Does the improper integral $\large\int_{\mathbb{R}^2} e^{-xy}\sin(x)\mathrm{d}x\mathrm{d}y$ converge?
I know the when the integral is only on $\mathbb{R}^2_+ \times \mathbb{R}^2_+$ then it converges to $\frac{\pi}{2}$.
However here the boundaries are different. Also it is not a positive function so I need to show it on the absolute value. It means I need to show that $$\int_{\mathbb{R}^2} e^{-xy}|\sin(x)|\mathrm{d}x\mathrm{d}y$$ converges.
I'm pretty sure it doesn't converge, since I can't bound it from the right by a converge-bale function (I tried with $e^{-xy}$, also, Wolfram Alpha said so too).
However I can't think of a function that won't converge and will bound it from the left.
Help would be appreciated.
Best Answer
To begin, the function $(x,y) \mapsto e^{-xy} \sin x$ is not absolutely integrable. It is enough to show this for the region $[0,\infty)^2$.
If it were then we could apply Fubini's theorem to the iterated integrals and conclude
$$\int_0^\infty \int_0^\infty e^{-xy} | \sin x| \, dx \, dy = \int_0^\infty \int_0^\infty e^{-xy} | \sin x| \, dy \, dx $$
However,
$$\int_0^\infty e^{-xy} | \sin x| \, dy= \frac{|\sin x|}{x} $$
and this is neither Lebesgue nor improperly Riemann integrable over $[0,\infty]$.
The question seems to be is the improper Riemann integral $\int_{\mathbb{R}^2} f$ convergent, where $f(x,y) = e^{-xy} \sin x$.
For integrals over $\mathbb{R}$ the definition of an improper integral as $\lim_{a \to -\infty, b \to +\infty}\int_a^b f$ is unambiguous. For multiple integrals the definition of improper integral requires more care. The standard is
$$\tag{*}\int_{\mathbb{R}^2}f = \lim_{n \to \infty}\int_{A_n}f,$$
where $A_n$ is an admissible sequence, that is a sequence of compact Jordan measurable sets such that $A_n \subset A_{n+1}$ for all $n$ and $\cup_{n=1}^\infty A_n = \mathbb{R}^2$.
For the improper integral to be well defined, (*) must hold for any admissible sequence with convergence to a unique value. This has an important consequence:
See here for a proof.
Since $\int_{\mathbb{R}^2} e^{-xy} |\sin x|$ does not converge it follows that the improper integral $\int_{\mathbb{R}^2} e^{-xy} \sin x$ does not converge in the sense of (*). In fact, it can take on different values depending on exactly how the limiting process is defined. This is analogous to the problem with rearrangements of conditionally convergent infinite series. An example is given here.