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I'm trying to prove that $$\int_{-\infty}^{\infty}\frac{\sin{ax}\sin{bx}}{x^2}dx=\pi \min{(a,b)}$$ using $$\widehat{f*g}=\widehat{f}\widehat{g}$$

I didn't know were to start really so kind of out of desperation I tried a bunch of functions on my calculator to see whether $$\int_{-\infty}^{\infty}f(x)g(x)dx=\int_{-\infty}^{\infty}f(x)dx\int_{-\infty}^{\infty}g(x)dx$$ and for those I tried it seemed to work. But is this really correct? Feels weird since the following of course isn't true:

$$\int_{a}^{b}f(x)g(x)dx=\int_{a}^{b}f(x)dx\int_{a}^{b}g(x)dx$$

Could it be perhaps that my calculator (TI-nspire cx) just says that the integrals I tried was equal to each other because all integrals I tried perhaps diverged and therefore it perhaps interpreted it as the same though it isn't?

PS I know this is probably one of the most stupid questions posted here :P

Erik Sahlin
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    You're on the right track with the first property. It's helpful to realize that $$\int_{-\infty}^\infty f(x)g(x)dx = \int_{-\infty}^\infty f(x)g(x)e^{-i \cdot 0 \cdot x}dx $$ Can you take it from here? – Ninad Munshi Oct 19 '20 at 19:24
  • Proven here another way. – J.G. Oct 19 '20 at 19:31

1 Answers1

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This is badly false but of course you need to pick examples where the integrals actually converge. A simple example is to take $f, g$ to be step functions ("square waves") with disjoint supports, say $f$ the indicator function of $[0, 1]$ and $g$ the indicator function of $[2, 3]$; then $\int fg = 0$ but $(\int f)(\int g) = 1$.

More conceptually, no result of this form could possibly be true, because the LHS and the RHS behave differently with respect to rescaling the variable $x$. Said another way, the LHS and RHS have different units. Explicitly, if we perform a $u$-substitution $x = sy$ on both sides, then on the LHS we get

$$\int_{-\infty}^{\infty} f(sy) g(sy) \, s \, dy = s \int_{-\infty}^{\infty} f(sy) g(sy) \, dy = s \left( \int_{-\infty}^{\infty} f(sy) \, dy \right) \left( \int_{-\infty}^{\infty} g(sy) \, dy \right)$$

(here we are applying the identity to the pair of functions $f(sy), g(sy)$), but on the RHS we get

$$\left( \int_{-\infty}^{\infty} f(sy) \, s \, dy \right) \left( \int_{-\infty}^{\infty} g(sy) \, s \, dy \right) = s^2 \left( \int_{-\infty}^{\infty} f(sy) \, dy \right) \left( \int_{-\infty}^{\infty} g(sy) \, dy \right).$$

These two different scaling behaviors are incompatible, and this argument shows more generally that not only are the LHS and the RHS not equal but there can't even be a nontrivial linear inequality relating them.

In terms of units, if we take $x$ to have units of time and $f, g$ to have units of distance / time (velocity), then the LHS has units of $\frac{ \text{distance}^2}{ \text{time} }$ but the RHS has units of $\text{distance}^2$.

As for the question you're trying to answer, write the integrand as $\frac{\sin ax}{x} \frac{\sin bx}{x}$. What do you know about the Fourier transform of $\frac{\sin ax}{x}$?

Qiaochu Yuan
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