.

 

[arkajad]

You can use spherical coordinates for both b and r. Then use trygonometrical identities.

 

[poonintoon]

You can use spherical coordinates for both b and r. Then use trygonometrical identities.

Hey, could I pester you again? I am still banging my head against the wall with this question. I have
[tex]
\int_{0}^{2\pi} d\phi \int_{0}^{\pi} d\theta \int_{0}^{\infty} dr \, r \sin(\theta) \exp[r\{a+ib\left[\sin(\theta)\cos(\phi)\sin(b_{\theta})\cos(b_{\phi}))+\sin(\theta)\sin(\phi)\sin(b_{\theta})\sin(b_{\phi}))+\cos(\theta)\cos(b_{\theta}\right]\}]
[/tex]

I just cannot find a trig identity that helps though. I tried integrating out r using[tex]
\int_{0}^{\infty}xe^{-ax}=1/a^{2}[/tex]but couldn't get anywhere that way either

 

[arkajad]

First notice that the value of the integral is rotation independent (it depends only on absolute value of r and on the scalar product of b and r, both invariant under rotations. Therefore you can choose your coordinate system the way you like - the result will not change. So, choose it so that the vector b, which is kept constant during the integration, has coordinates (0,0,b), b positive.

Then the trigonometry part will be almost trivial.