How to calculate the series ##\sum_{x = 1}^{\infty} \frac{sin(x)}{x}##

[shrub_broom]

Homework Statement

I have known how to calculate the integral$\int_{0}^{\infty} \frac{sin(x)}{x}$, however I find it hard to calculate this one since the similar technique(by parametric integral) cannot be simply applied on series.
Any approach or hint is welcomed.

Relevant Equations

$\sum_{x = 1}^{\infty} \frac{sin(x)}{x} = \frac{\pi - 1}{2}$

Maybe introduce a parametric factor can be help.

 

[BvU]

Is this homework ? Some activity on your part is required in PF guidelines

 

[shrub_broom]

No, just for fun.

 

[bobob]

Problem Statement: I have known how to calculate the integral$\int_{0}^{\infty} \frac{sin(x)}{x}$, however I find it hard to calculate this one since the similar technique(by parametric integral) cannot be simply applied on series.
Any approach or hint is welcomed.
Relevant Equations: $\sum_{x = 1}^{\infty} \frac{sin(x)}{x} = \frac{\pi - 1}{2}$

Maybe introduce a parametric factor can be help.

You could try replacing $\sin(x)$ with it's representation as an infinite product.

 

[vela]

Express the sine in terms of complex exponentials.

 

[BvU]

Similar thread:

Homework Statement

Determine whether the series Ʃ(1 to infinity) sinx / x converges or diverges.

Homework Equations

This question appears in the integral test section, but as far as i know the integral test can only be used for decreasing functions, right?

The Attempt at a Solution

Using the ratio test, limx->infinity sin(x+1)/(x+1)*(sinx/x)=limx->infinity x*sin(x+1)/((x+1)(sinx))
This is where i got stuck-this limit oscillates between positive infinity and negative infinity.
Using the root test, i need to find the limit of (sinx)^(1/x) as x approaches infinity, which also gets me nowhere.

we have not done taylor series yet so I'm sure there is a relatively simple approach to this question...please help?

 

[Fred Wright]

Your integral is egregiously improper. You must normalize it by redefining the function $sinc(x)=1$ when $x=0$ and $sinc(x)= \frac{sin(x)}{x}$ when $x\neq 0$. Please see wikipedia page on sinc function.

 

[haruspex]

Similar thread:

Yes, but that is not looking for the value. @vela's method works.