# Evaluate $\lim_{n \to \infty} \int_{1}^{2}\frac{\sin(nx)}{x}dx$

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

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 $n\to\infty$.

• 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? Jun 6, 2016 at 22:44
• @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 $0$. Jun 6, 2016 at 22:49
• @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 $0$. Jun 6, 2016 at 23:02

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, $$f(2n) - f(n) \xrightarrow[n\to\infty]{} \ell - \ell = 0.$$

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.

• Thanks but we have not covered that lemma Jun 6, 2016 at 22:54
• Oh, then in that case André Nicolas provided a perfect answer that you can see :)
– Hmm.
Jun 6, 2016 at 22:57

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