I have to compute lim
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
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.
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