Answer
$$\int_{1}^{\infty} \int_{0}^{1 / x} y \ \ d y \ \ d x=\frac{1}{2}$$
Work Step by Step
Given $$\int_{1}^{\infty} \int_{0}^{1 / x} y \ \ d y \ \ d x$$
So, we get
\begin{align}
\int_{1}^{\infty} \int_{0}^{1 / x} y \ \ d y \ \ d x&=\int_{1}^{\infty}\left[\frac{y^{2}}{2}\right]_{0}^{1 / x} d x\\
&=\frac{1}{2} \int_{1}^{\infty}\left( (\frac{1}{x})^{2}-0 \right)d x\\
&=\frac{1}{2} \int_{1}^{\infty} \frac{1}{x^{2}} d x\\
&=\frac{1}{2}\left[\frac{x^{-2+1}}{-1}\right]_{1}^{\infty}\\
&=\left[-\frac{1}{2 x}\right]_{1}^{\infty}\\
&=0+\frac{1}{2} \\
&=\frac{1}{2}
\end{align}
