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

Question

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? – Angina Seng Sep 25 '18 at 15:05 — Angina Seng, Sep 25 '18 at 15:05
By Lebesgue-Riemann Lemma, the limit is $0$ – Jakobian Sep 25 '18 at 15:06 — Jakobian, Sep 25 '18 at 15:06
@LordSharktheUnknown $\infty$ – jasmine Sep 25 '18 at 15:06 — jasmine, Sep 25 '18 at 15:06
Hint: the integral is $\int_{na}^{nb}\frac{\sin x,dx}{x}$. – J.G. Sep 25 '18 at 15:08 — J.G., Sep 25 '18 at 15:08

Angina Seng · Accepted Answer · 2018-09-25T15:12:15.363

7

$$\int_a^b\frac{\sin nx}x\,dx=\left[-\frac{\cos nx}{nx}\right]_a^b- \int_a^b\frac{\cos nx}{nx^2}\,dx =\frac{\cos na}{na}-\frac{\cos nb}{nb}-\frac1n\int_a^b\frac{\cos nx}{x^2}\,dx.$$ There are some very convenient denominators here!

edited Sep 25 '18 at 15:12

answered Sep 25 '18 at 15:09

Angina Seng

158,341

1

Second equation is wrong. Should be $\frac{\cos(na)}{na}-\frac{\cos(nb)}{nb}$. Good answer. +1 – Jakobian Sep 25 '18 at 15:11

Robert Z · Answer 2 · 2018-09-25T15:14:22.737

3

Hint. We assume that $0<a<b$. 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?).

edited Sep 25 '18 at 15:14

answered Sep 25 '18 at 15:08

Robert Z

145,942

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

2 Answers2