# Solve $\int_{[0,\infty[}(\int_{[0,\infty[} e^{-xy}\sin{x} \sin{y} d\lambda(x)) d\lambda(y)$

Hint Required for :

$$\int_{[0,\infty[}(\int_{[0,\infty[} e^{-xy}\sin{x} \sin{y} d\lambda(x)) d\lambda(y)$$

My ideas:

Looking at the first inetgral $$\int_{[0,\infty[} e^{-xy}\sin{x} \sin{y} d\lambda(x)$$

I would say $$\int_{[0,\infty[} e^{-xy}\sin{x} \sin{y} d\lambda(x)=\sin{y}\int_{[0,\infty[}e^{-xy}\sin{x}d\lambda(x)$$

And we know $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$

So $$\sin{y}\int_{[0,\infty[}e^{-xy}\sin{x}d\lambda(x)=\sin{y}\int_{[0,\infty[}e^{-xy}\frac{e^{ix}-e^{-ix}}{2i}d\lambda(x)$$

and then I do not know what to do. Should I use this route or is partial integration better suited?

I've just started with multivariable integration, so any pointers would be of great assistance.

$$\int_{[0,\infty)}e^{-xy}(\cos x+i\sin x)\, dx=\int_{[0,\infty)} e^{-x(y-i)}\, dx=-\frac{e^{-x(y-i)}}{y-i}\bigg|_0^\infty=\frac1{y-i}$$
where the result is taken assuming that $$y> 0$$ (the integral doesn't exists for $$y=0$$, however we can ignore this point because the set $$\{0\}$$ have measure zero). In any case it is easy to check that the above integral is absolutely Lebesgue integrable for $$y>0$$, hence Lebesgue integrable.
Of course you can also use the identity $$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$ as you was doing, and you will find the same answer.
If you did not check any sort of absolute convergence for the multivariate integral, I strongly suggest integrating wrt $$x$$ first, then $$y$$ second. Note that $$\sin(x)$$ is the imaginary part of $$e^{ix}$$, that should help you compute your integral better (you know the integral from $$0$$ to $$\infty$$ of $$t \longmapsto e^{-pt}$$ for reasonable complex numbers $$p$$, right?)