# Help following this solution

So a couple of minutes ago I asked help on how to prove the following:

$$\displaystyle \lim_{n \to \infty} \displaystyle \int_0 ^{2\pi} \dfrac{\sin nx}{x^2 + n^2} dx = 0$$

I got some answers which are generally the same as the answer provided in my book which I don't understand, this is the (bulk of the) answer:

$$| \displaystyle \int_0 ^{2\pi} \dfrac{\sin nx}{x^2 + n^2} dx| \leq \displaystyle \displaystyle \int_0 ^{2\pi} |\dfrac{\sin nx}{x^2 + n^2}| dx \leq \displaystyle \int_0 ^{2\pi} \dfrac {dx}{n^2} = \dfrac{2\pi}{n^2}$$

What I don't understand is

• $\displaystyle \int_0 ^{2\pi} |\dfrac{\sin nx}{x^2 + n^2}| dx \leq \displaystyle \int_0 ^{2\pi} \dfrac {dx}{n^2}$, How did they find this?

• $\displaystyle \int_0 ^{2\pi} \dfrac {dx}{n^2} = \dfrac{2\pi}{n^2}$ , I also don't understand how they got this integral.

1) The modulus of Sine is always at most $1$ (for real arguments), the denominator is always greater or equal to $n^2$ because $x^2$ is positive.
We have $$0\leq\frac{1}{x^2+n^2}\leq \frac{1}{1+x^2}, \quad\forall n\geq 1,$$
and by the Riemann–Lebesgue lemma we have $$\lim_{n\to \infty}\int_0^{2\pi}\frac{\sin nx}{1+x^2}\mathrm dx=0,$$