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I am trying to show that the following integral is convergent but not absolutely.

$$\int_0^\infty\frac{\sin x}{x}dx.$$

My attempt:

I first obtained the taylor series of $\int_0^x\frac{sin x}{x}dx$ which is as follows: $$x-\frac{x^3}{3 \times 3!}+\frac{x^5}{5\times5!}-\frac{x^7}{7 \times 7!}+\cdots = \sum_{n=0}^\infty (-1)^n\frac{x^{(2n+1)}}{(2n+1) \times (2n+1)!} $$

Now $\int_0^\infty\frac{\sin x}{x}dx=\lim_{x\to \infty} \sum_{n=0}^\infty (-1)^n\frac{x^{(2n+1)}}{(2n+1) \times (2n+1)!}$

and I got stuck here! What is the next step?

Majid
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1 Answers1

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I don't see how Taylor expansions would help. Since $\sin(x)/x$ is bounded on $\mathbb R\setminus\{0\}$, the only place to check for absolute convergence is towards $+\infty$, where you can't take Taylor expansions. To test for absolute convergence, one can see that

\begin{align}\int_0^\infty\frac{|\sin(x)|}x~\mathrm dx&=\sum_{n=0}^\infty\int_{n\pi}^{(n+1)\pi}\frac{|\sin(x)|}x~\mathrm dx\\&=\sum_{n=0}^\infty\int_0^\pi\frac{\sin(x)}{x+n\pi}~\mathrm dx\\&=\sum_{n=0}^\infty\left[\frac1{(n+1)\pi}+\frac1{n\pi}-\int_0^\pi\frac{\cos(x)}{(x+n\pi)^2}~\mathrm dx\right]\tag{ibp}\end{align}

What can you conclude about the convergence of this?

Simply Beautiful Art
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  • Thanks for your help! However, from https://math.stackexchange.com/questions/793595/convergence-i-int-0-infty-frac-sin-xxsdx?rq=1 we can see that my case is convergence but not absolutely. So, there would be some solution to it which I am looking for. – Majid May 17 '19 at 00:20
  • It's unclear what you're trying to ask. – Simply Beautiful Art May 17 '19 at 00:27
  • @Majid: Your integral converges but not absolutely; do you have a question about one of the solutions posted there? – Clayton May 17 '19 at 00:47
  • @SimplyBeautifulArt Actually, I was trying to find if it is possible to solve that integral somehow by using Taylor series, however I found many solutions in this post https://math.stackexchange.com/questions/5248/evaluating-the-integral-int-0-infty-frac-sin-x-x-dx-frac-pi-2/8350#8350 but no solution by using Taylor series which may show there is no such a way to solve that integral. Moreover, these series are valid only around zero not going to infinity. – Majid May 17 '19 at 00:59
  • @Clayton No, thanks Clayton! – Majid May 17 '19 at 00:59
  • @Majid As I said in my answer, I don't see how Taylor is relevant here. – Simply Beautiful Art May 17 '19 at 01:01