Definite Integral of Fourier Series at Ends of Interval
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Theorem
Let $f: \R \to \R$ be a real function defined in the open interval $\openint {-\pi} \pi$.
Let $f$ fulfil the Dirichlet conditions in $\openint {-\pi} \pi$.
Let $a_0, a_1, \dotsc; b_1, \dotsc$ be the Fourier coefficients of $f$ in $\openint {-\pi} \pi$.
Consider the real function:
- $\map F x = \ds \int_{-\pi}^x \map f t \rd t - \dfrac {a_0} 2 x$
Then:
- $\map F \pi = \map F {-\pi} = \dfrac {a_0 \pi} 2$
Proof
From Definite Integral on Zero Interval:
\(\ds \map F {-\pi}\) | \(=\) | \(\ds \int_{-\pi}^{-\pi} \map f t \rd t - \dfrac {a_0} 2 \paren {-\pi}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0 - \dfrac {a_0} 2 \paren {-\pi}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {a_0 \pi} 2\) |
Then:
\(\ds \map F \pi\) | \(=\) | \(\ds \int_{-\pi}^\pi \map f t \rd t - \dfrac {a_0} 2 \pi\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_{-\pi}^\pi \paren {\dfrac {\map {f^+} t + \map {f^-} t} 2} \rd t - \dfrac {a_0} 2 \pi\) | Fourier's Theorem |
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Sources
- 1961: I.N. Sneddon: Fourier Series ... (previous) ... (next): Chapter Three: Properties of Fourier Series: $1$. Integration of Fourier Series