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
Best Answer
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 $n\to\infty$.