Fourier's Theorem/Lemma 2/Mistake

Source Work

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

Mistake

 * we find that:
 * $\displaystyle \int_0^a \psi \left({u}\right) \frac {\sin N u} u \rd u = \psi \left({0^+}\right) \int_0^a \frac {\sin N u} u + \int_0^a \phi \left({u}\right) \sin N u \rd u$
 * where:
 * $\phi \left({u}\right) = \dfrac {\psi \left({u}\right) - \psi \left({0^+}\right)} u$.

The variable of integration is missing from the middle integral.

It should be:


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