Does $ \int_0^{\infty}\frac{\sin x}{x}dx $ have an improper Riemann integral or a Lebesgue integral?

Question

In this wikipedia article for improper integral, $$ \int_0^{\infty}\frac{\sin x}{x}dx $$ 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?

Answer 1

$\displaystyle \int_0^a\frac{\sin x}xdx$ 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 tdt\\ &=\frac 2{\pi}\sum_{k=1}^N\frac 1{k+1}, \end{align*} and we can conclude since the harmonic series is not convergent.

Answer 2

New try:

To see that it has an improper Riemann integral argue that the function is continuous in $0$.

Now,

$$\int_0^\infty \frac{\sin x}{x} \, \text{d}x = \lim_{y \to \infty} \left (\sum_{k = 1}^{[y/\pi]} \int_{k \pi}^{(k + 1) \pi} \frac{\sin x}{x} \, \text{d}x + \int_{[y/\pi]\pi}^y \frac{\sin x}{x} \, \text{d}x \right ).$$ This in its turn is equal to $$\lim_{y \to \infty} \left (\sum_{k = 1}^{[y/\pi]} \int_{0}^{\pi} (-1)^k \frac{\sin x}{x + k \pi} \, \text{d}x + \int_{[y/\pi]\pi}^y \frac{\sin x}{x} \, \text{d}x \right ).$$ Now the first term converges by the alternating series test (Cauchy test?) and the second one converges to $0$.

Answer 3

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 $2\pi+\dfrac1\pi$.

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Another general test is the Dirichlet test (Theorem 17.5). It says that if $$ \left|\int_a^xf(t)\;\mathrm{d}t\right|<M $$ independent of $x\in[a,\infty)$, and $g(x)$ monotonically decreases to $0$ as $x\to\infty$, then $$ \int_a^\infty f(t)g(t)\;\mathrm{d}t $$ converges.

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.

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In fact, contour integration yields that $$ \int_{-\infty}^\infty\frac{\sin(t)}{t}\mathrm{d}t=\pi $$ which, by symmetry, tells us that $$ \int_0^\infty\frac{\sin(t)}{t}\mathrm{d}t=\frac{\pi}{2} $$

Answer 4

In addition to @David's answer, I should have noticed that the answer to the questions are partially in that wiki article which also gives a comparison between improper Riemann integrals and Lebesgue integrals.

What's more, the second question is somehow incorrect: $\frac{\sin x}{x}$ is bounded on $(0,\infty)$:

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Thanks to the following two questions:

• Show that $\int^{\infty}_{0} x^{-1} \sin x dx = \frac\pi2$
• Solving the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$?

one can actually come up with $$ \int_0^{\infty}\frac{\sin x}{x}dx = \frac{\pi}{2}. $$ It's worth knowing that this is also called Dirichlet integral.

Answer 5

You can succeed in proving $\frac{\sin x}{x}$ is not Lebesgue integrable over $[0,\infty[$ also by using some street-fighting mathematics.

Actually, you only have to show that: $$\int_\pi^\infty \frac{|\sin x|}{x}\ \text{d} x =\infty\; ,$$ for the integral $\int_0^\pi \frac{|\sin x|}{x}\ \text{d}x$ is finite (due to $\lim_{x\to 0^+} \frac{\sin x}{x} =1$ and continuity of $\frac{\sin x}{x}$ in $]0,\pi]$).

Let $f(x):=\frac{|\sin x|}{x}$ for sake of simplicity. Then $f$ is nonnegative and concave in each interval of the type $[k\pi, (k+1)\pi]$ and it attains its global minimum (i.e. $0$) in $x_k:=k\pi$, with $k\in \mathbb{N}$; moreover, $f$ attains local maximum in $\xi_k \in ]k\pi,(k+1)\pi[$, where $\xi_k$ is the unique solution of: $$\sin x=x\ \cos x$$ in $[k\pi,(k+1)\pi]$.

The triangle $\mathfrak{T}_k$ having vertices in $A_k:=(k\pi ,0)$, $B_k:=((k+1)\pi ,0)$ and $C_k:=(\xi_k,f(\xi_k))$ lies in the trapezoid $\mathfrak{R}_k:=\{(x,y)\in \mathbb{R}^2|\ k\pi\leq x\leq (k+1)\pi,\ 0\leq y\leq f(x)\}$ by concavity, hence for each index $k$: $$\int_{k\pi}^{(k+1)\pi} f(x)\ \text{d} x =\operatorname{Area}(\mathfrak{R}_k) \geq \operatorname{Area}(\mathfrak{T}_k)=\frac{\pi}{2}\ f(\xi_k)$$ and: $$\tag{1} \int_{\pi}^{(k+1)\pi} f(x)\ \text{d} x\geq \sum_{n=1}^k \frac{\pi}{2}\ f(\xi_n)\; .$$

Now, you win if you prove that the RHside of (1) is the $k$-th partial sum of a positively divergent series.

You can prove that: $$\xi_k = \frac{\pi}{2} +k\pi -\varepsilon_k = \frac{\pi}{2} (2k+1)-\varepsilon_k$$ with $0<\varepsilon_k<\pi/2$ and $\varepsilon \to 0$ as $k\to \infty$ (cfr. Mahajan, Street-fighting Mathematics, 6.4), thus you get: $$\begin{split} f(\xi_k) & = \frac{|\sin \xi_k|}{\xi_k} \\ &= \frac{|\sin (\pi/2 +k\pi -\varepsilon_k)|}{\frac{\pi}{2} (2k+1)-\varepsilon_k} \\ &= \frac{\sin (\pi/2 -\varepsilon_k)}{\frac{\pi}{2} (2k+1)-\varepsilon_k} &\qquad \text{(} \sin t \text{ is periodic)} \\ &\geq \frac{2}{\pi}\ \frac{\sin (\pi/2 -\varepsilon_k)}{2k+1} &\qquad \text{(denominator increased + algebra)} \\ & = \frac{2}{\pi}\ \frac{\cos \varepsilon_k}{2k+1} &\qquad \text{(trigonometric trick)} \\ & \geq \frac{2}{\pi}\ \frac{1- \frac{1}{2} \varepsilon_k^2}{2k+1}\; , \end{split}$$ and the latter inequality holds because of the elementary inequality $\cos t \geq 1-\frac{1}{2}\ t^2$. Therefore you find: $$\sum_{n=1}^k \frac{\pi}{2}\ f(\xi_n) \geq \sum_{n=1}^k \frac{1- \frac{1}{2} \varepsilon_n^2}{2n+1}$$ and the RHside diverges in the positive sense when $k$ goes to $\infty$ (for the summand $\frac{1- \frac{1}{2} \varepsilon_n^2}{2n+1}$ is asymptotically equivalent to that of a harmonic series). Finally you can pass to the limit in (1) to get $\int_\pi^\infty f(x)\ \text{d} x =\infty $ as you claimed.