What is the correct integral for rexp(-ar^2) using cylindrical coordinates?

[francescopadormo]

Hi! Can someone help me with this integral? I can't seem to get my head around it!

Integrate r^2.exp(-ar^2) imperfectly.

Thanks!

 

[StatusX]

Do you mean indefinitely? If so, this will involve erf, a non-elementary function. That is, unless you are doing an area integral in polar coordinates, in which case the actual function you integrate will be r^3e^(-ar^2), and this can be solved by substitution.

 

[HallsofIvy]

I imagine it would be easy to integrate it "imperfectly"!

As StatusX said: \int x^2 e^{-x^2}dx, like \int e^{-x^2}dx has no expression in terms of elementary functions, only in terms of the "error" function Erf(x).

HOWEVER, since your variable is "r", it is possible that you are really attempting to integrate over polar coordinates, in which case you need to remember that the "differential of area" in polar coordinates is rdrd\theta. And the "r" in that let's you make the substitution u= r².

 

[francescopadormo]

Sorry everyone! I was mistaken in what the integral should actually be: it should be the integral of rexp(-ar^2). I should be using cylindrical co-ords, not spherical.

Thanks for the help though!