# Showing $\frac{\sin x}{x}$ is NOT Lebesgue Integrable on $\mathbb{R}_{\ge 0}$

Let $$f(x) = \frac{\sin(x)}{x}$$ on $\mathbb{R}_{\ge 0}$.

$$\int |f| < + \infty\quad\text{iff}\quad \int f^+ < \infty \color{red}{\wedge \int f^- < \infty}$$

EDIT: So what I'm trying to do is to show that in fact $\int f^+ > \infty$ so that therefore $\int |f| = \infty$

Now consider the assertion that: $$\int f+ = \sum_{k=1}^\infty \int_{[2\pi k , 2\pi k + \pi]} \left( \frac{\sin(x)}{x} \right) dx \le \sum_{k=1}^\infty\int_{[2\pi k , 2\pi k + \pi]} \left( \frac{\sin(x)}{2 \pi k + \pi}\right) dx$$

Two Questions:

(1) Is the first step of asserting that

$$\int f^+ = \sum_{k=1}^\infty \int_{[2\pi k , 2\pi k + \pi]} \left( \frac{\sin(x)}{x} \right) dx$$

correct, in the sense that you can partition a Lebesgue integral into an INFINITE series of integrals being added together (whose individual term domains cover all of the overall domain of the original integral s.t. they are also pairwise disjoint)?

(2) Is there a well known lower bound of $\sum_{k=1}^\infty\int_{[2\pi k , 2\pi k + \pi]} \left( \frac{\sin(x)}{2 \pi k + \pi}\right) dx$ that diverges whose existence establishes that $f^+$ is in fact not integrable?

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BTW, $\int|f|=\int f^- +\int f^+$, so $$\int|f|<\infty \qquad\text{ iff}\qquad\int f^+<\infty\quad\text{AND }\quad\int f^-<\infty$$ –  AD. Nov 8 '12 at 16:28

## 4 Answers

You might want to start by showing that $f$ is conditionally convergent. This would show that $f$ is not Lebesgue integrable. Use the fact that $$\int_{(n - 1)\pi}^{n\pi} \frac{\left| \sin x \right|}{x} dx \geq \frac{1}{n\pi}\int_{(n - 1)\pi}^{n\pi} \left| \sin x \right| dx = \frac{2}{n\pi}$$ to get $$\int_0^{N\pi} \frac{\left| \sin x \right|}{x} dx \geq \frac{2}{\pi}\sum_{n = 1}^N \frac{1}{n}$$ which shows that $$\int_0^\infty \frac{\left| \sin x \right|}{x} dx$$ is divergent.

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That's exactly what I'm trying to show by showing that $\int f^+$ diverges (so that $\int |f|$ must also diverge). –  user1770201 Oct 30 '12 at 19:34

Let $f(x)=\frac{\sin x}{x}$. Surely the function is measurable, so we need to prove that $$\int|f(x)|dx=+\infty$$ In order to do that we need to look closer to the function. To this end, note that it is almost like $x\mapsto\frac{1}{x}$, however our function has zeros so we cannot compare them directly. A better approach is to use that the local maximums are attained at $x=\frac{\pi}{2}+\pi \cdot n$, in fact we have $$|f(x)| \geq \frac{1}{\sqrt{2}x}, \quad\text{for  \left|\frac{(2n+1)\pi}{2}-x\right|\leq \frac{\pi}{4}}$$ Now, do you see how to estimate the function from below?

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Doing a change of variables, it's enough to show that

$$\int_\mathbb{R^+} |\sin(2\pi y)/y |dy$$ diverges.

You can estimate this by

$$\int_\mathbb{R^+} |\sin(2\pi y)/y | dy \geq \sum_{k=0}^\infty \int_{k+3/4}^{k+5/4}|\sin(2\pi y)/y | \geq 1/2\sum_{k=0}^\infty \frac{\sin \pi/4}{k + 5/4}$$ which diverges as it is essentially the harmonic series.

I may have made a few mistakes on the constants but the idea is that you can fit rectangles underneath the curve to compare it to the harmonic series

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It is well known that the integral $\int_{0}^{\infty}\left|\frac{\sin(x)}{x}\right|dx$ does not converge. Now use the theorem "a function $f$ is Lebesgue integrable iff |f| is" to show, that $f(x)=\frac{\sin(x)}{x}$ is not Lebesgue integrable..

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In case my post was not clear, this is exactly what I'm trying to do. –  user1770201 Oct 30 '12 at 19:35