Fourier's Theorem/Lemma 3

Lemma for Fourier's Theorem
Let $\psi$ be a real function defined on an open interval $\left({a \,.\,.\, b}\right)$.

Let $\psi$ and its derivative $\psi'$ be piecewise continuous with one-sided limits on $\left({a \,.\,.\, b}\right)$.

Let $\psi \left({u}\right)$ have both right-hand derivative and left-hand derivative at a point $u = x$ where $x \in \left({a \,.\,.\, b}\right)$.

Then:
 * $\displaystyle \lim_{N \mathop \to \infty} \int_a^b \psi \left({u}\right) \frac {\sin N \left({u - x}\right)} {u - x} \rd u = \frac \pi 2 \left({\psi \left({x^+}\right) + \psi \left({x^-}\right)}\right)$

where:
 * $\psi \left({x^+}\right)$ denotes the limit of $\psi$ at $x$ from the right
 * $\psi \left({x^-}\right)$ denotes the limit of $\psi$ at $x$ from the left.

Proof
From Sum of Integrals on Adjacent Intervals for Integrable Functions, we have:
 * $\displaystyle \int_a^b \psi \left({u}\right) \frac {\sin N \left({u - x}\right)} {u - x} \rd u = \int_a^x \psi \left({u}\right) \frac {\sin N \left({u - x}\right)} {u - x} \rd u + \int_x^b \psi \left({u}\right) \frac {\sin N \left({u - x}\right)} {u - x} \rd u$

Let $u = x - \xi$.

Then by Integration by Substitution:
 * $\displaystyle \int_a^x \psi \left({u}\right) \frac {\sin N \left({u - x}\right)} {u - x} \rd u = \int_0^{x - a} \phi \left({\xi}\right) \frac {\sin N \xi} \xi \rd \xi$

where:
 * $\phi \left({\xi}\right) = \psi \left({u - \xi}\right)$

By Fourier's Theorem: Lemma 2:

Similarly, substituting $u = x + \eta$:


 * $\displaystyle \int_x^b \psi \left({u}\right) \frac {\sin N \left({u - x}\right)} {u - x} \rd u = \int_0^{b - x} \chi \left({\eta}\right) \frac {\sin N \eta} \eta \rd \eta$

where:
 * $\chi \left({\xi}\right) = \psi \left({x + \eta}\right)$

By Fourier's Theorem: Lemma 2:

The result follows by adding the two limits.