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I'm wondering if the series $$ \sum^{\infty}_{i=1}\int_{i\pi}^{(i+1)\pi} \dfrac{\sin x}{x} \ \ \ \ \ i\in \mathbb N$$
is convergent or not.

This is my attempt:

I consider the partial sum of it as a sequence, so I let $$I_n=\sum^{n-1}_{i=1}\int_{i\pi}^{(i+1)\pi} \dfrac{\sin x}{x}$$

By the property of integration, this is equal to $$ I_n= \int_{\pi}^{n\pi} \dfrac{\sin x}{x} \ dx $$

Then, by using integration by parts, I can get
$$ I_n=\int_{\pi}^{n\pi} \dfrac{\sin x}{x} \ dx=x^{-1} \cos x\bigg|_{n\pi}^\pi -\int ^{n\pi}_{\pi}\dfrac{\cos x}{x^2} \ dx$$

Thus the series is $$ \lim_{n\to \infty } \int_{\pi}^{n\pi} \dfrac{\sin x}{x} \ dx=\lim_{n\to \infty } x^{-1} \cos x\bigg|_{n\pi}^\pi -\int ^{n\pi}_{\pi}\dfrac{\cos x}{x^2} \ dx$$

Then, the first term is zero, the second term is an improper integration. And since $\left|\dfrac{\cos x}{x^2}\right| \le \dfrac{1}{x^2}$, the second term is convergent.

Thus the series is convergent.

Is this proof right? I strongly doubt my proof. Could you please check this? Thank you very much!

Angelo
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M_k
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  • You may search Dirichlet integral $\int_{0}^{\infty}\frac{\sin x}{x}dx$. It is well-known that the integral exists in improper sense: $\lim_{t\rightarrow+\infty}\int_{0}^{t}\frac{\sin x}{x}dx$ exists and can be computed using residue theory in complex analysis. However, $\int_{0}^{\infty}\left|\frac{\sin x}{x}\right|dx=+\infty$. – Danny Pak-Keung Chan Nov 26 '22 at 23:10
  • The first piece of your integration by parts is not correct but you can easily fix it. – Anne Bauval Nov 26 '22 at 23:13
  • @Anne Bauval. Yeah, you are right, I fixed it. Are there any other problems with this proof? Thanks – M_k Nov 26 '22 at 23:21
  • @Danny Pak-Keung Chan. I see. I think that means my conclusion that this series is convergent is right but is my proof okay? – M_k Nov 26 '22 at 23:22
  • @amWhy, in titles is it possible to use $\lim\limits_{n\to\infty}$? Here seems that it is accepted. Please, read it and tell me if you agree. – Angelo Nov 26 '22 at 23:46
  • @Angelo Why wouldn't it be possible to use $\lim_{n\to\infty}$ in the title? It is supported in inline MathJax. – FShrike Nov 26 '22 at 23:49
  • @FShrike, $\lim_{n\to\infty}$ looks ugly, so if it is possible I would prefer to use $\lim\limits_{n\to\infty}$, moreover, if you click on the link in my previous comment, it seems that it is accepted by this community. – Angelo Nov 26 '22 at 23:57
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    This type of question has been discussed in MSE many times. Here is on instance I am trying to show $\int^\infty_0\frac{\sin(x)}{x}dx=\frac{\pi}{2}$ – Mittens Nov 27 '22 at 00:41

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