Chapter 15: Multiple Integrals - Section 15.2 - Double Integrals over General Regions - Exercises 15.2 - Page 883: 71

$\pi^2$

Consider $I= \int_{-\infty}^{\infty} \dfrac{dx}{x^2+1} \int_{-\infty}^{\infty} \dfrac{dy}{y^2+1}$ or, $=[ \arctan x]_{-\infty}^{\infty} [ \arctan y]_{-\infty}^{\infty}$ or, $=[\arctan x (-\infty) -\arctan x (\infty)] [\arctan y (-\infty) -\arctan y (\infty)] $ Thus, we have $I=\pi \cdot \pi =\pi^2$