Full Question: Consider $a>0$ and $b\in R$ constants. Use a Tonelli condition to justify the change of order of integration in $\int^\infty_0 (\int^1_0 e^{(-ay)} \sin(2bxy)dx)dy$
and then prove $\int^\infty_0 e^{(-ay)} \frac{1-\cos(2by)}{2y} dy = \int^\infty_0 e^{(-ay)} \frac{\sin^2(by)}{y} dy = \frac{1}{4} \ln(\frac{a^2+4b^2}{a^2})$.
Prove that this formula is also correct for $\alpha=0$, as $\infty=\infty$.
I'm on the first part of the question so let's focus on that.
The Tonelli condition I think we're supposed to use is:
$\int^\infty_{-\infty}[\int^\infty_{-\infty} |f(x,y)|dx]dy < \infty$
I attempted solve this by saying $e^{(-ay)}\sin(2bxy)\leq \sin(2bxy)$ and then plug $\sin(2bxy)$ into the Tonelli condition to find that it is convergent which would then mean that the full thing was convergent but I ended up getting does not exist. I also tried it but with $\leq e^{(-ay)}$ but that doesn't work either. Perhaps this is not the right route at all?