Fourier's Theorem/Lemma 2/Mistake

Source Work

 * : Chapter Two: $\S 2$. Some Important Limits

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:


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