DiGMi

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A Gaussian function (named after one of the greatest mathematician, Carl Friedrich Gauss) is a function of the form: \[ f\left(x\right)=ae^{-\frac{\left(x-b\right)^{2}}{2c^{2}}} \]

We would have like to examine the following integral: \[ I=\intop_{-\infty}^{\infty}e^{-x^{2}}\mathrm{d} x\]

But we are facing a problem, what is the antiderivative function of $e^{-x^{2}}$?

Well, You won’t be able to find one, at least not an elementary one. The integral of a Gaussian is the Gauss error function, but in this special case, there is a cool way to calculate this integral.

At first, Instead of looking at $I$, lets take a look at $I^{2}$. Meaning:
\[ I^{2}=\intop_{-\infty}^{\infty}e^{-x^{2}}\mathrm{d} x\intop_{-\infty}^{\infty}e^{-y^{2}}\mathrm{d} y \]

Note: In the second integral, I’ve changed the variable of integration to y. I can do that, because it’s a completely separated integral.

Now, notice that the following integral is equivalent to the following:\[
\intop_{-\infty}^{\infty}\intop_{-\infty}^{\infty}e^{-x^{2}-y^{2}}\mathrm{d} x\mathrm{d} y
\]
Well, That’s a bit weird. Now we have a double integral! We’ve made the problem much more complex! Or did we?

Well, yes, now we have a double integral, but there is a good reason for that. Now we can change our coordinate system to the polar coordinate system (Meaning: $x=r\cos\theta, y=r\sin\theta$) and remember to multiply by the Jacobian determinant! In our case, the Jacobian determinant is $r$. So, we got:
\[ \intop_{0}^{2\pi}\intop_{0}^{\infty}e^{-r^{2}}r\mathrm{d}r\mathrm{d}\theta \]
But wait a minute, this looks familiar! Notice that:
\[ =-\frac{1}{2}\intop_{0}^{2\pi}\intop_{0}^{\infty}\frac{\mathrm{d}}{\mathrm{d}r}\left(e^{-r^{2}}\right)\mathrm{d}r\mathrm{d}\theta=-\frac{1}{2}2\pi\intop_{0}^{\infty}\frac{\mathrm{d}}{\mathrm{d}r}\left(e^{-r^{2}}\right)\mathrm{d}r=\] \[ =- \pi \left[e^{-r^{2}}\right]_{0}^{\infty}= -\pi\left[0-1\right]=\pi\]
Therefore:
\[ I^{2}=\pi\Rightarrow I=\intop_{-\infty}^{\infty}e^{-x^{2}}\mathrm{d} x = \sqrt{\pi} \]