# Does ∫∞0sinxxdx \int_0^{\infty}\frac{\sin x}{x}dx have an improper Riemann integral or a Lebesgue integral?

In this wikipedia article for improper integrals,

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?

$$∫a0sinxxdx\displaystyle \int_0^a\frac{\sin x}xdx$$ converges since we can extend the function $$x↦sinxxx\mapsto \frac{\sin x}x$$ by continuity at $$00$$ (we give the value $$11$$ at $$00$$). To see that the second integral converges, integrate by parts $$∫Aasinxxdx\displaystyle\int_a^A\frac{\sin x}x dx$$. Indeed, we get
$$∫Aasinxxdx=[−cosxx]Aa+∫Aa−cosxx2dx=cosaa−cosAA−∫Aacosxx2dx,\int_a^A\frac{\sin x}xdx =\left[-\frac{\cos x}x\right]_a^A+\int_a^A-\frac{\cos x}{x^2}dx = \frac{\cos a}a-\frac{\cos A}A-\int_a^A\frac{\cos x}{x^2}dx,$$
and $$lim\displaystyle\lim_{A\to +\infty}\frac{\cos A}A=0$$, and the fact that $$\displaystyle\int_a^{+\infty}\frac{dx}{x^2}\displaystyle\int_a^{+\infty}\frac{dx}{x^2}$$ is convergent gives use the convergence of $$\displaystyle\int_a^{+\infty}\frac{\sin x}xdx\displaystyle\int_a^{+\infty}\frac{\sin x}xdx$$
. But $$f(x):=\frac{\sin x}xf(x):=\frac{\sin x}x$$ has not a Lebesgue integral, since the integral $$\displaystyle\int_0^{\infty}\left|\frac{\sin x}x\right| dx\displaystyle\int_0^{\infty}\left|\frac{\sin x}x\right| dx$$ is not convergent (but it’s not a consequence of the fact that $$ff$$ is not bounded, first because $$ff$$ is bounded, and more generally consider $$g(x)=\frac 1{\sqrt x}g(x)=\frac 1{\sqrt x}$$ for $$0 and $$g(x)=0g(x)=0$$ for $$x>1x>1$$). To see that the integral is not convergent, note that for $$N\in\mathbb NN\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*}\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*}