real analysis - Does $ int_0^{infty}frac{sin x}{x}dx $ have an improper Riemann integral or a Lebesgue integral?

In this wikipedia article for improper integrals,
$$
\int_0^{\infty}\frac{\sin x}{x}dx
$$
is given as an example for the integrals that have an improper Riemann integral but do not have a (proper) Lebesgue integral. Here are my questions:

• Why does this one have an improper Riemann integral? (I don't see why $\int_0^a\frac{\sin x}{x}dx$ and $\int_a^{\infty}\frac{\sin x}{x}dx$ converge.)


• Why doesn't this integral have a Lebesgue integral? Is it because that $\frac{\sin x}{x}$ is unbounded on $(0,\infty)$ and Lebesgue integral doesn't deal with unbounded functions?

Answer

$\displaystyle \int_0^a\frac{\sin x}xdx$ converges since we can extend the function $x\mapsto \frac{\sin x}x$ by continuity at $0$ (we give the value $1$ at $0$). To see that the second integral converges, integrate by parts $\displaystyle\int_a^A\frac{\sin x}x dx$. Indeed, we get
$$\int_a^A\frac{\sin x}xdx =\left[-\frac{\cos x}x\right]_a^A+\int_a^A-\frac{\cos x}{x^2}dx = \frac{\cos a}a-\frac{\cos A}A-\int_a^A\frac{\cos x}{x^2}dx,$$
and $\displaystyle\lim_{A\to +\infty}\frac{\cos A}A=0$, and the fact that $\displaystyle\int_a^{+\infty}\frac{dx}{x^2}$ is convergent gives use the convergence of $\displaystyle\int_a^{+\infty}\frac{\sin x}xdx$
. But $f(x):=\frac{\sin x}x$ has not a Lebesgue integral, since the integral $\displaystyle\int_0^{\infty}\left|\frac{\sin x}x\right| dx$ is not convergent (but it's not a consequence of the fact that $f$ is not bounded, first because $f$ is bounded, and more generally consider $g(x)=\frac 1{\sqrt x}$ for $0 and $g(x)=0$ for $x>1$). To see that the integral is not convergent, note that for $N\in\mathbb N$
\begin{align*}
\int_{\pi}^{(N+1)\pi}\left|\frac{\sin x}x\right|dx&=\sum_{k=1}^N\int_{k\pi}^{(k+1)\pi}\left|\frac{\sin x}x\right|dx\\\
&=\sum_{k=1}^N\int_0^{\pi}\frac{|\sin(t+k\pi)|}{t+k\pi}dt\\\
&=\sum_{k=1}^N\int_0^{\pi}\frac{|\sin t|}{t+k\pi}dt\\\

&\geq \sum_{k=1}^N\frac 1{(k+1)\pi}\int_0^{\pi}\sin t\,dt\\\
&=\frac 2{\pi}\sum_{k=1}^N\frac 1{k+1},
\end{align*}
and we can conclude since the harmonic series is not convergent.