Chapter 7.12, Problem 34P

Show that if (12.2) is written with the factor $1/\sqrt{2\pi }$ multiplying each integral, then the corresponding form of Parseval’s theorem ( 12.24 ) is ${\displaystyle {\int }_{-\infty }^{\infty }|}f(x){|}^{2}dx={\displaystyle {\int }_{-\infty }^{\infty }|}g(\alpha ){|}^{2}d\alpha$.