Complex - How to approach improper integral

Question

I'm trying to solve this integral

$$ \int_{-\infty}^{\infty} \frac{\sin(at) \sin(b(u-t))}{t(u-t)} dt $$

where $a$ and $b$ are positive.

Any ideas how to approach this?

Answer 1

Let $f_k(x)=\text{sinc}(kx)$. We want to compute $(f_a * f_b)(u)$. Let we consider the Fourier transform of $g_{a,b}=f_a*f_b$, that is just $\widehat{f_a}\cdot \widehat{f_b}$, i.e. a multiple of the product between the characteristic function of $(-a,a)$ and the characteristic function of $(-b,b)$, i.e. a multiple of the characteristic function of $(-\min(a,b),\min(a,b))$. By Fourier inversion, $g_{a,b}$ is a multiple of $\text{sinc}(\min(a,b)x)$, namely:

$$ \int_{-\infty}^{+\infty}\frac{\sin(at)\sin(b(u-t))}{t(u-t)}\,dt = \color{red}{\frac{\pi}{u}\cdot\sin(u\min(a,b))}.$$

Answer 2

Hint: Use partial fractions and use $2\sin(x)\sin(y) = \cos(x-y)-\cos(x+y)$ To split the integral in four parts.