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

We have that $$\int_1^{\infty}\frac{\sin x}{x}dx$$ $$u=\frac{1}{x}$$ $$du=-\frac{1}{x^2}dx$$ $$dv=\sin xdx$$ $$v=-\cos x$$ $$\int_1^{\infty}\frac{\sin x}{x}dx=-\cos(1)-\int_1^{\infty}\frac{\cos x}{x^2}dx$$ $$\int_1^{\infty}\frac{\cos x}{x^2}dx<\int_1^{\infty}\frac{1}{x^2}dx$$ So it converges. Now I need to find out if $\int_1^{\infty}|\frac{\sin x}{x}|dx$ 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 '13 at 22:39
  • $\begingroup$ Also, it should be $\cos 1$ and not $-\cos x/x$ in your formula. $\endgroup$ – Julien May 13 '13 at 22:53
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    $\begingroup$ @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 '15 at 5:03
  • $\begingroup$ Yes, technically the function that is being integrated should be positive in the integration domain for the comparison test. $\endgroup$ – user1242967 Jun 5 '15 at 20:59

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.

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    $\begingroup$ @julien Thanks. This trick is often useful when dealing with integrals of periodic functions over a neighborhood of $\infty$. $\endgroup$ – Ayman Hourieh May 14 '13 at 0:03
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    $\begingroup$ could someone please explain the first inequality? $\endgroup$ – Rubenz Jul 8 '16 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 '16 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 '18 at 12:26
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    $\begingroup$ @MaryStar Exactly, $\sin$ is periodic. $\endgroup$ – Ayman Hourieh Apr 7 '18 at 17:41

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 positive and negative.


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