# Fourier's Theorem/Lemma 2/Mistake

## Mistake

we find that:
$\ds \int_0^a \map \psi u \frac {\sin N u} u \rd u = \map \psi {0^+} \int_0^a \frac {\sin N u} u + \int_0^a \map \phi u \sin N u \rd u$
where:
$\map \phi u = \dfrac {\map \psi u - \map \psi {0^+} } u$.

The variable of integration is missing from the middle integral.

It should be:

$\ds \int_0^a \map \psi u \frac {\sin N u} u \rd u = \map \psi {0^+} \int_0^a \frac {\sin N u} u \rd u + \int_0^a \map \phi u \sin N u \rd u$