Starting with the symmetrized integrals as in Problem 23, make the substitutions α=2πp/h (where p is the new variable, h is a constant), f(x)=ψ(x), g(α)=h/2πϕ(p); show that then
ψ(x)ϕ(p)∫−∞∞∣ψ(x)∣2dx=h1∫−∞∞ϕ(p)e2πipx/hdp,=h1∫−∞∞ψ(x)e−2πipx/hdx,=∫−∞∞∣ϕ(p)∣2dp.
This notation is often used in quantum mechanics.