Russell Gillen

2022-01-04

How do I show that:
${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}\left(ax\right)\mathrm{sin}\left(bx\right)}{{x}^{2}}dx=\pi min\frac{a,b}{2}$

Kindlein6h

In several steps:
Trigonometric relation
$\left(\frac{\mathrm{sin}ax}{x}\right)\left(\frac{\mathrm{sin}bx}{x}\right)=\frac{1-\mathrm{cos}\left(a+b\right)x}{2{x}^{2}}-\frac{1-\mathrm{cos}\left(a-b\right)x}{2{x}^{2}}$
Dirchlet integral
${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}\alpha t}{t}dt=\frac{\pi }{2}sgn\left(\alpha \right)$
Integration by parts
${\int }_{0}^{\mathrm{\infty }}\frac{1-\mathrm{cos}\left(\alpha \right)t}{{t}^{2}}=\alpha {\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}\alpha t}{t}dt=\frac{\pi }{2}|\alpha |$
Combine
${\int }_{0}^{\mathrm{\infty }}\left(\frac{\mathrm{sin}ax}{x}\right)\left(\frac{\mathrm{sin}bx}{x}\right)dx=\frac{\pi }{4}\left(|a+b|-|a-b|\right)$
$=\frac{\pi }{2}min\left(a,b\right)$

Ella Williams

One way is to to this by residues. Another way to integrate once by parts to get
$I={\int }_{0}^{\mathrm{\infty }}\frac{b\mathrm{sin}ax\mathrm{cos}bx+a\mathrm{cos}ax\mathrm{sin}bx}{x}dx$
then to use the formula $2\mathrm{sin}\alpha \mathrm{cos}\beta =\mathrm{sin}\left(\alpha +\beta \right)+\mathrm{sin}\left(\alpha -\beta \right)$ and the mentioned integral (note that your formula needs to be corrected on the left and on the right)
${\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}xy}{x}dx=\frac{\pi }{2}sgn\left(y\right)$
This gives
$I=\frac{\pi }{4}\left[bsign\left(a+b\right)+bsign\left(a-b\right)+asign\left(a+b\right)+asign\left(b-a\right)\right]=\frac{\pi }{4}\left(|a+b|-|a-b|\right)$
For $a,b>0$ the last expression is obviously equal to $\pi min\frac{a,b}{2}$

star233