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I need to study the convergence of the following improper integral:

$$\int_{0}^{\infty} \dfrac{\sin(x)}{x+1}\, \mathrm dx$$

I did the following:

$$ -1 \leq \sin(x) \leq 1 \\ \implies \dfrac{-1}{x+1} \leq \dfrac{\sin(x)}{x+1} \leq \dfrac{1}{x+1} \\ \implies \left|\dfrac{\sin(x)}{x+1}\right| \leq \dfrac{1}{x+1} \\ \implies \int_{0}^{\infty} \left|\dfrac{\sin(x)}{x+1}\right| \, \mathrm dx \leq \int_{0}^{\infty}\dfrac{1}{x+1}\, \mathrm dx = \infty $$

I planned to use the comparison criterion and then the absolute convergence criterion. However, the idea did not work for me.

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    $\begingroup$ The convergence of this integral hinges on cancellation by the oscillation of the sine function, like how an alternating series converges conditionally. One of the best way of making use of this behavior is to perform integration by parts. $\endgroup$ – Sangchul Lee Jul 24 '20 at 21:06
  • $\begingroup$ Your integral is close to the integral of the sine-cardinal function. I suggest you to look at how people prove this integral exist to get inspired how you'll prove this one exists too. Also it's pointless to try to prove that this function is $L^1$, it's not. $\endgroup$ – BlueCharlie Jul 24 '20 at 21:11
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Notice that $$\int_0^\infty \frac{\sin x}{x+1}\,dx = \frac{-\cos x}{x+1}\Bigg|_0^\infty - \int_0^\infty \frac{\cos x}{(x+1)^2}\,dx = 1 - \int_0^\infty \frac{\cos x}{(x+1)^2}\,dx$$

and the last integral converges absolutely since $$\int_0^\infty \frac{\left|\cos x\right|}{(x+1)^2}\,dx \le \int_0^\infty \frac{dx}{(x+1)^2} = \int_1^\infty \frac{dx}{x^2} < +\infty.$$

However the original integral does not converge absolutely. Namely, we have $$x \in \bigcup_{k \in \mathbb{N}_0} \left[\frac\pi6+k\pi,\frac{5\pi}6+k\pi\right] \implies \left|\sin x\right| \ge \frac12$$ so $$\int_0^\infty \frac{\left|\sin x\right|}{x+1}\,dx \ge \frac12\sum_{k=0}^\infty \int_{\frac\pi6+k\pi}^{\frac{5\pi}6+k\pi} \frac{dx}{x+1} = \frac12\sum_{k=0}^\infty \ln \frac{\frac{5\pi}6+k\pi+1}{\frac\pi6+k\pi+1} = +\infty.$$

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  • $\begingroup$ To use the comparison test on line 4, shouldn't we do ${\left|\int_{0}^{\infty}\frac{\cos(x)}{(x+1)^2}dx\right|\leq \int_{0}^{\infty}\left|\frac{\cos(x)}{(x+1)^2}\right|dx < \int_{0}^{\infty}\frac{1}{(x+1)^2}dx...}$? Since to use comparison test we require both functions be positive on the interval $\endgroup$ – Riemann'sPointyNose Jul 24 '20 at 21:25
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    $\begingroup$ @Riemann'sPointyNose You're right, updated. $\endgroup$ – mechanodroid Jul 24 '20 at 21:30
  • $\begingroup$ Awesome :) (+1)'d your answer, super elegant $\endgroup$ – Riemann'sPointyNose Jul 24 '20 at 21:30
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The Cauchy criterion for improper integrals is:

An improper integral $\int_0^\infty f(x) \, dx$ is convergent if and only if for any $\epsilon > 0$ there exists $C_\epsilon > 0$ such that $\left|\int_a^b f(x) \, dx \right| < \epsilon$ for all $b > a> C_\epsilon.$

Since $x \mapsto \frac{1}{1+x}$is decreasing, by the second mean value theorem for integrals, there exists $\xi \in (a,b)$ such that

$$\left|\int_a^b \frac{\sin x}{1+x} \, dx\right| = \left|\frac{1}{1+a}\int_a^\xi \sin x\, dx\right| = \frac{|\cos a - \cos \xi|}{1+a}\leqslant \frac{2}{1+a}$$

For all $b > a > C_\epsilon = \frac{2}{\epsilon}-1$ we have the RHS less than $\epsilon$ and the Cauchy criterion is satisfied.

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Granted, the integral does not converge in the sense of Lebesgue. As a proper Riemann integral it does.

Here is another solution based which uses elementary facts about alternating series.

  • The sequence $a_n=\Big|\int^{(n+1)\pi}_{n\pi}\frac{\sin x}{x+1}\,dx\Big|$ is non decreasing and $a_n\xrightarrow{n\rightarrow\infty}0$. This is because on $[\pi n,\pi(n+1)]$, $\sin x=(-1)^n|\sin x|$, and so $$ \begin{align} a_{n+1}&=\int^{(n+2)\pi}_{(n+1)\pi}\frac{|\sin x|}{x+1}\,dx=\int^{(n+1)\pi}_{n\pi}\frac{|\sin(x+\pi)|}{x+\pi+1}\,dx\\ &\leq \int^{(n+1)\pi}_{n\pi}\frac{|\sin x|}{x+1}=a_n\leq\frac{\pi}{\pi n +1}\xrightarrow{n\rightarrow\infty}0 \end{align}$$

  • The series $s=\sum_{n\geq0}(-1)^na_n$ has partial sums $s_n=\int^{n\pi}_0\frac{\sin x}{1+x}\,dx$. Being a nice alternating series, $s_n$ converges.

  • In general, for $T>0$, let $[T]$ be its integer part. Then

$$ \Big|\int^{T\pi}_0\frac{\sin x}{x+1}\,dx - \int^{[T]\pi}_0\frac{\sin x}{x+1}\,dx\Big|\leq \int^{\pi T}_{[T]\pi}\frac{|\sin x|}{x+1}\leq \frac{\pi}{[T]\pi+1}\xrightarrow{T\rightarrow\infty}0$$

Therefore $\lim_{A\rightarrow\infty}\int^{A}_0\frac{\sin x}{x+1}\,dx$ exists and equal $s$.

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  • $\begingroup$ (+1) Very original answer! $\endgroup$ – Jean L. Jul 24 '20 at 21:50
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Let $$ a_n = \int_{\pi n}^{\pi(n+1)}\frac{|\sin x|}{x+1}dx. $$ Note that $$ \int_0^\infty \frac{\sin x}{x+1}dx = \sum_{n=0}^\infty (-1)^n a_n. $$

So, if the series converges, the integral must also converge. For any positive integer $n$, we can see that $a_n$ is positive, and we can rewrite it the following way: $$ \begin{align} a_n = \int_{\pi n}^{\pi(n+1)}\frac{|\sin x|}{x+1}dx &= \int_{\pi n}^{\pi(n+1)}\frac{\sin (x - \pi n)}{x+1}dx\\ &= \int_0^\pi\frac{\sin x}{x+1+\pi n}dx. \end{align} $$ This makes it clear that the denominator of $a_{n+1}$ is larger than the denominator of $a_n$ over the entire interval integrated, so $a_n$ must be decreasing. Furthermore, it is easy to see that $\lim_{n\to\infty} a_n=0$. Therefore, by the alternating series test, the integral converges.

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  • $\begingroup$ I don't think that solution was there when I started writing. Thanks for the nice helpful feedback. $\endgroup$ – Polygon Jul 24 '20 at 23:21
  • $\begingroup$ @Polygon: The convergence of $\sum_n(-1)^na_n$ implies convergence of $\int^{n\pi}_0\frac{\sin t}{t+1}\,dt$. The convergence of $\int^A_0\frac{\sin t}{t+1}\,dt$ requires a little proof. $\endgroup$ – Oliver Diaz Jul 24 '20 at 23:25
  • $\begingroup$ @OliverDiaz I'm not sure I'm understanding what you're saying. Are you saying I need to support the claim that $\int^A_0\frac{\sin t}{t+1}\,dt$ converges? Or are you saying there is simpler reasoning that I missed? Edit: Nevermind, I understand now. $\endgroup$ – Polygon Jul 24 '20 at 23:33
  • $\begingroup$ The former. Although this will be the case, in general convergence of $f(a_n)$ along a paericulr sequence $a_n\rightarrow\infty$, does not imply convergence of $f(b_n)$ along any other sequence $b_n\rightarrow\infty$. See my solution. $\endgroup$ – Oliver Diaz Jul 24 '20 at 23:42

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