Dirichlet Integrals

We will now begin to discuss a very special type of integral known as a Dirichlet integral. We will see their importance later on. We first define this type of integral below.

Definition: A Dirichlet Integral is an integral of the form $\displaystyle{\int_0^b g(t) \frac{\sin \alpha t}{t} \: dt}$ where $g$ is a function defined on $[0, b]$.

Assume that the righthand limit $\displaystyle{g(0+) = \lim_{t \to 0+} g(t)}$ exists and consider the following limit:

(1)

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

Suppose that $g$ is constant on $[0, b]$. Then $g(0+) = \lim_{t \to 0+} g(t) = g(t)$ for all $t \in [0, b]$. Therefore we have that:

(2)

\begin{align} \quad \lim_{\alpha \to \infty} \int_0^b g(t) \frac{\sin \alpha t}{t} \: dt &= \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+) \frac{\pi}{2} \end{align}

So if $g$ is constant on $[0, b]$ we have the following formula:

(3)

\begin{align} \quad \lim_{\alpha \to \infty} \frac{2}{\pi} \int_0^b g(t) \frac{\sin \alpha t}{t} \: dt = g(0+) \quad (*) \end{align}

Now suppose instead that $g \in L([0, b])$ is not a constant function on $[0, b]$. Then for $h$ such that $0 < h < b$ we have that:

(4)

\begin{align} \quad \lim_{\alpha \to \infty} \int_0^b g(t) \frac{\sin \alpha t}{t} \: dt &= \lim_{\alpha \to \infty} \int_0^h g(t) \frac{\sin \alpha t}{t} \: dt + \lim_{\alpha \to \infty} \int_h^b g(t) \frac{\sin \alpha t}{t} \: dt \end{align}

Since $g \in L([0, b])$ we have that $g \in L([h, b])$ and since $\displaystyle{\frac{1}{t} \in L([h, b])}$ since $\displaystyle{\frac{1}{t}}$ is continuous on the closed and bounded interval $[h, b]$ and $h > 0$ we see that $\displaystyle{\frac{g(t)}{t} \in L([h, b])}$ and so by The Riemann-Lebesgue Lemma we see that:

(5)

\begin{align} \quad \lim_{\alpha \to \infty} \int_h^b g(t) \frac{\sin \alpha t}{t} \: dt =0 \end{align}

But this holds true for all $h$ with $0 < h < b$ which suggests that the validity of $(*)$ for general functions $g$ depends on $g$ at $0$ from the right. We will look at two important theorems that gives us sufficient conditions for when $(*)$ holds. They can be found on the Jordan's Theorem for Dirichlet Integrals and Dini's Theorem for Dirichlet Integrals pages.