What is this fourier identity from? $\int_{-\infty}^\infty f(x)g(x)dx = \frac1{2\pi}\int_{-\infty}^\infty\hat{f}(y)\hat{g}(y)dy$

Question

In this question, Bonrey uses the identity $$\int_{-\infty}^\infty f(x)g(x)dx = \frac1{2\pi}\int_{-\infty}^\infty\hat{f}(y)\hat{g}(y)dy$$ to attempt to integrate $$\int_{-\infty}^{\infty}\frac{\sin(x-3)}{x-3}e^{-|x|}dx$$ where he defines the fourier transform of a function to be $$\mathcal{F}\left[f(x)\right](y)=\hat{f}(y)=\int_{-\infty}^\infty f(x)e^{ixy}dx$$

Where is this identity from? The closest formula I could find through intensive searching was Parseval's theorem, which states that for two square integrable functions $f$ and $g$, $$\int_{-\infty}^\infty f(x)\overline{g(x)}dx=\int_{-\infty}^\infty\hat{f}(y)\overline{\hat{g}(y)}dy$$ where $\overline{f(x)}$ is $f$'s complex conjugate. Due to this conjugation, I do not believe this is the same theorem that Bonrey applied.

Any help is appreciated! :)

Answer 1

I would just like to note a few things in regards to this. In my case with $g(x)=e^{-|x|}$, this is a real, even function.

We know a few things

1. The Fourier transform of an even integrable function $f\colon \mathbb{R}\to\mathbb{R}$ is even real-valued function.
2. For some $f:\mathbb{R}\mapsto\mathbb{R}$, $f^*(x)=f(-x)$. ($f^*$ is the conjugate of $f$)

If $f(x)$ and its fourier transform are real (which is the case here), then we can say that $$\int_{-\infty}^{\infty}\hat{f}(s)\hat{g}(-s)ds=\int_{-\infty}^{\infty}g(y)f(-y)dy$$

If in addition to this, our function is real valued and even (which is also the case here), then we can reduce this to the identity in question $$\int_{-\infty}^{\infty}\hat{f}(s)\hat{g}(s)ds=\int_{-\infty}^{\infty}g(y)f(y)dy$$

Answer 2

You are correct in questioning the result because it's just not true in general. $$ \int_{-R}^{R}\hat{f}(s)\hat{g}(s)ds = \frac{1}{2\pi}\int_{-R}^{R}\int_{-\infty}^{\infty}f(x)e^{-isx}dx\int_{-\infty}^{\infty}g(y)e^{-isy}dy ds \\ = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x)g(y)\left[\frac{1}{2\pi}\int_{-R}^{R}e^{-is(x+y)}ds\right]dxdy \\ =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x)g(y)\frac{e^{-iR(x+y)}-e^{iR(x+y)}}{-2\pi i(x+y)}dxdy \\ =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x)g(y)\frac{\sin(R(x+y))}{\pi(x+y)}dxdy \\ =\int_{-\infty}^{\infty}g(y)\int_{-\infty}^{\infty}f(x)\frac{\sin(R(x+y))}{\pi(x+y)}dxdy $$ As $R\rightarrow\infty$, the inner integral on the far right plucks out the value of $f$ at $-y$. That's the standard Dirichlet kernel result. Therefore, $$ \int_{-\infty}^{\infty}\hat{f}(s)\hat{g}(s)ds=\int_{-\infty}^{\infty}g(y)f(-y)dy. $$ If either $g$ or $f$ happens to be even, then you get what you want, but not otherwise, at least not in general.