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Let $f(x)=\frac{\sin x}{x}$ if $x\neq 0$ and $f(x)=0$ if $x=0$. Is $f$ Lebesgue integrable? Also, is $\sin^2x/x^2$ Lebesgue integrable?

Note that $\lim_{x\to 0}\frac{\sin x}{x}=1=f(0)$. So $f$ is measurable. We have $$ \int_\mathbb{R} |f(x)|dx=2\int_0^\infty f(x)dx $$

My question is that I saw one approach: $$ \int_\mathbb{R} |f(x)|dx\ge \int_{[0,\infty)} |f(x)|dx\ge \int_{\cup_{n\ge 1}[2n\pi, 2n\pi+\pi]} |f(x)|dx=\sum_{n=0}^\infty \int_{[2n\pi, 2n\pi+\pi]} |f(x)|dx $$ $$ \ge \sum_{n=0}^\infty \int_{[2n\pi, 2n\pi+\pi]} \frac{|\sin x|}{2\pi(n+1)}=\infty $$ So $f$ is not Lebesgue integrable.

However, in this answer: $\int_{0}^{\infty}\frac{\sin x}{x}dx $ converges, we know that $$ \int_0^\infty f(x)dx=\pi/2 $$

So why do we get two different answers?

To find $\int_\mathbb{R} \sin^2x/x^2dx$, I repeat the above argument: $$ \int_\mathbb{R} \sin^2x/x^2dx=2\int_{[0,\infty)}\sin^2x/x^2dx=\int_{\cup_{n\ge 1}[2n\pi, 2n\pi+\pi]}\sin^2x/x^2dx=\sum_{n=0}^\infty \int_{[2n\pi, 2n\pi+\pi]} \frac{|\sin^2 x|}{x^2}dx $$ $$ \le \sum_{n=0}^\infty \int_{[2n\pi, 2n\pi+\pi]} \frac{|\sin^2 x|}{(2n\pi)^2}dx<\infty $$

Christian E. Ramirez
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H.Y Duan
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    $\int_0^\infty\frac{\sin x}xdx$ is conditionally convergent, not absolutely convergent. – user14111 Apr 11 '24 at 06:27
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    This function has a conditionally convergent Riemann improper integral. The integral of the absolute value diverges though. In Lebesgue integration there is only absolute convergence, so this function doesn't have a convergent Lebesgue integral. – Mark Apr 11 '24 at 06:28
  • Why is $ \int_\mathbb{R} |f(x)|dx=2\int_0^\infty f(x)dx $. What is true is $ \int_\mathbb{R} |f(x)|dx=2\int_0^\infty |f(x)|dx $ – Kavi Rama Murthy Apr 11 '24 at 06:40
  • @geetha290krm Because $f(x)=\sin x/x$ is a even function ($f(-x)=f(x)$), then $\int_{-\infty}^\infty f(x)dx=2\int_0^\infty f(x)dx$? – H.Y Duan Apr 11 '24 at 07:07
  • No, that is where your basic consfusion had arisen. $f(x)$ is not non-negative, so you have to keep the absolute value sign. – Kavi Rama Murthy Apr 11 '24 at 07:14

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For $\int_0^\infty \frac{\sin x}{x} d x$, we get two different answers because we are using two different definitions of the integral. With the definition of Lebesgue, the integral diverges because $\int_0^\infty |\frac{\sin x}{x}| d x = +\infty$. As an improper Riemann integral, the integral converges because $\lim_{A\to\infty} \int_0^A\frac{\sin x}{x } d x$ exists and is equal to $\frac{\pi}{2}$.

For a nonnegative measurable function $f$ such as $\frac{\sin^2 x}{x^2}$, the two definitions give the same answer, that is to say \begin{equation}\int_0^\infty f(x) dx = \lim_{A\to\infty} \int_0^A f(x) d x\end{equation}

Gribouillis
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  • Thank you! So is the proof for convergence of $\int_0^\infty \sin^2 x/x^2dx$ ok? – H.Y Duan Apr 11 '24 at 07:09
  • No it is not OK for two reasons: the union of the intervals $[2n\pi, 2n\pi + \pi]$ does not cover $[0,\infty[$ and also when $n=0$ you can't write $\frac{1}{(2n \pi)^2}$ – Gribouillis Apr 11 '24 at 07:12
  • So how to prove that? Can we prove it by partitioning $[0,\infty)$ by $[0,\delta)\cup [\delta, \infty)$? Then $\int_0^\delta f(x)dx$ is bounded and $\int_\delta^\infty f(x)$ is convergent by $p$-integral. – H.Y Duan Apr 11 '24 at 17:21
  • @H.YDuan Yes you can use $\frac{\sin^2 x}{x}\le \frac{1}{x^2}$ on $[\delta, \infty)$ and the rest of the proof is easy. – Gribouillis Apr 11 '24 at 18:52