Chapter 15, Problem 5PP

The double integral ${\displaystyle {\int }_0^1{\displaystyle {\int }_0^1\frac{1}{1-xy}dxdy}}$is an improper integral and could be defined as the limit of double integrals over the rectangle [0, t] × [0, t] as t → 1⁻. But if we expand the integrand as a geometric series, we can express the integral as the sum of an infinite series. Show that ${\displaystyle {\int }_0^1{\displaystyle {\int }_0^1\frac{1}{1-xy}dxdy}}={\displaystyle \underset{n-1}{\overset{\infty }{\sum }}\frac{1}{n^2}}$