How to proof $\int_{(0,\infty)} \int_{(0,\infty)} |\sin x| \,\, e^{-xy} \,\, dx \,\, dy < \infty$

Question

How can I proof this?

$\int_0^\infty \int_0^\infty |\sin x| \,\, e^{-xy} \,\, dx \,\, dy < \infty$

---

I tried to say $\int_0^\infty \int_0^\infty |\sin x| \,\, e^{-xy} \,\, dx \,\, dy \leq \int_0^\infty \int_0^\infty e^{-xy} \,\, dx \,\, dy$ but that didn't help since the right side is $\infty$. All other ideas failed as well.

---

My "main idea" was to use Tonelli Fubini:

$\int_0^\infty \int_0^\infty |\sin x| \,\, e^{-xy} \,\, dy \,\,dx = \int_0^{\infty} \frac{1}{x} |\sin(x)| \,\,dx$

But I don't know how to continue here now. I tried again to say $\int_0^{\infty} \frac{1}{x} |\sin(x)| \,\,dx \leq \int_0^{\infty} \frac{1}{x} \,\,dx$ but the right side is $\infty$ again.

For Tonelli it would be also enough to check $\int_0^\infty \int_0^\infty |\sin x| \,\, e^{-xy} \,\, dx \,\,dy$ about finity, but here I get no better result.

Answer 1

Note that the function $(x,y)\mapsto\left|\sin x\right|e^{-xy}$ is non-negative and measurable, so Tonelli Theorem is applicable. We have that \begin{eqnarray*} & & \int_{0}^{\infty}\int_{0}^{\infty}\left|\sin x\right|e^{-xy}dydx\\ & = & \int_{0}^{\infty}\left|\sin x\right|\left(-\frac{1}{x}e^{-xy}\mid_{y=0}^{y=\infty}\right)dx\\ & = & \int_{0}^{\infty}\left|\frac{\sin x}{x}\right|dx\\ & = & \infty. \end{eqnarray*} To prove that $\int_{0}^{\infty}\left|\frac{\sin x}{x}\right|dx=\infty$, we note that $|\sin x|\geq\frac{1}{2}$ for $x\in[2n\pi+\frac{\pi}{6},2n\pi+\frac{5\pi}{6}]$. Therefore, \begin{eqnarray*} \int_{0}^{\infty}\left|\frac{\sin x}{x}\right|dx & \geq & \sum_{n=0}^{\infty}\int_{2n\pi+\frac{\pi}{6}}^{2n\pi+\frac{5\pi}{6}}\left|\frac{\sin x}{x}\right|dx\\ & \geq & \sum_{n=0}^{\infty}\frac{1}{2}\cdot\frac{1}{2n\pi+\frac{5\pi}{6}}\cdot(\frac{2\pi}{3})\\ & = & \infty. \end{eqnarray*} It is impossible to prove that $\int_{0}^{\infty}\int_{0}^{\infty}\left|\sin x\right|e^{-xy}dydx<\infty$. Is there any typo ?