In this wikipedia article for improper integrals,

∫∞0sinxxdx

is given as an example for the integrals that have an improper Riemann integral but do not have a (proper) Lebesgue integral. Here are myquestions:

- Why does this one have an improper Riemann integral? (I don’t see why ∫a0sinxxdx and ∫∞asinxxdx converge.)
- Why doesn’t this integral have a Lebesgue integral? Is it because that sinxx is unbounded on (0,∞) and Lebesgue integral doesn’t deal with unbounded functions?

**Answer**

∫a0sinxxdx converges since we can extend the function x↦sinxx by continuity at 0 (we give the value 1 at 0). To see that the second integral converges, integrate by parts ∫Aasinxxdx. Indeed, we get

∫Aasinxxdx=[−cosxx]Aa+∫Aa−cosxx2dx=cosaa−cosAA−∫Aacosxx2dx,

and lim, 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.

**Attribution***Source : Link , Question Author : Community , Answer Author : Kelvin Lois*