## Calculus (3rd Edition)

The integral $\int_{0}^{\infty} (x) p(x) \ dx$ diverges because the mean value is infinite.
We have: $p(x)=\dfrac{2}{\pi(x^2+1)}$ The mean value of $p(x)$ (probability density function) on $[0, \infty)$ is given by: $\mu=\int_{0}^{\infty} (x) p(x) \ dx \\=\int_{0}^{\infty} \dfrac{2x}{\pi(x^2+1)}\\=\dfrac{1}{\pi} \int_0^{\infty} \dfrac{2x}{(x^2+1)} dx$ Suppose that $a=x^2+1 \implies da=2x dx$ Therefore, we have: $\mu=\dfrac{1}{\pi} \int_0^{\infty} \dfrac{2x}{(x^2+1)} dx\\=\dfrac{1}{\pi} \int_1^{\infty} \dfrac{da}{a}\\=\dfrac{1}{\pi} [\ln a]_1^{\infty}\\=\infty$ So, the integral $\int_{0}^{\infty} (x) p(x) \ dx$ diverges because the mean value is infinite.