How Laplace Solved The Gaussian Integral!

[00:00:00]Today, we will be dealing with the famous Gaussian integral, where we have to find the exact value of the integral of E raised to minus X squared from minus infinity to plus infinity. This integral is one of the most important integrals in all of mathematics. It appears in probability, statistics, physics, engineering, and many other fields. The surprising thing is that nobody has found an elementary integral formula for E raised to minus X squared.

[00:00:34]So, the usual integration techniques, such as substitution, integration by parts, and partial fractions do not work here.

[00:00:43]Most textbooks solve this problem using polar coordinates. However, long before that method became popular, Pierre-Simon Laplace discovered a beautiful solution that avoids polar coordinates completely.

[00:00:57]Let us see how he solved it. For that, we let the value of the Gaussian integral be called I. Now, since this is an even function, minus infinity to plus infinity becomes two times integral zero to plus infinity of E raised to minus X squared dx. Great.

[00:01:18]Now, since we do not know how to evaluate this integral I directly, Laplace decided to look at I squared instead. When we square I, we are multiplying the integral by an identical copy of itself.

[00:01:32]But, we will make one small change. For the second copy, we simply replace the variable X by another variable called Y.

[00:01:41]Since variables inside an integral are only placeholders, changing the letter does not change the value.

[00:01:49]Multiplying these two integrals together, we get four times double integral from zero to plus infinity of E raised to minus x squared minus y squared times dy times dx.

[00:02:05]Noise. At this stage, using the traditional approach, we would have switched to polar coordinates. But Laplace chose a completely different path. He introduced a new variable by writing y as t multiplied by x.

[00:02:21]Next, we differentiate both sides with respect to t.

[00:02:26]At this stage, x is treated as a constant because we are allowing only t to vary while keeping x fixed.

[00:02:33]Therefore, differentiating gives dy equals x multiplied by dt. Also, when y equals 0, we get t equals 0.

[00:02:45]Similarly, when y approaches infinity, t also approaches infinity, assuming x is a fixed positive value.

[00:02:53]Substituting y equals tx into the exponent gives e raised to minus x squared minus t squared multiplied by x squared. Both terms contain x squared, so we can factor it out.

[00:03:07]This gives minus x squared multiplied by 1 plus t squared.

[00:03:13]After replacing dy with x multiplied by dt, an extra factor of x appears in the integral.

[00:03:20]The next step is very important, which is about changing the order of integration. A theorem called Fubini's theorem allows us to interchange the order of integration. This works because this function is always positive and finite, so the total area under the surface remains the same regardless of whether we integrate with respect to x first or with respect to t first. So, instead of integrating with respect to t first, we integrate with respect to x first.

[00:03:52]So, we get 4 times the double integral of X times this exponent times DX first, and then we integrate with respect to DT.

[00:04:02]Now, we focus only on the inner integral.

[00:04:05]To evaluate it, we introduce a new variable called U.

[00:04:09]Let U be equal to minus X squared multiplied by 1 plus T squared.

[00:04:17]Next, we differentiate with respect to X, and we consider T as some constant because during the inner integration, only X is changing, while the value of T remains fixed.

[00:04:30]Differentiating U gives DU equals minus 2 multiplied by X multiplied by 1 plus T squared multiplied by DX.

[00:04:41]Rearranging this expression gives X multiplied by DX equals DU divided by minus 2 times 1 plus T squared.

[00:04:50]Now, when X equals 0, U becomes 0, and when X approaches positive infinity, X squared also approaches positive infinity.

[00:05:01]Since there is a negative sign in front, U approaches minus infinity. Therefore, on substitution, we get 1 divided by minus 2 times 1 plus T squared times the integral from 0 to minus infinity of E raised to U times DU.

[00:05:21]At this point, the integral becomes extremely simple.

[00:05:24]The integral of E raised to U is simply E raised to U evaluated from 0 to minus infinity.

[00:05:32]E raised to minus infinity approaches 0, and E raised to 0 equals 1.

[00:05:39]Substituting these values gives 0 minus 1, which equals minus 1.

[00:05:45]Both these minus sign cancels out, and we are left with 1 / 2 * 1 + t squared.

[00:05:54]This is the magical moment of the proof.

[00:05:56]Substitute this result back into the main integral.

[00:06:00]4 / 2 gives 2 * the integral from 0 to infinity of 1 / 1 + t squared.

[00:06:10]Fortunately, this is a standard integral that every calculus student eventually learns whose value is the inverse tangent of t. When t = 0, the inverse tangent of 0 = 0.

[00:06:24]As t approaches positive infinity, the inverse tangent approaches pi / 2.

[00:06:31]Therefore, the integral becomes 2 multiplied by pi / 2 - 0.

[00:06:38]The 2s cancel out leaving us with pi.

[00:06:42]Therefore, we have shown that I squared is equal to pi.

[00:06:46]Taking the square root of both sides gives I = the square root of pi.

[00:06:52]This gives the final value of the Gaussian integral.

[00:06:55]That was simply brilliant, wasn't it?

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