Evaluating $\lim_{a,b\to + \infty} \iint_{[0,a]\times[0,b]}e^{-xy} \sin x \,dx\,dy$ I'm trying to calculate the following: 
$$\lim_{a,b\to + \infty} \iint_{[0,a]\times[0,b]}e^{-xy} \sin x \,dx\,dy$$
Trying to calculate by definition didn't get me far. Any ideas how to attack this problem?

Let $I=\int_0^b\sin xe^{-xy}dx$. Let $u=\sin x,\,dv=e^{-xy}dx$. Then $$I=\int_0^b\sin xe^{-xy}dx=\frac{\sin xe^{-xy}}{-y}\bigg|_0^b+\frac{1}{y}\int_0^b\cos xe^{-xy}dx.$$
Now apply by parts once again with $u=\cos x,\, dv=e^{-xy}dx$
we get$$I=\int_0^b\sin xe^{-xy}dx=\frac{\sin xe^{-xy}}{-y}\bigg|_0^b-\frac{\cos xe^{-xy}}{y^2}\bigg|_0^b-\frac{1}{y^2}\int_0^b\sin xe^{-xy}dx.$$
Thus we find $$(1+\frac{1}{y^2})I=\frac{\sin xe^{-xy}}{-y}\bigg|_0^b-\frac{\cos xe^{-xy}}{y^2}\bigg|_0^b\Longrightarrow I=\frac{-y\sin xe^{-xy}}{y^2+1}\bigg|_0^b-\frac{\cos xe^{-xy}}{y^2+1}\bigg|_0^b.$$ The values at $b$ will go to zero as $b\to \infty$, so $I=\frac{1}{y^2+1}$. Putting this in the original integral and integrating w.r. to $y$ you get $\arctan y$. I think you can proceed from here. Your result will be $\pi/2$. 
P.S. Actually @MichaelHardy's comments complete the theoretical part of the solution as we require while interchangin limits and integrals.
