# Find $\lim_{n\rightarrow \infty}\int_{a}^{b} \frac{\sin (nx)}{x} dx$

For $$0 < a < b$$, find $$\lim_{n\rightarrow \infty}\int_{a}^{b} \frac{\sin (nx)}{x} dx.$$

My attempt : $$\int_{a}^{b} \frac{\sin (nx)}{x} dx= \sin x\int_{a}^{b}\frac{1}{x}dx -\int_{a}^{b}\cos x\cdot \log x dx$$ after that I'm not able to proceed further.

Pliz help me. Any hints/solution?

Thanks

• Where did the $n$ go to? Sep 25, 2018 at 15:05
• By Lebesgue-Riemann Lemma, the limit is $0$ Sep 25, 2018 at 15:06
• @LordSharktheUnknown $\infty$ Sep 25, 2018 at 15:06
• Hint: the integral is $\int_{na}^{nb}\frac{\sin x\,dx}{x}$.
– J.G.
Sep 25, 2018 at 15:08

$$\int_a^b\frac{\sin nx}x\,dx=\left[-\frac{\cos nx}{nx}\right]_a^b- \int_a^b\frac{\cos nx}{nx^2}\,dx =\frac{\cos na}{na}-\frac{\cos nb}{nb}-\frac1n\int_a^b\frac{\cos nx}{x^2}\,dx.$$ There are some very convenient denominators here!
• Second equation is wrong. Should be $\frac{\cos(na)}{na}-\frac{\cos(nb)}{nb}$. Good answer. +1 Sep 25, 2018 at 15:11
Hint. We assume that $$0. By letting $$t=nx$$, we have that $$\int_{a}^{b} \frac{\sin (nx)}{x} dx=\int_{na}^{nb} \frac{\sin (t)}{t} dt =\int_{0}^{nb} \frac{\sin (t)}{t} dt-\int_{0}^{na} \frac{\sin (t)}{t} dt.$$ Now note that the integral $$\int_{0}^{\infty} \frac{\sin (t)}{t} dt$$ is convergent (see for example $$\int_{0}^{\infty}\frac{\sin x}{x}dx$$ converges?).