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

Question

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

Answer 1

This can be cleaned up further. You've done almost everything already.

First, you mean that you recognize the expression to be integrated as the derivative of that quotient, not an equality. But after doing this, you can evaluate.

$$\sum_{n=1}^\infty \frac{\cos(2(n+1)\pi)}{2(n+1)\pi} - \frac{\cos 2n \pi}{2n\pi} =\frac{1}{2\pi}\sum_{n=1}^\infty \frac{1}{n+1} - \frac{1}{n}$$

This is a telescoping series with limit $1$, so you get $\frac{1}{2\pi}$.