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,
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 my questions:

  • 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 xsinxx 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+Aacosxx2dx=cosaacosAAAacosxx2dx,
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

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