I was playing around with the Plancherel's theorem and stumbled across a derivation of the Gaussian integral which I have never seen before. Could folks check to see if derivation is correct? If so, have you seen this before?
We start from the unitary definition of the Fourier Transform (FT) and the corresponding expression of the Plancherel's theorem. Assuming $f(t)$ is our time-domain function we have:
$$ \begin{align} &F(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(t) e^{-i\omega t} dt \\[0.2in] &\int_{-\infty}^\infty \vert f(t) \vert^2dt = \int_{-\infty}^\infty \vert F(\omega \vert^2d\omega \end{align} $$
Two points to note before we start:
- Proof of the Plancherel's theorem does not make any use of the Gaussian integral, so the derivation is not circular.
- Although we used a particular definition of FT pair, the derivation below holds for any definition provided that we use the matching form for the Plancherel's theorem. In other words, either $\sqrt{2\pi}$ will appear in the definition of the FT or $2\pi$ will appear in the Plancherel's theorem; take your pick. In either case, we get the same result.
Ok now on to the main part. Take $f(t) = e^{-t^2/2}$ and compute its FT:
$$ \begin{align} F(\omega) &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-t^2/2}e^{-i\omega t}dt \\ &= \frac{e^{-\omega^2/2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-(t + i\omega)^2/2}dt \\&= \frac{e^{-\omega^2/2}}{\sqrt{\pi}}\int_{-\infty}^\infty e^{-u^2}du \qquad\ldots\text{ put } t + i\omega = \sqrt{2} u\\ &= \frac{e^{-\omega^2/2}}{\sqrt{\pi}} \mathcal{J} \end{align} $$
where
$$ \mathcal{J} = \int_{-\infty}^{\infty}e^{-t^2}dt $$
Note that $\mathcal{J}$ is just a scalar.
Now we apply the Plancherel's theorem on $f(t)=e^{-t^2/2}$ and $F(\omega)$:
$$ \begin{align} \int_{-\infty}^\infty \vert f(t)\vert^2dt &= \int_{-\infty}^\infty \vert F(\omega)\vert^2d\omega\\ \int_{-\infty}^\infty \left\vert e^{-t^2/2}\right\vert^2dt &= \int_{-\infty}^\infty \left\vert \frac{e^{-\omega^2/2}}{\sqrt{\pi}}\mathcal{J}\right\vert^2d\omega\\ \int_{-\infty}^\infty e^{-t^2}dt &= \frac{\mathcal{J}^2}{\pi}\int_{-\infty}^\infty e^{-\omega^2}d\omega\\ \mathcal{J} &= \frac{\mathcal{J}^3}{\pi}\\ \Rightarrow \mathcal{J} &= \sqrt{\pi} \end{align} $$
I feel less confident about the step where I'm computing the expression for $F(\omega)$. I.e. as soon as I have a complex integrand I have to specify the contour of integration and cant just be making substitutions like $u = t + i\omega$ carelessly. Thanks in advance for any comments and critiques.
EDIT: I have since found that this technique can be used to solve a few other integrals as well. Starting from a top-hat, we can prove
$$ \int_{-\infty}^{\infty}\left(\frac{\sin u}{u}\right)^2\, du= \int_{-\infty}^{\infty}\frac{\sin u}{u}\, du = \pi $$
Trivially, if either the time-domain function or the frequency domain function can be integrated explicitly, then the Plancherel's theorem implies the value of the other.
I would say the derivation is mostly correct, except for the step where you formally make the substitution $u = t + i\omega$. However, you were also right about diagnosing the problem. The identity you write is true, after passing through some complex analysis to change the contour of integration. A rectangular contour with vertices $-a,a,-a+i\omega,a+i\omega$ with $a\to\infty$ and an application of the residue theorem should iron things out.