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

I'm doing the something that results in following integral: $$f(z) = \int_{-\infty}^\infty \frac{1}{2\pi}x\exp\left(\frac{-x^2}{2}\left(1+z^2\right)\right) dx$$ Then since $f(z)$ is even we get: $$f(z) = \int_{0}^\infty \frac{1}{\pi}x\exp\left(\frac{-x^2}{2}\left(1+z^2\right)\right) dx$$ Which gives $$ f(z) = \frac{1}{\pi} \frac{1}{1+z^2}$$

I can't seem to understand why would I get $0$ if I evaluate the first line directly and substituting in the limits as $\infty$ and ${-\infty}$ without using the fact that the function is even? Am I missing something trivial?

PS. This calculation is from calculating the ratio distribution of X/Y where X, Y IID standard normal. The working agrees with my lecturer's notes & the answer's supposed to Cauchy(0,1)

PPS. Thanks for the replies so far. I got as far as that $f(z)$ is even as a function of $z$ but the integrand is odd as a function of $x$. So why is the answer Cauchy (0,1) and not $0$ (according to my professor & online sources eg wiki)?

share|cite|improve this question
The integrand as a function of $x$ is odd, not even. Rewrite it as $ax\exp(-bx^2)$ and the oddness becomes transparent. – J. M. Dec 1 '11 at 17:09
So, your second displayed equation is not correct. – David Mitra Dec 1 '11 at 17:13
This is just $\mathrm{Const}\cdot\mathsf E(\xi)$ with $\xi\sim \mathscr N(0,1/(1+z^2))$ – S.D. Dec 1 '11 at 17:14

This is because the integrand is not even. The factor x makes it odd.

share|cite|improve this answer
$f$ is even. It's the integrand that's odd. – David Mitra Dec 1 '11 at 17:15
yes, sorry, I was referring to the integrand. Hence the integral is always zero, assuming it exists. – user20266 Dec 1 '11 at 17:19

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.