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
 A: 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 $n\to\infty$.
A: 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. 
A: 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, 
$$
f(2n) - f(n) \xrightarrow[n\to\infty]{} \ell - \ell = 0.
$$
A: 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 : $\lim \limits_{n\to +\infty} \int_{1}^{2}\frac{\sin(nx)}{x}\mathrm{d}x \neq \int_{1}^{2} \lim \limits_{n\to +\infty}\frac{\sin(nx)}{x}\mathrm{d}x$.
