How do I show that: \int_0^\infty\frac{\sin(ax)\sin(bx)}{x^2}dx=\pi min(a,b)/2

Russell Gillen

Russell Gillen

Answered question

2022-01-04

How do I show that:
0sin(ax)sin(bx)x2dx=πmina,b2

Answer & Explanation

Kindlein6h

Kindlein6h

Beginner2022-01-05Added 27 answers

In several steps:
Trigonometric relation
(sinaxx)(sinbxx)=1cos(a+b)x2x21cos(ab)x2x2
Dirchlet integral
0sinαttdt=π2sgn(α)
Integration by parts
01cos(α)tt2=α0sinαttdt=π2|α|
Combine
0(sinaxx)(sinbxx)dx=π4(|a+b||ab|)
=π2min(a,b)
Ella Williams

Ella Williams

Beginner2022-01-06Added 28 answers

One way is to to this by residues. Another way to integrate once by parts to get
I=0bsinaxcosbx+acosaxsinbxxdx
then to use the formula 2sinαcosβ=sin(α+β)+sin(αβ) and the mentioned integral (note that your formula needs to be corrected on the left and on the right)
0sinxyxdx=π2sgn(y)
This gives
I=π4[bsign(a+b)+bsign(ab)+asign(a+b)+asign(ba)]=π4(|a+b||ab|)
For a,b>0 the last expression is obviously equal to πmina,b2
star233

star233

Skilled2022-01-11Added 403 answers

Note that 

0sin(ax)xsin(bx)xdx=12bsin(ax)axsin([b/a]ax)(b/a)axadx=12bsin(x)xsin(x/μ)x/μdx, μab>0 (1)sin(x)xsin(x/μ)x/μdx|μ>0=(1211eikxdk)(1211eiqx/μdq)dx=12π1111ei(kq/μ)xdx2πdkdq=12π1111δ(kq/u)dkdq=12π11[1<qμ<1]dq=π01[q<μ]dq=π([μ<1]0μdq+[μ>1]01dq)=[a<b]πab+[a>b]π (2)With (1) and (2):0sin(ax)xsin(bx)xdx=12π([a<b]a+[a>b]b)=12πmin[a,b]

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