Why can I use Fubini' theorem on this function?

I used the fact that $\displaystyle \int_0^\infty\int_0^1 e^{-y}\sin(2xy)\,dxdy=\int_0^1\int_0^\infty e^{-y}\sin(2xy)\,dydx$ to solve $\displaystyle\int_0^\infty e^{-y}\frac{\sin^2(y)}{y}\,dy$. (The answer is $\frac{1}{4}\ln(5)$)

But why can I use Fubini's theorem in this case?

• See math.stackexchange.com/a/77507 -- does that help? – David K Aug 22 '15 at 2:36
• @DavidK It dind't help very much – Andre Gomes Aug 22 '15 at 2:42
• Could you be more specific about your question, then? Do you doubt the equation you wrote on the first line? If so, why? If not, where else is there a source of doubt? – David K Aug 22 '15 at 2:48
• Yes, I doubt the first equation. Because I don't know why I can change the order of the integrals. – Andre Gomes Aug 22 '15 at 3:01

Since $$\int_0^1\int_0^\infty| e^{-y}\sin(2xy)|\,dydx<\int_0^1\int_0^\infty e^{-y}dydx=1$$ By Fubini's theorem, you can switch the order of integration.
Skipping over details, all Fubini's theorem needs is that $$\int_A |f(x,y)| dx dy < \infty$$ for some region $A$ in the $xy$-plane. There is no requirement for $A$ to be a finite region (if that's what you may have mistakenly thought); in fact, a bounded $A$ is not even a sufficient condition for Fubini's theorem in general.
Here, the inner integral is bounded by $C \cdot e^{-y}$ for some constant $C$, which is finite when you later integrate wrt $y$ from $0$ to $\infty$, so Fubini's theorem works.