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I have to show that for $a,b > 0$

$$\int_\mathbb{R} \dfrac{\sin(at)\sin(bt)}{t^2}dt = \pi\min(a,b)$$

by using characteristic functions and the inversion formula. We do have the hint that we should start with uniform distributed random variables $X \sim \mathcal{U}([-a,a])$ and $Y \sim \mathcal{U}([-b,b])$, but I still don't know how to start. I tried some things already but nothing led me anywhere.

The inversion formula says, that when $\varphi(t)$ can be integrated, the following identity is true:

$f(x) = \dfrac{1}{2\pi} \int e^{-itx} \varphi(t) dt$.

Davide Giraudo
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Billy Klein
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1 Answers1

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The CF of $X$ is $\varphi_X(t) = \frac{e^{ita}-e^{-ita}}{2ita} = \frac{\sin(at)}{at}$. Similarly $\varphi_Y(t) = \frac{\sin(bt)}{bt}$.

Note that $\varphi_X \varphi_Y = \varphi_{X+Y}$ when $X$ and $Y$ are independent. By the inversion formula applied to $X+Y$, I think the remaining step is to compute the density* of $X+Y$ at $0$ , which I believe you can show to be $\frac{2 \min\{a,b\}}{4ab}$.

${}^*$e.g., using the convolution formula, or by working with the CDFs by reasoning about areas in the rectangle $[-a,a] \times [-b, b]$

angryavian
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