Dini's Theorem for Dirichlet Integrals

# Dini's Theorem for Dirichlet Integrals

Recall from the Dirichlet Integrals page that if $g$ is a function defined on $[0, b]$, $b > 0$, and $\alpha \in \mathbb{R}$ then a Dirichlet integral is of the following form:

(1)
\begin{align} \quad \int_0^b g(t) \frac{\sin \alpha t}{t} \: dt \end{align}

We will now investigate two theorems known as Jordan's Theorem for Dirichlet Integrals and Dini's Theorem (below) which give sufficient conditions for the following equality to hold (provided $g(0+)$ exists of course):

(2)
\begin{align} \quad \lim_{\alpha \to \infty} \frac{2}{\pi} \int_0^b g(t) \frac{\sin \alpha t}{t} \: dt = g(0+) \end{align}
 Theorem 1 (Dini's Theorem): Let $g$ be a function such that $g(0+) = \lim_{t \to 0+} g(t)$ exists, and suppose that there exists a $b > 0$ such that $\displaystyle{\int_0^b \frac{g(t) - g(0+)}{t} \: dt}$ exists as a Lebesgue integral. Then $\displaystyle{\lim_{\alpha \to \infty} \frac{2}{\pi} \int_0^b g(t) \frac{\sin \alpha t}{t} \: dt = g(0+)}$.
• Proof: Let $\displaystyle{g(0+) = \lim_{t \to 0+} g(t)}$ exist and assume that for some $b > 0$ we have that the following exists as a Lebesgue integral:
(3)
• Then:
(4)
\begin{align} \quad \lim_{\alpha \to \infty} \frac{2}{\pi} \int_0^b g(t) \frac{\sin \alpha t}{t} \: dt &= \lim_{\alpha \to \infty} \frac{2}{\pi} \underbrace{\int_0^b [g(t) - g(0+)] \frac{\sin \alpha t}{t} \: dt}_{(*)} + \underbrace{\lim_{\alpha \to \infty} \frac{2}{\pi} \int_0^b g(0+) \frac{\sin \alpha t}{t} \: dt}_{(**)} \quad (\dagger \dagger) \end{align}
• Notice that $(\dagger)$ holds we have that $\displaystyle{\frac{g(t) - g(0+)}{t}}$ is Lebesgue integrable on $[0, b]$. So by The Riemann-Lebesgue Lemma we have that:
(5)
\begin{align} \quad (*) : \quad \lim_{\alpha \to \infty} \int_0^b \frac{g(t) - g(0+)}{t} \sin \alpha t \: dt = 0 \end{align}
• Furthermore, the integral at $(**)$ can be evaluated as follows:
(6)
\begin{align} \quad \lim_{\alpha \to \infty} \int_0^b g(0+) \frac{\sin \alpha t}{t} \: dt &= \lim_{\alpha \to \infty} g(0+) \int_0^b \frac{\sin \alpha t}{t} \: dt \\ &= g(0+) \lim_{\alpha \to \infty} \int_0^{\alpha b} \frac{\sin t}{t} \: dt \\ &= g(0+) \int_0^{\infty} \frac{\sin t}{t} \: dt \\ &= g(0+) \cdot \frac{\pi}{2} \\ \end{align}
• Substituting $(*)$ and $(**)$ into $(\dagger \dagger)$ and we get:
(7)
\begin{align} \quad \lim_{\alpha \to \infty} \frac{2}{\pi} \int_0^b g(t) \frac{\sin \alpha t}{t} \: dt &= \frac{2}{\pi} \cdot g(0+) \cdot \frac{\pi}{2} = g(0+) \quad \blacksquare \end{align}