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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.

Sahan Manodya
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user1242967
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  • 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. – Julien May 13 '13 at 22:39
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    @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 $\geq 0 \forall x$ ? – S.Dan Jun 04 '15 at 05:03
  • Yes, technically the function that is being integrated should be positive in the integration domain for the comparison test. – user1242967 Jun 05 '15 at 20:59
  • 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? – Philipp Jan 05 '22 at 08:56
  • nice problem...........+1 – TShiong Jan 11 '23 at 18:17

4 Answers4

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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.

Ayman Hourieh
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  • Much nicer, +1. – Julien May 13 '13 at 23:16
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    @julien Thanks. This trick is often useful when dealing with integrals of periodic functions over a neighborhood of $\infty$. – Ayman Hourieh May 14 '13 at 00:03
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    could someone please explain the first inequality? – Rubenz Jul 08 '16 at 16:41
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    @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. – Michael Sep 06 '16 at 13:30
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    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? – Mary Star Apr 07 '18 at 12:26
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    @MaryStar Exactly, $\sin$ is periodic. – Ayman Hourieh Apr 07 '18 at 17:41
  • Ah ok! Thank you! @AymanHourieh – Mary Star Apr 07 '18 at 17:53
  • Can anyone explain how does the last sum diverges as $N\rightarrow\infty$? – Sher Afghan Jan 24 '19 at 19:57
  • @SherAfghan It's harmonic series: https://mathworld.wolfram.com/HarmonicSeries.html – S.H.W Aug 20 '20 at 01:48
  • @AymanHourieh sorry I know it is a very old answer but I have a doubt on your great answer...if I apply this idea without the absolute value I will obtain that $\int_1^{\infty}\frac{\sin{x}}{x}, dx$ is divergent and it is an error...where am I failing? – pawel Apr 27 '21 at 10:38
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    @pawel The inequality step is not true with the absolute values removed. In particular the inequality in Michael's comment above is false without the absolute values since $\sin(x)$ changes sign. – Jakob Streipel May 15 '21 at 01:27
  • @tissuepaper Ok thank you very much! – pawel May 17 '21 at 07:24
8

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.

Newton
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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 $\:\lim_{j\in\mathbf N}\:\pi(j+3/4)=\infty,$ $$\int_{1}^\infty{|\sin x|\over x^q}\:\text{d}x=\infty$$

budlight
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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

Surjeet Singh
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Ram
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