#gaussian integral

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∫ℝ e^(-x²) dx = √π

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The Gaussian Integral is a beautiful integral for which the area between the e^(-x^2) and the x-axis from negative infinity to positive infinity perfectly equals the square root of pi. Image sources: 1, 2.

\( \int_{-\infty}^{\infty} e^{-u} \frac{du}{2x} \). Suddenly, we have a dilemma on our hands.  This integral cannot be evaluated further simply because we are unable to cancel out the original variable, x.  If we attempt to perform another substitution, say p, a similar problem would arise and we would have no way to further our research.  Henceforth, we must discard this methodology and analyze this integral from another point of view. Note: Evaluating this problem through the IBP(integration by parts) method will immediately lead to infinite descent. We could assess this integral via series solution.  Indeed, the power series \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \) can be of immense use.  (Author’s note: I presented a paper at a conference which made use of this fact.  It went quite well, if I say so myself!  In the division I was in, I did reasonably well by placing first!) We could amend it accordingly to serve our purpose, but there’s an easier method and we shall evaluate it through such a method. An idea. Let \( I = \int_{-\infty}^{\infty} e^{-x^2} dx \).  Whatever the solution of this integral is, the variable clearly does not matter in the slightest.  Using this knowledge, we can make the following conclusion: \(I = \int_{-\infty}^{\infty} e^{-x^2} dx = \int_{-\infty}^{\infty} e^{-y^2} dy \). With this in mind, suppose we want to find what \( I^2 \) looks like?  Well, \( I^{2} = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2} e^{-y^2} dx \; dy = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2 - y^2} dx \; dy \). Note: The last line merely made use of exponential rules found in a typical HS algebra course. Conversion from rectangular coordinates to polar coordinates. Indeed, \( x^2 + y^2 = r^2 \), where r is the radius of a circle.  If you have a graphing calculator around, you could try and plug this into it.  Doing so, you will immediately see a circle form upon your screen.  Note that this is NOT a function since it fails the vertical line test. It is important now to note that the area of a rectangle with height y and width x is \( A = xy \).  Differentiating, we find that \( dA = dx dy \). Note: This could be expanded upon further, but doing so would detract from the solution. Converting to polar coordinates, we find that \( dA = dx \; dy = r \; dr \; d\theta \).  We must also convert the limits of our integral to fit our polar coordinate conversion.  Thus, \( I^2 = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2 - y^2} dx \; dy = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)} dA \) \( =  \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} r \; dr \; d\theta \). (1) Informal notes on the change of integration limits. This sometimes goes over the heads of many students who encounter this change.  Let's inquire why we change the limits of integration first. Remember, our conversion implies that the same 'rules' that we are bound by in the rectangular coordinate system do not follow in this current system.  As such, we must amend the limits accordingly.  When thinking of polar coordinates, think of degrees and radial length in a circle.  By the usual definition of a circle, we cannot have the radius extend from negative infinity to positive infinity; we must have a center.  We dictate this center as the origin of the plane; i.e, this is why the limits of the integration for r begin at 0.  Now theoretically, we can have a radial length of infinite length; as such, our radial length spans from 0 to infinity.  This explains our first limits of integration. Our second set of limits is a bit more intuitive, but not that much harder.  Clearly, if we think in terms of degrees, we know that a full revolution is from 0 to \( 2\pi \).  This is equivalent to spanning an entire coordinate plane from negative infinity to positive infinity. Conclusion. This is the easy part(or probably the hardest depending on who you speak to!). Setting \( u = r^2 \), we have \( \frac{du}{dr} = 2r \).  Solving for dr, we have \( dr = \frac{du}{2r} \).  Plugging this back into (1), we find that \( \int_{0}^{2\pi} \int_{0}^{\infty} e^{-u} \; r \; \frac{du}{2r} \; d\theta \) = \( \frac{1}{2} \; \int_{0}^{2\pi} \int_{0}^{\infty} e^{-u} \; du \; d\theta \) = \( -\frac{1}{2} \; \int_{0}^{2\pi} d\theta \; [e^{-u}|_{0}^{\infty}] \) Setting \( \beta = \infty \), we have: = \( -\frac{1}{2} \; \int_{0}^{2\pi} d\theta \; [ \frac{1}{e^{ \beta}}- \frac{1}{e^0} ] \) = \( -\frac{1}{2} \; \int_{0}^{2\pi} d\theta \; [0 - 1] \). = \( \frac{1}{2} \; \int_{0}^{2\pi} d\theta = \frac{1}{2} [ 2\pi - 0] \) = \( \frac{1}{2} \; [2\pi] \) = \( \pi \). Thus, \( I^2 = \pi \).  Taking the square root of both sides, we have \( I = \sqrt \pi \).  But, \( I = \int_{-\infty}^{\infty} e^{-x^2} dx \) as we defined earlier(see above).  Hence, \( \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt \pi \).  The proof is complete. .`, There you have it!  Voila. As I stated previously, please feel free to message me to inquire about anything.  Next time, I shall expand more upon my set theory series and tackle some common 'elementary' exercises in set theory. See you then!

  Now, can you solve it?