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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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4 Answers 4

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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? $\endgroup$
    – cwbrd
    Jun 6, 2016 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 $0$. $\endgroup$ Jun 6, 2016 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 $0$. $\endgroup$ Jun 6, 2016 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 $\endgroup$
    – cwbrd
    Jun 6, 2016 at 22:54
  • $\begingroup$ Oh, then in that case André Nicolas provided a perfect answer that you can see :) $\endgroup$
    – Hmm.
    Jun 6, 2016 at 22:57
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Here is another tool that you can use :

According to the Bonnet's theorem or the second form of the mean value theorem for definite integral with here the right hypothesis (the two functions are continuous on $[1,2]$ and one is monotonic), we can find $c \in ]1,2[$ such that :

$\int_{1}^{2}\frac{\sin(nx)}{x}\mathrm{d}x= \dfrac{1}{1}\int_{1}^{c}\sin(nx)\mathrm{d}x + \dfrac{1}{2}\int_{c}^{2}\sin(nx)\mathrm{d}x =\dfrac{\cos(x)-\cos(nc)}{n} + \dfrac{\cos(nc)-\cos(2x)}{2n}$.

Then for all $x\in [1,2]$ : $\left \vert \int_{1}^{2}\frac{\sin(nx)}{x}\mathrm{d}x \right \vert \le \dfrac{2}{n} + \dfrac{2}{2n} = \dfrac{3}{n}$.

Hence $\lim \limits_{n\to +\infty} \int_{1}^{2}\frac{\sin(nx)}{x}\mathrm{d}x = 0$.

$\underline{\textbf{NB}}$ : Notice that maybe at first, we wanted to use an interversion ($\lim / \int$) theorem. But here for the sequence of functions $g_{n}(x)=\dfrac{\sin(nx)}{x}$ defined on $[1,2]$ and for $n\in \mathbb{N}$, we need the uniform convergence of $(g_n(x))_{n\ge 0}$ on $[1,2]$. As you can see for several values of $x\in[1,2]$, $\lim\limits_{n\to +\infty} g_{n}(x)$ is not continuous so you cannot apply that theorem. It's a case where : $\lim \limits_{n\to +\infty} \int_{1}^{2}\frac{\sin(nx)}{x}\mathrm{d}x \neq \int_{1}^{2} \lim \limits_{n\to +\infty}\frac{\sin(nx)}{x}\mathrm{d}x$.

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