Derivation of Fourier Series over General Range

Theorem
Let $\alpha \in \R$ be a real number.

Let $\lambda \in \R_{>0}$ be a strictly positive real number.

Let $f: \R \to \R$ be a function such that $\displaystyle \int_{\mathop \to \alpha}^{\mathop \to \alpha + 2 \lambda} \map f x \rd x$ converges absolutely.

Let:
 * $\displaystyle f \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$

The Fourier coefficients for $f$ are calculated by:

Proof
By definition of Fourier series over the range of integration $\openint \alpha {\alpha + 2 \pi}$:


 * $(1): \quad \dfrac {a_0} 2 + \displaystyle \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$

where:

Let $\xi = \dfrac {\pi x} \lambda$.

Then let:
 * $\map \phi \xi \equiv \map f x$

We have that $\displaystyle \int_{\mathop \to \alpha}^{\mathop \to \alpha + 2 \lambda} \map f x \rd x$ converges absolutely.

Thus we are ensured of the existence of:
 * the limit from the right of $f$ at $\alpha$
 * the limit from the left of $f$ at $\alpha + 2 \lambda$

Thus $f$ is defined and bounded on $\openint \alpha {\alpha + 2 \lambda}$.

Then:

Setting $\beta = \dfrac {\pi \alpha} \lambda$, this allows us:

Thus we have that $\phi$ is defined and bounded on $\openint \beta {\beta + 2 \pi}$.

Then:
 * $\map \phi \xi \sim \displaystyle \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \xi + b_n \sin \xi}$

where:

We have that:
 * $\dfrac {\d \xi} {\d x} = \dfrac \pi \lambda$

and so:

and so:


 * $\displaystyle \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$