Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can I evalute this integral?

$$\psi(z)=\int\limits_{-\infty}^\infty\int\limits_{-\infty}^\infty [(x-a)^2+(y-b)^2+z^2]^{-3\over 2}f(x,y)\,\,\,dxdy\;.$$

I think we can treat $z$ as a constant and take it out of the integral or something. Maybe changing variables like taking $u={1\over z}[(x-a)^2+(y-b)^2]$? But then how do I change $f(x,y)$ which is some arbitrary function, etc?


share|cite|improve this question
I'm not saying I know how to compute this integral, but $\psi(x,y,z)$ doesn't make sense ; you should write $\psi(z)$ instead, since your double integral is just a function of $z$ ; the variables $x$ and $y$ in it could be replaced by any other symbol and the result would still be the same. (In french we have a term that says that the variables don't speak, they're... "mute"? I don't feel like this makes sense though. For those who speak French here, "les variables $x$ et $y$ sont muettes". If someone could translate this in mathematical terms it would be nice.) – Patrick Da Silva Dec 21 '11 at 9:31
@PatrickDaSilva: Thanks! You are right. My teacher call those variables "dummy variables". I have edited the question. – gareth Dec 21 '11 at 10:21

There's no general way to evaluate this integral, for if there were, you could integrate any function $g(x,y)$ by calculating this integral for $f(x,y)=g(x,y)[(x-a)^2+(y-b)^2+z^2]^{\frac32}$.

share|cite|improve this answer
+1. Ergo. $ $ $ $ – Did Dec 21 '11 at 11:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.