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\begin{align*} \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \exp \left\{- \frac{x^2}{2\sigma^2} \right\} \, dx &= \sigma \\ \end{align*}

I can verify that that is correct. Is there a relatively simple way to demonstrate this integral?

I see many other questions addressing similar integrals, but I'd like to see how to do this particular integral.

clay
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    Which "similar" integrals have you seen? In particular, by scaling it is likely you could reduce to one of those integrals. – Sarvesh Ravichandran Iyer Sep 14 '20 at 05:57
  • this one is close, but really not quite the same: https://math.stackexchange.com/questions/370518/evaluating-int-mathbbr-frac-exp-x21x2-mathrmdx – clay Sep 14 '20 at 06:04
  • here's another that is close but more complicated: https://math.stackexchange.com/questions/3793382/calculating-frac12-pi-int-infty-infty-dt-exp-left-frac12-l – clay Sep 14 '20 at 06:04
  • Take the result from $\int_{-\infty}^\infty e^{-x^2} dx$ and use a substitution to get that result. – Ninad Munshi Sep 14 '20 at 06:32
  • Have you seen the computation for $\int_{-\infty}^\infty e^{-x^2}dx$? (Usually done by polar coordinates) The problem you have reduces to this by a change of variable. – Sarvesh Ravichandran Iyer Sep 14 '20 at 06:48

2 Answers2

1

Let $A = (\int_{- \infty}^{\infty} \text{exp}(-x^{2})dx)^{2} = \int_{- \infty}^{\infty} \text{exp}(-x^{2})dx \cdot \int_{- \infty}^{\infty} \text{exp}(-y^{2}) dy = \int_{-\infty}^{\infty}\int_{- \infty}^{\infty} \text{exp}(-x^{2}-y^{2})dxdy$

$ \ \ $

Take $x = rcos(t), y = rsen(t)$, with $0<r< \infty$, $0≤t<2π$ (thus $x² + y² = r²$)

Then

$ \ \ $

$A = \int_{0}^{2π}\int_{0}^{\infty} r\cdot \text{exp}(-r²) dr dt $

$ \ \ $

We have that $\int r \cdot \text{exp}(-r²) dr = -\frac{ \text{exp}(-r²)}{2}$

Then $\int_{0}^{\infty} r \cdot \text{exp}(-r²) dr = -\frac{ \text{exp}(-\infty)}{2} + \frac{ \text{exp}(0)}{2} = \frac{1}{2}$

$ \ \ $

Thus $A = \int_{0}^{2π} \frac{1}{2} dt = π$

$ \ \ $

So $\sqrt{π} = \sqrt{A} = \int_{- \infty}^{\infty} \text{exp}(-x^{2})$

$ \ \ $

Now, go to the original problem..

$ \ \ $

$B = \int_{-\infty}^{\infty} \text{exp}(\frac{-x²}{2\sigma^{2}}) dx$

Doing change of variable $u = \frac{x}{\sqrt{2}\sigma} \implies du = \frac{dx}{\sqrt{2}\sigma} \implies \sqrt{2}\sigma du = dx$

Then we have

$ B = \sqrt{2}\sigma \int_{-\infty}^{\infty} \text{exp}(-u²) du = \sqrt{2} \sigma \sqrt{π}$

Thus $\frac{B}{\sqrt{2π}} = \sigma$

And we are done.

ZAF
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Using the Gaussian integral:

$$\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}$$ and then substituting $x=\sqrt{2}\sigma u$, so $dx=\sqrt{2}\sigma du$ we have

$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{x^2}{2\sigma^2}}dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-u^2}\sqrt{2}\sigma du$$ $$=\frac{\sigma}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-u^2}du=\frac{\sigma}{\sqrt{\pi}}\sqrt{\pi}=\sigma.$$

(Note in this case $\sigma>0,$ but if $\sigma<0$, then the result should be $-\sigma.$ Thus is general the result is $|\sigma|.$)

Alessio K
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