Chapter 7.12, Problem 35P

Starting with the symmetrized integrals as in Problem 34, make the substitutions $\alpha =2\pi p/h$ (where $p$ is the new variable, $h$ is a constant), $f(x)=\psi (x),g(\alpha )=$ $\sqrt{h/2\pi }\varphi (p);$ show that then $\psi (x)=\frac{1}{\sqrt{h}}{\displaystyle {\int }_{-\infty }^{\infty }\varphi }(p)e^{2\pi ipx/h}dp$, $\varphi (p)=\frac{1}{\sqrt{h}}{\displaystyle {\int }_{-\infty }^{\infty }\psi }(x)e^{-2\pi ipx/h}dx$, ${\displaystyle {\int }_{-\infty }^{\infty }|}\psi (x){|}^{2}dx={\displaystyle {\int }_{-\infty }^{\infty }|}\varphi (p){|}^{2}dp$.

This notation is often used in quantum mechanics.