In this wikipedia article for improper integrals, $$ \int_0^{\infty}\frac{\sin x}{x}dx $$ 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?
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    $\begingroup$ $\int_0^a\frac{\sin x}xdx$ 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 $\int_a^A\frac{\sin x}x dx$. But $f(x):=\frac{\sin x}x$ has not a Lebesgue integral, since the integral $\int_0^{\infty}\left|\frac{\sin x}x\right| dx$ is not convergent (but it's not a consequence of the fact that $f$ is not bounded, for example that $g(x)=\frac 1{\sqrt x}$ for $0<x\leq 1$ and $g(x)=0$ for $x>1$). $\endgroup$ – Davide Giraudo Sep 24 '11 at 14:48
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    $\begingroup$ This question has some more information about this function, its Riemann integrability, and the relationship with the Lebesgue integral. $\endgroup$ – Rachel Sep 24 '11 at 15:24
  • $\begingroup$ Another closely related question: math.stackexchange.com/questions/25586/… $\endgroup$ – Jonas Meyer Jan 5 '12 at 2:15
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    $\begingroup$ Personally, I find it weirder that $\int_0^{\infty}\frac{\cos(x)}{\sqrt{x}}dx$ converges. $\endgroup$ – Meow Jun 10 '13 at 20:30

$\displaystyle \int_0^a\frac{\sin x}xdx$ 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 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.

  • $\begingroup$ Maybe I made a mistake, I think $f=\frac{\sin x}{x}$ is indeed bounded, right? $\endgroup$ – user9464 Sep 24 '11 at 15:21
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    $\begingroup$ @jack yes, you're right (hence a fortiori the divergence of the integral can't be caused by unboundedness, since the function is bounded). $\endgroup$ – Davide Giraudo Sep 24 '11 at 15:23
  • $\begingroup$ @DavideGiraudo You're welcome. I knew it was ;) could you help me out with this question? $\endgroup$ – Pedro Tamaroff Feb 10 '12 at 20:39
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    $\begingroup$ @DavideGiraudo all you showed was that $-\int_a^{\infty} \frac{dx}{x^2} \leq \int_a^{\infty} \frac{\cos x}{x^2}\, dx\leq \int_a^{\infty} \frac{dx}{x^2}$ This doesn't necessarily show convergence of $\int_a^{\infty} \frac{\cos x}{x^2}\, dx$. $\endgroup$ – Glassjawed Nov 26 '14 at 1:02
  • $\begingroup$ Oh wait, this is since absolute convergence implies convergence in a regular sense. $\endgroup$ – Glassjawed Nov 26 '14 at 3:30

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 $2\pi+\dfrac1\pi$.

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.

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

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    $\begingroup$ One could calculate $\int_0^\infty\sin xdx/x$ elementarily. It's equivalent to $\lim_{n\to\infty}\int_0^{\pi/2}\sin2nxdx/x$. We're pleased to see that $g(x)=1/x-1/\sin x=O(1/x)$ and so $\lim_{n\to\infty}\int_0^{\pi/2}g(x)\sin2nxdx=0$ (Generally, Riemann-Lebesgue lemma; however, since $g$ is of $C^\infty$, we could integrate it by part and obtain the result). Since $\sin2nx/\sin x=\sum_{k=1}^n(\sin2kx-\sin2(k-1)x)$, we could easily determine $\int_0^{\pi/2}\sin2nxdx/\sin x$. $\endgroup$ – Yai0Phah Apr 29 '13 at 6:55

New try:

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


$$\int_0^\infty \frac{\sin x}{x} \, \text{d}x = \lim_{y \to \infty} \left (\sum_{k = 1}^{[y/\pi]} \int_{k \pi}^{(k + 1) \pi} \frac{\sin x}{x} \, \text{d}x + \int_{[y/\pi]\pi}^y \frac{\sin x}{x} \, \text{d}x \right ).$$ 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 $0$.

  • $\begingroup$ This is very sloppy! Firstly, your second equation is wrong (and can't be fixed just by changing $=$ to $\le$). Secondly, to show that an alternating series converges, it's not enough to show that its terms are less than $1/x$ in absolute value; you must also show that the terms decrease monotonically in absolute value. $\endgroup$ – TonyK Sep 24 '11 at 15:13
  • $\begingroup$ @TonyK: Oops, you're right. Got confused. $\endgroup$ – Jonas Teuwen Sep 24 '11 at 15:15
  • $\begingroup$ @TonyK: I have tried to correct it using the same method. $\endgroup$ – Jonas Teuwen Sep 24 '11 at 15:29
  • $\begingroup$ That's better! I removed my down-vote. $\endgroup$ – TonyK Sep 24 '11 at 16:44
  • $\begingroup$ The function isn't countinous in 0, but it can be continuously extended to $x=0$. $\endgroup$ – Jannik Pitt Mar 29 '18 at 19:47

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 $(0,\infty)$:

enter image description here

Thanks to the following two questions:

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.


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.


Recall that all Riemann-integrable functions are defined on a closed interval $[a,b]$ and are bounded there. Hence the integral $\int_0^a \frac{\sin(x)}{x} dx$ has a problem with the definition of integral. However, the integrand can be extended to $\mathbb{R}$ by defining the value $1$ at $0$. Now the extension is Riemann-integrable over $[0,a]$ because it is continuous by the properties of $\frac{\sin(x)}{x}$: it is continuous on $(0,a]$ as product of analytic functions. In addition it has the limit $1$ at $0$ (recall derivation of derivatives of trigonometric functions). Hence it is continuous on the interval $[0,a]$. The extension can be expressed in terms of the sine cardinal function by $\textrm{sinc}(x/\pi)$, where

$\textrm{sinc}(x) = \Bigg\{\begin{eqnarray} & \frac{\sin(\pi x)}{\pi x} & , \ \ x \neq 0 \\ & 1 & , \ \ x = 0 \end{eqnarray}$.

Because of its existence on the interval $[0,a]$ the expression $\textrm{sinc}(x/\pi)$ is used here as the integrand. Because of the definitions in my studybooks there is left the $\pi$ in the definition of $\textrm{sinc}(x)$. It is cancelled in the integrand expression. Now $\int_0^a \textrm{sinc}(x/\pi) dx$ exists for every $a\in \mathbb{R}^+$.

Consider now the integral $\int_a^M \frac{\sin(x)}{x} dx$. We have \begin{eqnarray} \frac{|\cos(x)|}{x^2} \leq \frac{1}{x^2} \ , \ \ x \geq a . \end{eqnarray} The right hand side is a majorant to the non-negaitve left hand side. The right hand side is also integrable from $a$ to $\infty$. Hence also the left hand side is. This implies the existence of $\int_a^\infty \frac{\cos(x)}{x^2} dx$. We now have \begin{eqnarray} \int_a^M \frac{\sin(x)}{x} dx & = & \Bigg|_a^M \frac{-\cos(x)}{x}-\int_a^M -\frac{-\cos(x)}{x^2} dx \\ & = & \frac{\cos(a)}{a}-\frac{\cos(M)}{M}-\int_a^M \frac{\cos(x)}{x^2} dx \end{eqnarray} for $M \geq a$. All terms on the right hand side have a limit as $M \rightarrow \infty$. Hence the integral $\int_a^\infty \frac{\sin(x)}{x} dx$ exists for every $a \in \mathbb{R}^+$. Hence we can write \begin{eqnarray} \int_0^a \textrm{sinc}(x/\pi) dx + \int_a^\infty \frac{\sin(x)}{x} dx & = & \int_0^a \textrm{sinc}(x/\pi) dx + \int_a^\infty \textrm{sinc}(x/\pi) dx \\ & = & \int_0^\infty\textrm{sinc}(x/\pi) dx \ . \end{eqnarray} Now all terms on the left hand side exist. Hence also the right hand side.

Then we analyze the existence of the respective Lebesgue integral. Recall that by definition of the Lebesgue integral it is obtained by substraction from the integrals of the positive and the negative part. The claim is that the integral diverges and we have to show that at least one of the integrals of the positive and the negative part diverge. We choose the positive part \begin{eqnarray} f^+(t) = \chi_{\bigcup_{k=0}^\infty[2\pi k, 2\pi (k+1/2)]} \textrm{sinc}(x/\pi) \ . \end{eqnarray} Note that we can write a diverging lower limit estimate for Lebesgue integral because it accepts the value of $\infty$. We calculate \begin{eqnarray} \int_{\mathbb{R} \textrm\ \mathbb{R}^-} f^+(x) d\mu & = & \int_{\mathbb{R} \textrm\ \mathbb{R}^-} \chi_{\bigcup_{k=0}^\infty [2\pi k, 2\pi (k+1/2)]} \textrm{sinc}(x/\pi) d\mu \\ & \geq & \int_{\mathbb{R} \textrm\ \mathbb{R}^-} \chi_{\bigcup_{k=0}^n [2\pi k,2\pi (k+1/2)]} \textrm{sinc}(x/\pi) d\mu \\ & = & \sum_{k=0}^n \int_{[2\pi k, 2\pi(k+1/2)]} \textrm{sinc}(x/\pi) d\mu \\ & = & \sum_{k=0}^n \int_{2\pi k}^{2\pi (k+1/2)} \textrm{sinc}(x/\pi) dx \\ & = & \int_0^\pi \textrm{sinc}(x/\pi) dx + \sum_{k=1}^n \int_{2\pi k}^{2\pi (k+1/2)} \textrm{sinc}(x/\pi) dx \\ & = & \int_0^\pi \textrm{sinc}(x/\pi) dx + \sum_{k=1}^n \int_0^\pi \textrm{sinc}((x+2\pi k)/\pi) dx \\ & = & \int_0^\pi \textrm{sinc}(x/\pi) dx + \sum_{k=1}^n \int_0^\pi \frac{\sin(x+2\pi k)}{x+2\pi k} dx \\ & \geq & \int_0^\pi \textrm{sinc}(x/\pi) dx + \sum_{k=1}^n \int_0^\pi \frac{\sin(x)}{\pi + 2\pi k} dx \\ & = & \int_0^\pi \textrm{sinc}(x/\pi) dx + \sum_{k=1}^n \frac{1}{\pi+2\pi k} \Bigg|_0^\pi -\cos(x) dx \\ & \geq & \int_0^\pi \textrm{sinc}(x/\pi) dx + \sum_{k=1}^n \frac{2}{2\pi(k+1)} \\ & = & \int_0^\pi \textrm{sinc}(x/\pi) dx + \frac{1}{\pi} \sum_{k=2}^{n+1} \frac{1}{k} \ . \end{eqnarray} for every $n > 0$. The harmonic sum diverges as $n \rightarrow \infty$. Hence the left hand side equals to $\infty$. This shows the claim. Similar analysis shows that also the integral of the negative part diverges. Hence the Lebesgue integral has the form $\infty - \infty$ that is not defined, that is, the Lebesgue integral of $\textrm{sinc}(x/\pi)$ doesn't exist. However, the Lebesgue integral of $|\textrm{sinc}(x/\pi)|$ exists and equals to $\infty$.

The function $\textrm{sinc}(x/\pi)$ is bounded because it is continuous on $[0,a]$ and hence bounded there and has an upper limit estimate $\frac{1}{a}$ on $[a,\infty)$. The divergence is not due to any singularities of $\textrm{sinc}(x/\pi)$ but because of the too slowly decreasing factor $\frac{1}{x}$ of the integrand and periodicity of $\sin(x)$.


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