Fourier's Theorem/Lemma 2/Mistake
< Fourier's Theorem | Lemma 2
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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