Chapter 15.3, Problem 40E

(a) We define the improper integral (over the entire plane ℝ²)

$\begin{array}{l}I={\displaystyle \underset{R^2}{\iint }{e}^{-(x^2+y^2)}dA}\\ ={\displaystyle {\int }_{-\infty }^{\infty }{\displaystyle {\int }_{-\infty }^{\infty }{e}^{-(x^2+y^2)}dydx}}\\ =\underset{a\to \infty }{\lim }{\displaystyle \underset{Da}{\iint }{e}^{-(x^2+y^2)}dA}\end{array}$

where D_a is the disk with radius a and center the origin. Show that

${\displaystyle {\int }_{-\infty }^{\infty }{\displaystyle {\int }_{-\infty }^{\infty }{e}^{-(x^2+y^2)}dA=\pi }}$

(b) An equivalent definition of the improper integral in part (a) is

${\displaystyle \underset{R^2}{\iint }{e}^{-(x^2+y^2)}dA=\underset{a\to \infty }{\lim }{\displaystyle \underset{Sa}{\iint }{e}^{-(x^2+y^2)}}}dA$

where S_a is the square with vertices (±a, ±a). Use this to show that

${\displaystyle {\int }_{-\infty }^{\infty }{e}^{-x^2}dx{\displaystyle {\int }_{-\infty }^{\infty }{e}^{-y^2}dy=\pi }}$

(c) Deduce that

${\displaystyle {\int }_{-\infty }^{\infty }{e}^{-x^2}dx=\sqrt{\pi }}$

(d) By making the change of variable $t=\sqrt{2}x,$ show that

${\displaystyle {\int }_{-\infty }^{\infty }{e}^{-x^2/2}dx=\sqrt{2\pi }}$

(This is a fundamental result for probability and statistics.)