Show not Lebesgue integrable using step functions I have shown that conditionally $\int_0^\infty \int_0^\infty f(x,y) dx dy = \int_0^\infty \int_0^\infty f(x,y) dx dy = \frac{\pi}{2}$ where $f(x,y) = e^{-xy} \sin(x).$ This part is relatively easy because 
$$\int_0^\infty e^{-xy} \sin(x) dy = \frac{\sin(x)}{x} $$
Now I want to show that $f \not\in L^1(\mathbb{R}^{+2})$.  I know we can show $f$ is not Lebesgue integrable by finding a sequence of simple / step functions $\phi_n \leq |f|$ where $\int_{\mathbb{R}^{+2}} \phi_n \to \infty$, but I need some help here.

$\int |f(x,y)| dy=|\frac {\sin x} x|$ so $\int \int |f(x,y)| dy dx=\int |\frac {\sin x} x| dx=\infty$. By Tonelli's Theorem $f$ is not integrable. 
