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

Question

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.

Answer 1

HINT: note that the inner integral is equivalent to the imaginary part of the integral

$$\int_{[0,\infty)}e^{-xy}(\cos x+i\sin x)\, dx=\int_{[0,\infty)} e^{-x(y-i)}\, dx=-\frac{e^{-x(y-i)}}{y-i}\bigg|_0^\infty=\frac1{y-i}$$

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.

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However the outer integral doesn't have a closed form solution.

Answer 2

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?)