Gaussian Integral

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:

    \[ =-\frac{1}{2}\intop_{0}^{2\pi}\intop_{0}^{\infty}\frac{\mathrm{d}}{\mathrm{d}r}\left(e^{-r^{2}}\right)\mathrm{d}r\mathrm{d}\theta=-\frac{1}{2}2\pi\intop_{0}^{\infty}\frac{\mathrm{d}}{\mathrm{d}r}\left(e^{-r^{2}}\right)\mathrm{d}r=\]

    \[ =- \pi \left[e^{-r^{2}}\right]_{0}^{\infty}= -\pi\left[0-1\right]=\pi\]


    \[ I^{2}=\pi\Rightarrow I=\intop_{-\infty}^{\infty}e^{-x^{2}}\mathrm{d} x = \sqrt{\pi} \]

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One Response to “Gaussian Integral”

  • Thiago Says:

    We could try coinectnng the abstract to the real. If its just taught in a vacuum, math is very abstract and dry. It seems like a silly game. But if you connect it to solving real problems that people encounter, then one tends to grasp its utility. I remember when I was in high school and I was introduced to the concept of i –the square root of negative one. I thought the whole concept was useless–until I found out that certain things, including modeling the flow of air around an airplane’s wings, couldn’t be done without it. That made all the difference in the world to me at the time.The very fact that this girl couldn’t see HOW she would ever need algebra is an indictment of the people who were entrusted with teaching her math. If its just taught as an abstract game with no practical application to the real world you can’t really blame students for considering it to be useless. For them, it is useless.

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