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I found a lot of videos online showing how to calculate the Fourier transform of

$f(x) = \exp(-ax^2)$, but they all seem to have a mistake or so I think. (I'm not interested in the alternative solution. I'm trying to understand this.)

In these proofs we get by quadratic completion something of the form

$$\int_{-\infty}^\infty \exp\left(-a\left(x + i\frac{y}{2a}\right)^2 \right) \, dx$$

The next argument is to just substitute $z = x + i\frac{y}{2a}$ and get the known integral

$$\int_{-\infty}^\infty \exp(-az^2) \, dz,$$

but can I just blindly say that the boundaries stay the same even though I now integrate from a shifted place in $\mathbb{C}$?

Why should this be the same as

$$\lim_{b\rightarrow \infty}\int_{-b + i\frac{y}{2a}}^{b + i\frac{y}{2a}} \exp(-ax^2) \, dx,$$ which would be my solution?

Are those the same integrals?

Michael Hardy
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melon_lord
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    It is best to format the exponential function as $\exp$, not $exp$ (just type \exp.) – K.defaoite May 14 '20 at 20:27
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    Unfortunately, you're going to need to know some complex analysis to be able to understand why we're allowed to do this. Perhaps tagging your post with "complex-analysis" might yield you some more help, but I'm not exactly qualified to answer this as I have not yet studied the subject. – K.defaoite May 14 '20 at 20:31
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    You need to do a contour integral, probably a box contour and show that the edges go to zero at infinity. It is very important that the Gaussian extends to an entire function. – Cameron L. Williams May 14 '20 at 20:48
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    Whoever called this a duplicate should read the two questions more carefully. This one is centered on an issue that is not raised in the other one. – Michael Hardy May 14 '20 at 21:01

1 Answers1

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$$\int_{-\infty}^\infty \exp\left(-a\left(x + i\frac{y}{2a}\right)^2 \right) \, dx = \lim_{M\,\to\,+\infty} \int_{-M}^M \cdots\cdots $$ Here I will follow up on the suggestion made by Cameron Williams in a comment above. Consider the integral $$ \int_{-M}^M + \int_M^{M+ iy/(2a)} + \int_{M+iy/(2a)}^{-M + iy/(2a)} + \int_{-M+iy/(2a)}^{-M} $$ where each integral is along a straight line segment. If the function being integrated is an entire function, i.e. differentiable everywhere in $\mathbb C,$ which in particular means there are no singularities inside the region bounded by the path along which we are integrating, (and Gaussian functions are indeed entire functions) then this sum is $0.$

Now suppose you can somehow show that the second and fourth terms above approach $0$ as $M\to+\infty.$ Then you can draw a conclusion about the sum of the first and third terms.

Michael Hardy
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