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I have to compute $$\lim_{n \to \infty} \int_{1}^{2}\frac{\sin(nx)}{x}dx$$ I have tried to tackle it in different ways but I'm getting nowhere. In particular, I used substitution to obtain $$\lim_{n \to \infty} \int_{1}^{2}\frac{\sin(nx)}{x}dx = \lim_{n \to \infty} \int_{n}^{2n}\frac{\sin(u)}{u}du$$

But from here I'm not sure about what to do. I've found information about $\int_0^\infty\frac{\sin(nx)}{x}dx$, but I don't see if and how I could relate my integral with that one.

Any hints? Thanks

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Integrate by parts, letting $u=\frac{1}{x}$ and $dv=\sin(nx)\,dx$. Then $du=-\frac{1}{x^2}\,dx$ and we can take $v=-\frac{\cos nx}{n}$.

Our integral is equal to $$\left. -\frac{1}{x}\cdot \frac{\cos(nx)}{n}\right|_1^2 -\int_1^2 \frac{\cos nx}{nx^2}\,dx.$$ Both parts $\to 0$ as $n\to\infty$.

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  • $\begingroup$ I tried that, what I wasn't sure about is if I can conclude that the integral on the right tends to 0...? I mean, am I not left to deal with the evaluation of \int cos(nx)/x^2? $\endgroup$ – Generalbrus Jun 6 '16 at 22:44
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    $\begingroup$ @Generalbrus: We don't need to evaluate, estimation is enough. For $\left|\frac{\cos nx}{x^2}\right|$ is bounded above (by $1$) on our interval, so the integral has absolute value $\le \frac{1}{n}$, and therefore has limit $0$. $\endgroup$ – André Nicolas Jun 6 '16 at 22:49
  • $\begingroup$ @Generalbrus: Your approach will also work nicely. One can show (for example by an integration by parts argument!) that $\int_n^{2n}\frac{\sin x}{x}\,dx$ has limit $0$. $\endgroup$ – André Nicolas Jun 6 '16 at 23:02
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Cheating a bit, as this invokes another result as a blackbox, at the place marked $(\dagger)$.

Do the substitution $u=nx$, as you started: $$ \int_1^2 dx\frac{\sin nx}{x} = \int_{n}^{2n} du\frac{\sin u}{u} = \int_{0}^{2n} du\frac{\sin u}{u} - \int_{0}^{n} du\frac{\sin u}{u} $$ and now use the fact$^{(\dagger)}$ that the (improper) integral $\int_{0}^{\infty} du\frac{\sin u}{u}$ converges, i.e. the function $$ f(x) \stackrel{\rm def}{=} \int_{0}^{x} du\frac{\sin u}{u} $$ converges to a finite limit $\ell$ when $x\to \infty$. So by theorems of operations on limits, $$ f(2n) - f(n) \xrightarrow[n\to\infty]{} \ell - \ell = 0. $$

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Edit: Although detailed calculation is preferred, one may use the following : The function $f(x)= ~\frac{1}{x}$ is absolutely Riemann integrable on $[1,2]$, and hence Riemann-Lebesgue Lemma implies the desired result.

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  • $\begingroup$ Thanks but we have not covered that lemma $\endgroup$ – Generalbrus Jun 6 '16 at 22:54
  • $\begingroup$ Oh, then in that case André Nicolas provided a perfect answer that you can see :) $\endgroup$ – Hmm. Jun 6 '16 at 22:57

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