3
$\begingroup$

This question already has an answer here:

How do I evaluate the following integral $$\int_{-\infty}^{\infty} \exp\left(-\frac{\sigma^2 x^2}{2}\right) \mathrm dx\;?$$

How is it even possible to find an antiderivative?

The integral is evaluated "silently" in a book leading to a theorem.

Using Wolfram Alpha (after trying to evaluate on my own) I get

enter image description here

and this is not what I want, since at my level we've never worked with such a function.

Hoping someone can clarify.

$\endgroup$

marked as duplicate by Jack D'Aurizio, Lucian, TrueDefault, Davide Giraudo, Aditya Hase Dec 23 '14 at 10:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ By change of variable, you can reduce the integrand to $e^{-x^2}$ form. Then, calculate $\sqrt{\int e^{-x^2}dx\int e^{-y^2}dy}$. This link shows the way. en.wikipedia.org/wiki/Gaussian_integral $\endgroup$ – Math.StackExchange Dec 23 '14 at 8:54
  • $\begingroup$ Try searching this site for Gaussian integral. You can't find an expression for the antiderivative, by the way, other than the one with erf (full name: error function, which see). $\endgroup$ – Harald Hanche-Olsen Dec 23 '14 at 8:56
  • $\begingroup$ I will try, thank you. It might be a little out of my league, since I've only had Calculus. $\endgroup$ – Shuzheng Dec 23 '14 at 8:57
5
$\begingroup$

This integral is not evaluated finding a primitive. The trick is squaring, applying Fubini's theorem and changing to polar coordinates:

$$\left(\int_{-\infty}^\infty e^{-x^2}dx\right)^2=\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)}dxdy=\int_0^{2\pi}\int_0^\infty\rho e^{-\rho^2}d\rho d\theta$$

Can you take it form here?

$\endgroup$
2
$\begingroup$

\begin{align} \int_{-\infty}^{\infty} \exp\left(-\frac{\sigma^2 x^2}{2}\right) \mathrm dx&=2\int_{0}^{\infty} \exp\left(-\frac{\sigma^2 x^2}{2}\right) \mathrm dx\\[7pt] &=2\int_{0}^{\infty} e^{-t}\, \frac{\mathrm dt}{\sigma\sqrt{2t}}\qquad\implies\qquad t=\frac{\sigma^2 x^2}{2}\\[7pt] &=\frac{\sqrt{2}}{\sigma}\int_{0}^{\infty} t^{-1/2}\;e^{-t}\, \mathrm dt\\[7pt] &=\frac{\sqrt{2}}{\sigma}\Gamma\left(\frac{1}{2}\right)\qquad\implies\qquad\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi} \\[7pt] &=\bbox[5pt,border:3px #FF69B4 solid]{\color{red}{\large\frac{\sqrt{2\pi}}{\sigma}}}\tag{$\color{red}{❤}$} \end{align}

$\endgroup$

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