Improper integral of $\sin(x)/x $ converges absolutely, conditionally or diverges?

We have $$\int_1^{\infty}\frac{\sin x}{x}\text{d}x$$

Integrating by parts $$u=\frac{1}{x}$$ $$\text{d}u=-\frac{1}{x^2}\text{d}x$$ $$\text{d}v=\sin x\;\text{d}x$$ $$v=-\cos x$$ $$ \begin{aligned} \int_1^{\infty} \frac{\sin x}{x} \text{d}x & = \frac{-\cos x}{x} \Big|_1^{\infty} - \int_1^{\infty} \frac{\cos x}{x^2} \text{d}x \\ & = \cos 1 - \int_1^{\infty} \frac{\cos x}{x^2} \text{d}x \end{aligned} $$

$\int_1^{\infty} \frac{\cos x}{x^2} \text{d}x$ converges absolutely (using the Comparison Test For Improper Integrals):

$$ \int_1^{\infty} \frac{|\cos x|}{x^2} \text{d}x < \int_1^{\infty} \frac{1}{x^2} \text{d}x $$

So $\int_1^{\infty} \frac{\sin x}{x} \text{d}x$ converges.

Now I need to find out if $\int_1^{\infty} |\frac{\sin x}{x}| \text{d}x$ converges or diverges.

  • $\begingroup$ Nicely done (note you should take the absolute value, though, in $\int \frac{|\cos x|}{x^2}dx\leq \int \frac{1}{x^2}dx$). Now try to compare with the harmonic series regarding integrability. Yes, it is not integrable. $\endgroup$
    – Julien
    May 13, 2013 at 22:39
  • 2
    $\begingroup$ @user1242967 In your last step you have $\int_1^{\infty}\frac{\cos x}{x^2}dx<\int_1^{\infty}\frac{1}{x^2}dx$ and state it converges. Is it by comparison test? For that doesnt $\int_1^{\infty}\frac{\cos x}{x^2}dx$ need to be $\geq 0 \forall x$ ? $\endgroup$
    – S.Dan
    Jun 4, 2015 at 5:03
  • $\begingroup$ Yes, technically the function that is being integrated should be positive in the integration domain for the comparison test. $\endgroup$ Jun 5, 2015 at 20:59
  • $\begingroup$ What does $$u=\frac{1}{x}$$ $$du=-\frac{1}{x^2}dx$$ $$dv=\sin xdx$$ $$v=-\cos x$$ exactly mean? Is this a convention when applying integration by parts? It looks a bit confusing. Why haven't you simply said that you are applying integration by parts and omitted those $4$ equations? $\endgroup$
    – Philipp
    Jan 5, 2022 at 8:56
  • $\begingroup$ nice problem...........+1 $\endgroup$
    – TShiong
    Jan 11 at 18:17

4 Answers 4


Let $N \in \Bbb N, N > 1$, we have:

\begin{align} \int_0^{2\pi N} \left|\frac{\sin x}{x}\right|\,dx &= \sum_{n=0}^{N-1} \int_{2\pi n}^{2\pi(n+1)} \left|\frac{\sin x}{x}\right|\,dx \\ &\ge \sum_{n=0}^{N-1} \frac{1}{2\pi (n+1)} \int_{2\pi n}^{2\pi(n+1)} \left|\sin x\right|\,dx \\ &= \sum_{n=0}^{N-1} \frac{1}{2\pi (n+1)} \int_{0}^{2\pi} \left|\sin x\right|\,dx \\ &= \sum_{n=0}^{N-1} \frac{2}{\pi (n+1)} \end{align}

The last sum diverges as $N \to \infty$, and so does the original integral.

Your integral is on $[1, \infty]$, but it also diverges because $\left|\frac{\sin{x}}{x}\right|$ is continuous on $[0, 1]$. My proof is on $[0, \infty]$ because it makes managing the summation slightly easier.

  • 9
    $\begingroup$ @julien Thanks. This trick is often useful when dealing with integrals of periodic functions over a neighborhood of $\infty$. $\endgroup$ May 14, 2013 at 0:03
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    $\begingroup$ could someone please explain the first inequality? $\endgroup$
    – Rubenz
    Jul 8, 2016 at 16:41
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    $\begingroup$ @Rubenz $1/x:[2\pi n, 2\pi(n+1)]\to \mathbb{R}$ is decreasing. We have $\frac{|\sin(x)|}{x} \geq \frac{|\sin(x)|}{2\pi(n+1)}$ in $[2\pi n, 2\pi(n+1)]$, hence the expression. $\endgroup$
    – Michael
    Sep 6, 2016 at 13:30
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    $\begingroup$ Hello! Could you explain to me the equality $\sum_{n=0}^{N-1} \frac{1}{2\pi (n+1)} \int_{2\pi n}^{2\pi(n+1)} \left|\sin x\right|\,dx = \sum_{n=0}^{N-1} \frac{1}{2\pi (n+1)} \int_{0}^{2\pi} \left|\sin x\right|\,dx $ ? Do we use here that the sine function is periodic? $\endgroup$
    – Mary Star
    Apr 7, 2018 at 12:26
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    $\begingroup$ @MaryStar Exactly, $\sin$ is periodic. $\endgroup$ Apr 7, 2018 at 17:41

I would like to provide an intuitive graphic explanation.

enter image description here

Since $\left|\frac{\sin(x)}{x}\right|$ is continuous on $[0, \infty)$, thus it suffices to show the absolute divergence of the tail, namely $\int_{\pi}^{\infty} \left|\frac{\sin(x)}{x}\right| dx$.

Consider bounding this integral below by the infinite sum of area of triangles, namely the n-th triangle has width $\pi$ and height $\frac{1}{(n +1/2) \pi}$, thus the total area (of triangles) = $$\sum_{n = 1}^{\infty} \frac{1}{n + 1/2} = \infty$$ By a comparison test to the harmonic series $\sum_{k = 2}^{\infty} \frac{1}{k}$, thus the original integral diverges.


If one's allowed to use the Absolute Divergent Theorem for Improper Integrals, then one could use what follows:

$\:|\sin x|>\frac{\:1}{\sqrt2}\:,\:\:\:\forall x\in\left]\pi(j+\frac{1}{4}),\pi(j+\frac{3}{4})\right[=I_j\:,\:\:\:j\in\mathbf N.$

$\:\:\forall x\in I_j,\: \forall q\in\:]0,1]\:$ we have $$\sum_{j\in\mathbf N}\int_{I_j}{{\text{d}x}\over x^q}\:\le\:\int_1^\infty{{\text{d}x}\over x^q}=\infty\:,$$by comparison.

So with $\{I_j\}$ being an increasing sequence and $|I_j|=\pi/2\:$ with $\:\lim_{j\in\mathbf N}\:\pi(j+3/4)=\infty,$ $$\int_{1}^\infty{|\sin x|\over x^q}\:\text{d}x=\infty$$


Ayman's proof shows the original improper integral is not absolutely convergent. But, working without absolute values, we can show that the series is conditionally convergent. Work with the integral on $ [2 \pi, \infty)$ , and break up the integral into regions where the integrand is $+$ve and $-$ve


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