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I am reviewing a homework problem and I came upon the following statement, which is only a part of what I am trying to solve:

Given that $\frac{\sin x}{x}$ is not Lebesgue integrable on $[0,\infty)$, then $e^{-xy}\sin x$ is not integrable on $[0,\infty) \times [0,\infty)$ (but it is integrable on $[0,b)\times [0,\infty)$.)

Now, I understand the proof that $\frac{\sin x}{x}$ is not Lebesgue integrable on $[0,\infty)$. However, I fail to see how this relates to $e^{-xy}\sin x$. The Taylor series of $e^{-x}$ is alternating, so that doesn't seem useful. Furthermore, $\frac{1}{x} > e^{-x}$ on $(a,\infty)$ for some $a$, so I can't just do a simple comparison.

Finally, using the same harmonic series style argument on $e^{-xy}$ doesn't seem to work, either.

How is this conclusion arrived at?

Edit: Bah, nevermind, I have it.

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Emily
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Consider $$\int_0^\infty \int_0^\infty e^{-xy}\sin x\, dy\, dx = \int_0^\infty \frac{\sin x}{x}\, dx.$$

Since the right-hand side is not integrable, neither is the left-hand side. Apply Tonelli to flavor.

Emily
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