# Fourier's Theorem/Lemma 3

## Lemma for Fourier's Theorem

Let $\psi$ be a real function defined on an open interval $\openint a b$.

Let $\psi$ and its derivative $\psi'$ be piecewise continuous with one-sided limits on $\openint a b$.

Let $\map \psi u$ have both right-hand derivative and left-hand derivative at a point $u = x$ where $x \in \openint a b$.

Then:

$\displaystyle \lim_{N \mathop \to \infty} \int_a^b \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u = \frac \pi 2 \paren {\map \psi {x^+} + \map \psi {x^-} }$

where:

$\map \psi {x^+}$ denotes the limit of $\psi$ at $x$ from the right
$\map \psi {x^-}$ denotes the limit of $\psi$ at $x$ from the left.

## Proof

$\displaystyle \int_a^b \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u = \int_a^x \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u + \int_x^b \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u$

Let $u = x - \xi$.

Then by Integration by Substitution:

$\displaystyle \int_a^x \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u = \int_0^{x - a} \map \phi \xi \frac {\sin N \xi} \xi \rd \xi$

where:

$\map \phi \xi = \map \psi {u - \xi}$

 $\ds \lim_{N \mathop \to \infty} \int_0^{x - a} \map \phi \xi \frac {\sin N \xi} \xi \rd \xi$ $=$ $\ds \frac \pi 2 \map \phi {0^+}$ $\ds$ $=$ $\ds \frac \pi 2 \map \psi {x^-}$

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

$\displaystyle \int_x^b \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u = \int_0^{b - x} \map \chi \eta \frac {\sin N \eta} \eta \rd \eta$

where:

$\map \chi \xi = \map \psi {x + \eta}$
 $\ds \lim_{N \mathop \to \infty} \int_0^{b - x} \map \chi \eta \frac {\sin N \eta} \eta \rd \eta$ $=$ $\ds \frac \pi 2 \map \phi {0^+}$ $\ds$ $=$ $\ds \frac \pi 2 \map \psi {x^+}$

The result follows by adding the two limits.

$\blacksquare$