MHB Solving Improper Integral Challenge Problem

[polygamma]

My first post on the new forums is going to be a challenge problem.$\displaystyle \int_{0}^{\infty} \frac{\sin ax \ \sin bx}{x^{2}} \ dx \ , \ a > b \ge 0$

 

[AlexYoucis]

My first post on the new forums is going to be a challenge problem.$\displaystyle \int_{0}^{\infty} \frac{\sin ax \ \sin bx}{x^{2}} \ dx \ , \ a > b \ge 0$

It's the old standby, first let our integral be $I$ then $$\begin{aligned}2I &=\int_0^{\infty}\frac{\cos((a-b)x)-\cos((a+b)x)}{x^2}\\ &= \int_0^{\infty}\int_{a-b}^{a+b}\frac{\sin(xy)}{x}\\ &=\int_{a-b}^{a+b}\int_0^{\infty}\frac{\sin(xy)}{x}\\ &=\int_{a-b}^{a+b}\frac{\pi}{2}\\ &=\pi b\end{aligned}$$

 

[polygamma]

Fubini's theorem is not satisfied. What's the justification for changing the order of integration?

And I probably should have just said that $a,b \ge 0$ to make the problem slightly more interesting.

Anyways, my idea was to integrate by parts, use a trig product-to-sum identity, and then use the fact that $\displaystyle \int_{0}^{\infty} \frac{\sin \alpha x}{x} \ dx = \frac{\pi}{2}\text{sgn}(\alpha) $

 

[Markov2]

Most of this cases are tricky to justify, and everything works because is hidden, so the only trick here is to observe that $\dfrac{1-\cos ((a+b)x)-\left[ 1-\cos ((a-b)x) \right]}{{{x}^{2}}},$ so now use $\displaystyle\frac1{x^2}=\int_0^\infty te^{-tx}\,dt$ and this absolutely justifies the integration order by using Tonelli's Theorem.

 

[oasi]

My first post on the new forums is going to be a challenge problem.$\displaystyle \int_{0}^{\infty} \frac{\sin ax \ \sin bx}{x^{2}} \ dx \ , \ a > b \ge 0$

how can you solve this hard problems ,i'm lovin it

 

[alyafey22]

By symmetry :

\frac{1}{2} \int_0^{\infty}\frac{\cos((a-b)x)-\cos((a+b)x)}{x^2}\, dx \,=\, \frac{1}{4} \int_{-\infty}^{\infty}\frac{\cos((a-b)x)-\cos((a+b)x)}{x^2}\, dx

\mathcal {Re} ( \frac{1}{4} \int_0^{\infty}\frac{e^{i(a-b)z}-e^{i(a+b)z}}{x^2} \, dx )

can be solved using a contour in the upper half plane that doesn't include zero

\frac{1}{4}\int_{-\infty }^{\infty}\frac{e^{i(a-b) x}-e^{i(a+b)x}}{x^2}= \frac{\pi i}{4}\mathcal {Rez}(f(z) , 0)=\frac{\pi i}{4} (i(a-b) - i(a+b) )= -\frac{\pi}{4}(a-b-a-b)= \frac{\pi b}{2}