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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.

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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.

2) Integrating a constant function over an interval just gives the value of the function times the length of the interval.

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  • $\begingroup$ I see. About 2), it's unbelievable that my textbook doesn't mention this fact, does this rule have a formal name? Also, why do you need to get the modulus, doesn't it hold without the modulus? $\endgroup$ – Ylyk Coitus Mar 1 '13 at 16:44
  • $\begingroup$ It follows immediately from any reasobale definition of the integral. $\endgroup$ – Rasmus Mar 1 '13 at 16:45
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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 $$\lim_{n\to \infty}\int_0^{2\pi}\frac{\sin nx}{1+x^2}\mathrm dx=0,$$

thus we find the result.

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