So, during this lecture, Osgood shows that ; that is, this modified Gaussian function is “its own Fourier transform”.

I’ll sketch out what he did in the lecture at the end of this post. But just for fun (and to make a point) I’ll give a method that uses an elementary residue integral.

Both methods start by using the definition:

Method 1: combine the exponential functions in the integrand:

. Now complete the square to get:

Now factor out the factor involving alone and write as a square:

Now, make the substitution to obtain:

Now we show that the above integral is really equal to

To show this, we perform along the retangular path : and let

Now the integral around the contour is 0 because is analytic.

We wish to calculate the negative of the integral along the top boundary of the contour. Integrating along the bottom gives 1.
As far as the sides: if we fix we note that and the magnitude goes to zero as So the integral along the vertical paths approaches zero, therefore the integrals along the top and bottom contours agree in the limit and the result follows.

Method 2: The method in the video
This uses “differentiation under the integral sign”, which we talk about here.

Stat with and note

Now we do integration by parts: and the integral becomes:

Now the first term is zero for all values of as . The second term is merely:

.

So we have shown that which is a differential equation in which has solution (a simple separation of variables calculation will verify this). Now to solve for the constant note that .

The result follows.

Now: which method was easier? The second required differential equations and differentiating under the integral sign; the first required an easy residue integral.

By the way: the video comes from an engineering class. Engineers need to know this stuff!

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