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?
 A: 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.
A: \begin{align}
& \int e^{-xy} \Big( \sin x \, dx \Big) = \int u \, dv = uv - \int v\,du = -e^{-xy}\cos x - \int (-\cos x) ye^{-xy} \, dx \\[10pt]
= {} & -e^{-xy} \cos x + y\int e^{-xy}\Big( \cos x \, dx\Big) = -e^{-xy} \cos x + y \int u\,dw \\[10pt]
= {} & -e^{-xy} \cos x + y \left( uw - \int w\, du \right) = -e^{-xy} \cos x + y\left( e^{-xy}\sin x - \int (\sin x) ye^{-xy} \, dx \right) \\[10pt]
= {} & -e^{-xy} \cos x + ye^{-xy}\sin x - y^2\int (\sin x) e^{-xy} \, dx \\[10pt]
& \text{Therefore} \\
& \int e^{-xy}\sin x\, dx = -e^{-xy} \cos x + ye^{-xy}\sin x - y^2\int (\sin x) e^{-xy} \, dx \\[10pt]
\text{i.e. } & I = -e^{-xy} \cos x + ye^{-xy}\sin x - y^2 I \\[10pt]
\text{So } & (1+y^2) I = -e^{-xy} \cos x + ye^{-xy}\sin x \\[10pt]
& \text{and so } I = \frac {-e^{-xy} \cos x + ye^{-xy}\sin x} {1+y^2}.
\end{align}
Plugging in $x=0$ you get $\frac 1 {1+y^2}$ and plugging in $x=a$ you get something that approaches $0$ as $a\to+\infty.$
Then you have $\displaystyle \int_0^b \frac{dy}{1+y^2},$ and that is routine.
Second postscript (The first postscript appears below.): Above I changed
$$
\lim_{a,b \, \to\, +\infty} \int_{[0,b]} \left( \int_{[0,a]} f(x,y) \,dx\right) \, dy
$$
to
$$
\lim_{b\,\to\,+\infty} \int_{[0,b]} \left( \lim_{a\,\to\,+\infty} \int_{[0,a]} f(x,y) \, dx \right)\,dy.
$$
The question is: Why is that justified? When is it true that
$$
\lim_{a\,\to\,+\infty} \int_{[0,b]} g_a(x)\, dx \overset{\text{?}} = \int_{[0,b]} \lim_{a\,\to\,+\infty} g_a(x)\,dx \text{ ?}
$$
Lebesgue's dominated convergence theorem says it's true if there is a dominating function whose integral is finite, i.e. a function $h$ for which
$$
\text{for all } a>0, \text{ for all } x\in[0,b],\quad |g_a(x)| \le h(x),
$$
and
$$
\int_{[0,b]} h(x)\,dx < +\infty,
$$
where, as you see, $h(x)$ does not depend on $a.$
In this case you have
$$
\left| \frac 1 {1+y^2} - \frac {-e^{-ay} \cos a + ye^{-ay}\sin a} {1+y^2} \right| \le \text{what function of $y$ not depending on $a$?}
$$
Trigonometry tells us that $\left| \cos a + y\sin a \right| \le \sqrt{1+y^2},$ and we have $0 < e^{-ay} \le 1.$
Postscript: I have this criticism of your notation. If you write $$ \int_A \int_B f(x,y) \,dx\,dy, $$ that means $\displaystyle \int_A \left( \int_B f(x,y)\,dx\right)\,dy$ and $x$ runs through the set $B,$ but if you write $$ \iint_{B\times A} f(x,y) \, d(x,y) $$ then that does not mean an iterated integral; i.e. it is not one integral inside another, and $x$ again runs through the set $B$ since the pair $(x,y)$ runs through $B\times A.$ BUT you have $$ \iint_{B\times A} f(x,y)\,dx\,dy, $$ with $\text{“}dx \, dy\text{''}$ at the end rather than $d(x,y),$ and it's hard to be sure which variable was intended to run through which set.
