0
$\begingroup$

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

$\endgroup$
4
  • 1
    $\begingroup$ Where did the $n$ go to? $\endgroup$
    – Angina Seng
    Sep 25, 2018 at 15:05
  • $\begingroup$ By Lebesgue-Riemann Lemma, the limit is $0$ $\endgroup$
    – Jakobian
    Sep 25, 2018 at 15:06
  • $\begingroup$ @LordSharktheUnknown $\infty$ $\endgroup$
    – jasmine
    Sep 25, 2018 at 15:06
  • $\begingroup$ Hint: the integral is $\int_{na}^{nb}\frac{\sin x\,dx}{x}$. $\endgroup$
    – J.G.
    Sep 25, 2018 at 15:08

2 Answers 2

Reset to default
7
$\begingroup$

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

$\endgroup$
1
  • 1
    $\begingroup$ Second equation is wrong. Should be $\frac{\cos(na)}{na}-\frac{\cos(nb)}{nb}$. Good answer. +1 $\endgroup$
    – Jakobian
    Sep 25, 2018 at 15:11
3
$\begingroup$

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?).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .