# Finding $\sum^{\infty}_{n=1}\int^{2(n+1)\pi}_{2n\pi}\frac{x\sin x+\cos x}{x^2}$

Finding value of $\displaystyle \sum^{\infty}_{n=1}\int^{2(n+1)\pi}_{2n\pi}\frac{x\sin x+\cos x}{x^2}$

Try:$$\frac{\cos x}{x} = -\bigg(\frac{x\sin x+\cos x}{x^2}\bigg)$$

So $$\sum^{\infty}_{n=1}\bigg(\frac{\cos x}{x}\bigg)\bigg|^{2(n+1)\pi}_{2n\pi}$$

Could some help me to solve it,Thanks

• I have tried to straighten this post out. Feel free to edit further if I've changed your intended meaning. – Alfred Yerger Jan 16 '18 at 1:53
• Thanks Alfred Yerger. – DXT Jan 16 '18 at 1:54

$$\sum_{n=1}^\infty \frac{\cos(2(n+1)\pi)}{2(n+1)\pi} - \frac{\cos 2n \pi}{2n\pi} =\frac{1}{2\pi}\sum_{n=1}^\infty \frac{1}{n+1} - \frac{1}{n}$$
This is a telescoping series with limit $1$, so you get $\frac{1}{2\pi}$.