Fourier's Theorem/Lemma 2/Mistake

Source Work

  • 1961: I.N. Sneddon: Fourier Series: 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:

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


Sources

  • 1961: I.N. Sneddon: Fourier Series ... (previous) ... (next): Chapter Two: $\S 2$. Some Important Limits
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