I'm trying to prove the Riemann integral $$\int_0 ^ \infty \frac{\sin(x)} x \ dx = \frac{\pi}{2}$$ using tools that I, as a statistician, am likely not to forget. My general approach is
- Show that the Laplace transform $\psi(t) := \int_0 ^ \infty \frac{e^{-xt} \sin(x)} x \ dx$ is $-\arctan(t) + \frac \pi 2$ for $t \in (0, \infty)$ by differentiating under the integral (using dominated convergence stuff since the Lebesgue integral exists here) and solving the resulting differential equation.
- Argue that $\psi(t)$ is continuous at $0$.
I've hit a snag at step 2. So, under what conditions do we have that the Laplace transform of a function is continuous at $t = 0$? Any trivial garuntees that it should be for $\frac {\sin x} x$ in particular? I feel like this would be easier if the function was Lebesgue integrable since when I'm trying to bound $|\psi(0) – \psi(t)|$ I am having a hard time controlling the tail of the integral of $\psi(t)$ for $t$ uniformly in $t$ near $0$.
I'm also finding this somewhat interesting since it looks like $\frac{\sin x} x$ is sitting on a boundary of sorts: we are Lebesgue integrable up to a point, then Riemann integrable, and past that not integrable($t > 0, = 0, < 0$) So a bunch of Lebesgue integrable functions are converging to a function which has an improper Riemann integral but no Lebesgue integral. And yet interchanging limits produces the correct answer anyways when calculating the integral.
EDIT: We already have an answer below that suggests how to do this for $\frac {\sin x} x$ but I'd like to get something on the more general question of under what conditions we have right continuity of the Laplace transform at $0$. Is it always continuous provided that the function has a finite improper Riemann integral?
Best Answer
You can get right continuity at $s=0$ along the real axis, at least: We have $$I(s,n):=\int_{n\pi}^\infty \frac{e^{-st}\sin t}{t}\,dt=\sum_{k=n}^\infty\int_{k\pi}^{(k+1)\pi} \frac{e^{-st}\sin t}{t}\,dt$$ and use the alternating nature of the series to show that the entire sum lies between $0$ and its first term. Thus $I(s,n)\to0$ uniformly in $s\ge0$ as $n\to\infty$. On the other hand, you have unifom convergence of the integrand for $t\in[0,n\pi]$, so you don't even need to invoke Lebesgue theory to get the needed convergence.