# Fourier's Theorem/Lemma 2

## Lemma for Fourier's Theorem

Let $\psi$ be a real function defined on a half-open interval $\hointl 0 a$.

Let $\psi$ and its derivative $\psi'$ be piecewise continuous with one-sided limits on $\hointl 0 a$.

Let $\map \psi u$ have a right-hand derivative at $u = 0$.

Then:

$\displaystyle \lim_{N \mathop \to \infty} \int_0^a \map \psi u \frac {\sin N u} u \rd u = \frac \pi 2 \map \psi {0^+}$

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

## Proof

We have:

$\map \psi u = \map \psi {0^+} + \paren {\map \psi u - \map \psi {0^+} }$

from which:

$(1): \quad \displaystyle \int_0^a \map \psi u \frac {\sin N u} u \rd u = \map \psi {0^+} \int_0^a \frac {\sin N u} u \rd u + \int_0^a \map \phi u \sin N u \rd u$

where:

$\map \phi u = \dfrac {\map \psi u - \map \psi {0^+} } u$

Let $\xi = N u$.

Then:

 $\displaystyle \int_0^a \frac {\sin N u} u \rd u$ $=$ $\displaystyle \int_0^{N a} \frac {\sin \xi} \xi \rd \xi$ $\displaystyle$ $\to$ $\displaystyle \int_0^\infty \frac {\sin \xi} \xi \rd \xi$ $\displaystyle$ $=$ $\displaystyle \frac \pi 2$ Integral to Infinity of $\dfrac {\sin \xi} \xi$

We have that $\map \psi u$ is piecewise continuous with one-sided limits on $\hointl 0 a$.

Hence it follows that $\map \phi u = \dfrac {\map \psi u - \map \psi {0^+} } u$ is also piecewise continuous with one-sided limits on $\hointl 0 a$.

We also have that $\map \psi u$ has a right-hand derivative at $u = 0$.

It follows that $\map \phi u$ is piecewise continuous with one-sided limits on $\hointl 0 a$.

Thus from Lemma 1 for Fourier's Theorem:

$\displaystyle \lim_{N \mathop \to \infty} \int_0^a \map \phi u \sin N u \rd u = 0$

and letting $N \to \infty$ in $(1)$ above:

$\displaystyle \lim_{N \mathop \to \infty} \int_0^a \map \psi u \frac {\sin N u} u \rd u = \frac \pi 2 \map \psi {0^+}$

$\blacksquare$