Theorem: Derivation of Fourier Series over General Range

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 $\ds \int_{\mathop \to \alpha}^{\mathop \to \alpha + 2 \lambda} \map f x \rd x$ converges absolutely.
Let:
:$\ds f \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
The '''Fourier coefficients''' for $f$ are calculated by:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \dfrac 1 \lambda \int_{\mathop \to \alpha}^{\mathop \to \alpha + 2 \lambda} \map f x \cos \frac {n \pi x} \lambda \rd x
}}
{{eqn | l = b_n
      | r = \dfrac 1 \lambda \int_{\mathop \to \alpha}^{\mathop \to \alpha + 2 \lambda} \map f x \sin \frac {n \pi x} \lambda \rd x
}}
{{end-eqn}}

Proof:
By definition of Fourier series over the range of integration $\openint \alpha {\alpha + 2 \pi}$:
:$(1): \quad \dfrac {a_0} 2 + \ds \sum_{n \mathop = 1}^\infty \paren {a_n \cos n x + b_n \sin n x}$
where:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} \map f x \cos n x \rd x
}}
{{eqn | l = b_n
      | r = \dfrac 1 \pi \int_\alpha^{\alpha + 2 \pi} \map f x \sin n x \rd x
}}
{{end-eqn}}
Let $\xi = \dfrac {\pi x} \lambda$.
Then let:
:$\map \phi \xi \equiv \map f x$
We have that $\ds \int_{\mathop \to \alpha}^{\mathop \to \alpha + 2 \lambda} \map f x \rd x$ 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:
{{begin-eqn}}
{{eqn | l = \map f {\alpha^+}
      | r = \map \phi {\paren {\frac {\pi \alpha} \lambda}^+}
      | c = {{Defof|Limit from Above}}
}}
{{eqn | l = \map f {\paren {\alpha + 2 \lambda}^-}
      | r = \map \phi {\paren {\dfrac {\pi \alpha + \pi 2 \lambda} \lambda}^-}
      | c = 
}}
{{eqn | r = \map \phi {\paren {\frac {\pi \alpha} \lambda + 2 \pi}^-}
      | c = 
}}
{{end-eqn}}
Setting $\beta = \dfrac {\pi \alpha} \lambda$, this allows us:
{{begin-eqn}}
{{eqn | l = \map f {\alpha^+}
      | r = \map \phi {\beta^+}
      | c = 
}}
{{eqn | l = \map f {\paren {\alpha + 2 \lambda}^-}
      | r = \map \phi {\paren {\beta + 2 \pi}^-}
      | c = 
}}
{{end-eqn}}
Thus we have that $\phi$ is defined and bounded on $\openint \beta {\beta + 2 \pi}$.
Then:
:$\map \phi \xi \sim \ds \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \xi + b_n \sin \xi}$
where:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \dfrac 1 \pi \int_{\mathop \to \pi}^{\mathop \to \alpha + 2 \lambda} \map \phi \xi \cos \xi \rd \xi
}}
{{eqn | l = b_n
      | r = \dfrac 1 \pi \int_{\mathop \to \pi}^{\mathop \to \alpha + 2 \lambda} \map \phi \xi \sin \xi \rd \xi
}}
{{end-eqn}}
We have that:
:$\dfrac {\d \xi} {\d x} = \dfrac \pi \lambda$
and so:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \dfrac 1 \lambda \int_{\mathop \to \beta}^{\mathop \to \beta + 2 \lambda} \map f x \cos \frac {n \pi x} \lambda \rd x
}}
{{eqn | l = b_n
      | r = \dfrac 1 \lambda \int_{\mathop \to \beta}^{\mathop \to \beta + 2 \lambda} \map f x \sin \frac {n \pi x} \lambda \rd x
}}
{{end-eqn}}
and so:
:$\ds \map f x \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty \paren {a_n \cos \frac {n \pi x} \lambda + b_n \sin \frac {n \pi x} \lambda}$
{{qed}}