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?

$\displaystyle \int_0^a\frac{\sin x}xdx$ 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 $0<x\leq 1$ 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.

This answer is a modified version of an answer to a closed question.

The integral is not absolutely convergent. Because $$ \int_{k\pi}^{(k+1)\pi}|\sin(t)|\;\mathrm{d}t=2\tag{1} $$ we have $$ \frac{2}{(k+1)\pi}\le\int_{k\pi}^{(k+1)\pi}\left|\frac{\sin(t)}{t}\right|\;\mathrm{d}t\le\frac{2}{k\pi}\tag{2} $$ Since the harmonic series diverges, so does the integral of the absolute value. Therefore, the Lebesgue integral does not exist.

However, the improper Riemann integral $$ \int_0^\infty\frac{\sin(t)}{t}\mathrm{d}t\tag{3} $$ does exist. To see this, note that $$ \int_{2k\pi}^{2(k+1)\pi}\sin(t)\;\mathrm{d}t=0\tag{4} $$ With $a=\frac12\left(\frac{1}{2k\pi}+\frac{1}{2(k+1)\pi}\right)$, and using $(1)$ and $(4)$, we get $$ \begin{align} \left|\int_{2k\pi}^{2(k+1)\pi}\frac{\sin(t)}{t}\mathrm{d}t\right| &=\left|\int_{2k\pi}^{2(k+1)\pi}\sin(t)\;\left(\frac1t-a\right)\;\mathrm{d}t\right|\\ &\le\int_{2k\pi}^{2(k+1)\pi}\left|\sin(t)\right|\;\mathrm{d}t\;\max_{[2k\pi,2(k+1)\pi]}\left(\frac1t-a\right)\\ &=4\cdot\frac12\left(\frac{1}{2k\pi}-\frac{1}{2(k+1)\pi}\right)\\ &=\frac{1}{k(k+1)\pi}\tag{5} \end{align} $$ Furthermore, we have the telescoping series $$ \begin{align} \sum_{k=1}^\infty\frac{1}{k(k+1)} &=\sum_{k=1}^\infty\frac{1}{k}-\frac{1}{k+1}\\ &=1\tag{6} \end{align} $$ Thus, $(2)$, $(5)$, and $(6)$ guarantee that $$ \int_{2\pi}^\infty\frac{\sin(t)}{t}\mathrm{d}t\tag{7} $$ converges to a value no greater than $\dfrac1\pi$.

Since $\left|\dfrac{\sin(t)}{t}\right|\le1$, $$ \int_0^{2\pi}\frac{\sin(t)}{t}\mathrm{d}t\tag{8} $$ has a value no greater than $2\pi$.

$(7)$ and $(8)$ guarantee that $$ \int_0^\infty\frac{\sin(t)}{t}\mathrm{d}t\tag{9} $$ converges to a value no greater than $2\pi+\dfrac1\pi$.

In this case, $$ \left|\int_0^N\sin(t)\;\mathrm{d}t\right|\le2 $$ and $\dfrac1t$ is monotonically decreasing to $0$ on $(0,\infty)$. Thus, by Dirichlet, $$ \int_0^\infty\frac{\sin(t)}{t}\mathrm{d}t $$ converges.