I evaluated the Gaussian integral using polar substitution, and got that it is $\sqrt{\pi}$.

But my professor also asked us to compute the integral from $\int_{-\infty}^{\infty} e^{\frac{-x^2}{2}} dx$ and the integral from $\int_{0}^{\infty} x^2 e^{-x^2} dx$ using the evaluation of the Gaussian integral. How do I do that using my answer for the first part?

  • 1
    $\begingroup$ integration by parts. $\endgroup$
    – Doug M
    Dec 4, 2017 at 23:05
  • $\begingroup$ I believe you computed the integral from $0$ to $+\infty$. Observe that $e^{-x^2/2}$ is an even function. What can you say about its integration from $-\infty$ to $0$? $\endgroup$
    – Math Lover
    Dec 4, 2017 at 23:06
  • $\begingroup$ Math Lover- I don't get how that helps me. The only thing that changed for the first integral I have to compute is that e^-x^2 became e^(-x^2/2) $\endgroup$
    – Melanie
    Dec 7, 2017 at 4:16
  • $\begingroup$ Doug M-I know how to do integration by parts but I don't know what would be a really convenient and easy way to do it that uses what I computed in the first part $\endgroup$
    – Melanie
    Dec 7, 2017 at 4:17

1 Answer 1


A nice trick is to compute $$f(t) = \int_{0}^{\infty} e^{-tx^2} dx = \frac{1}{2} \sqrt{\frac{\pi}{t}}$$ Using the same trick with polar coordinates.

Taking the derivative of Now just differentiate both sides and plug in $t=1$

  • $\begingroup$ He told us to not use the polar coordinates method again, only to, in one or two lines of work, use the result from part 1. $\endgroup$
    – Melanie
    Dec 7, 2017 at 4:18

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