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