Jordan Theorem for Dirichlet Integrals Examples 1

# Jordan Theorem for Dirichlet Integrals Examples 1

Recall from the Jordan's Theorem for Dirichlet Integrals page that if we consider the Dirichlet integral, $\displaystyle{\int_0^b g(t) \frac{\sin \alpha t}{t} \: dt}$ then if $g$ is of bounded variation on $[0, b]$ for some $b > 0$ then we saw that:

(1)
\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}

We will now look at some examples of applying Jordan's theorem for Dirichlet integrals.

## Example 1

Use Jordan's theorem to evaluate $\displaystyle{\lim_{\alpha \to \infty} \int_0^b (t^3 - 2t^2 + 1) \frac{\sin \alpha t}{t} \: dt}$.

Let $g(t) = t^3 - 2t^2 + 1$. Then $\displaystyle{\lim_{\alpha \to \infty} \int_0^b (t^3 - 2t^2 + 1) \frac{\sin \alpha t}{t} \: dt = \lim_{\alpha \to \infty} g(t) \frac{\sin \alpha t}{t} \: dt}$.

Note that $g$ is of bounded variation for all $b > 0$ since $g$ is a polynomial and every polynomial is of bounded variation on any closed and bounded interval $[0, b]$. So, by Jordan's test for Dirichlet integrals we have that:

(2)
\begin{align} \quad \lim_{\alpha \to \infty} \int_0^b (t^3 - 2t^2 + 1) \frac{\sin \alpha t}{t} \: dt = \frac{\pi}{2} g(0+) \end{align}

We have that:

(3)
\begin{align} \quad g(0+) = \lim_{t \to 0+} g(t) = \lim_{t \to 0+} (t^3 - 2t^2 + 1) = 1 \end{align}

Hence:

(4)
\begin{align} \quad \lim_{\alpha \to \infty} \int_0^b (t^3 - 2t^2 + 1) \frac{\sin \alpha t}{t} \: dt = \frac{\pi}{2} \end{align}

## Example 2

Use Jordan's theorem to evaluate $\displaystyle{\lim_{\alpha \to \infty} \int_0^2 (\mid t \mid + 3\cos^2 t) \frac{\sin \alpha t}{t} \: dt}$.

Let $g(t) = \mid t \mid + 3 \cos^2 t$. Then $g$ is clearly of bounded variation on $[0, 2]$. We have that:

(5)
\begin{align} \quad g(0+) = \lim_{t \to 0+} \mid t \mid + 3 \cos^2 t = 3 \end{align}

Therefore by Jordan's theorem for Dirichlet integrals we have that:

(6)
\begin{align} \quad \lim_{\alpha \to \infty} \int_0^2 (\mid t \mid + 3\cos^2 t) \frac{\sin \alpha t}{t} \: dt = \frac{6}{\pi} \end{align}