Fourier's Theorem/Lemma 2

Lemma for Fourier's Theorem
Let $\psi$ be a real function defined on a half-open interval $\left({0 \,.\,.\, a}\right]$.

Let $\psi$ and its derivative $\psi'$ be piecewise continuous with one-sided limits on $\left({0 \,.\,.\, a}\right]$.

Let $\psi \left({u}\right)$ have a right-hand derivative at $u = 0$.

Then:
 * $\displaystyle \lim_{N \mathop \to \infty} \int_0^a \psi \left({u}\right) \frac {\sin N u} u \rd u = \frac \pi 2 \psi \left({0^+}\right)$

where $\psi \left({0^+}\right)$ denotes the limit of $\psi$ at $0$ from the right.

Proof
We have:
 * $\psi \left({u}\right) = \psi \left({0^+}\right) + \left({\psi \left({u}\right) - \psi \left({0^+}\right)}\right)$

from which:
 * $\displaystyle \int_0^a \psi \left({u}\right) \frac {\sin N u} u \rd u = \psi \left({0^+}\right) \int_0^a \frac {\sin N u} u \rd u + \int_0^a \phi \left({u}\right) \sin N u \rd u$

where:
 * $\phi \left({u}\right) = \dfrac {\psi \left({u}\right) - \psi \left({0^+}\right)} u$

Let $\xi = N u$.

Then:

We have that $\psi \left({u}\right)$ is piecewise continuous with one-sided limits on $\left({0 \,.\,.\, a}\right]$.

Hence it follows that $\phi \left({u}\right) = \dfrac {\psi \left({u}\right) - \psi \left({0^+}\right)} u$ is also piecewise continuous with one-sided limits on $\left({0 \,.\,.\, a}\right]$.

We also have that $\psi \left({u}\right)$ has a right-hand derivative at $u = 0$.

It follows that $\phi \left({u}\right)$ is piecewise continuous with one-sided limits on $\left[{0 \,.\,.\, a}\right]$.

Thus from Lemma 1 for Fourier's Theorem:
 * $\displaystyle \lim_{N \mathop \to \infty} \int_0^a \psi \left({u}\right) \frac {\sin N u} u \rd u = \frac \pi 2 \psi \left({0^+}\right)$