2
$\begingroup$

Let $$a_n=\int\limits_0^{\frac{\pi}{2}}\sin(2n x)\,\cot x\,\mathrm{d}x ~~~\textrm{ and} ~~~ b_n=\int\limits_0^{\frac{\pi}{2}}\frac{\sin(2n x)}{x}\,\mathrm{d}x.$$ Prove that $a_n-b_n \to 0.$

Attempt. We have that $$a_n-b_n = \int\limits_0^{\frac{\pi}{2}}\sin(2n x)\,\left(\cot x-\frac{1}{x}\right)\,\mathrm{d}x.$$ Integration by parts would give: $$a_n-b_n=\left[\log\left(\frac{\sin x}{x}\right)\sin(2nx)\right]_0^{\frac{\pi}{2}}-2n\int\limits_0^{\frac{\pi}{2}}\cos(2n x)\,\log\left(\frac{\sin x}{x}\right)\,\mathrm{d}x$$ $$=-2n\int\limits_0^{\frac{\pi}{2}}\cos(2n x)\,\log\left(\frac{\sin x}{x}\right)\,\mathrm{d}x$$ but this doesn't seem to go any further.

On the other hand, from Integral $\int_0^\pi \cot(x/2)\sin(nx)\,dx$ we have $a_n=\frac{\pi}{2}$, so:

$$b_n-a_n=\int\limits_0^{\frac{\pi}{2}}\left(\frac{\sin(2n x)}{x}-1\right)\,\mathrm{d}x,$$ so: $$|b_n-a_n|\leqslant \int\limits_0^{\frac{\pi}{2}}\left|\frac{\sin(2n x)}{x}-1\right|\,\mathrm{d}x,$$ but I also didn't manage to get this any further.

Thanks in advance for the help.


Edit. Consequence of the above limit is the evaluation of the Dirichlet integral: $$\int\limits_0^{+\infty}\frac{\sin x}{x}\,\mathrm{d}x= \lim_{n \to +\infty}b_n=\lim_{n \to +\infty}a_n=\frac{\pi}{2}.$$

$\endgroup$
6
  • 3
    $\begingroup$ Why don't you apply the Riemann-Lebesgue lemma? en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma $\endgroup$
    – Botond
    Nov 24 '19 at 2:00
  • $\begingroup$ I had in my mind an elementary proof, if possible of course, that would use only tools such as integration by parts, boundness arguments etc. $\endgroup$ Nov 24 '19 at 2:04
  • 1
    $\begingroup$ You've made a mistake somewhere, as $\log (-\sin x/x)$ is not defined on this interval. $\endgroup$
    – zhw.
    Nov 24 '19 at 2:25
  • 1
    $\begingroup$ You can prove the Riemann-Lebesgue lemma for $C^1$ functions by very elementary means (integration by parts). $\endgroup$ Nov 24 '19 at 2:52
  • $\begingroup$ @zhw of course, just edited. $\endgroup$ Nov 24 '19 at 11:29
2
$\begingroup$

N.B.: I think it's good, but I'm tired, so it might be wrong. I'm going to check it after I wake up; read it carefully until then.

Let $f(x)=\cot(x)-\frac{1}{x}$ with $f(0)=0$.
We have that \begin{align} \left|\int_0^{\pi/2}f(x)\sin(2nx)\mathrm{d}x\right| &=\left|-\int_0^{\pi/2}f'(x)\frac{\cos(2nx)}{2n}\mathrm{d}x+\left[\frac{f(x)\cos(2nx)}{2n}\right]_0^{\pi/2}\right|\\ &=\left|-\int_0^{\pi/2}f'(x)\frac{\cos(2nx)}{2n}\mathrm{d}x+\left(-\frac{2}{\pi}\right)\frac{\cos(n \pi)}{2n}\right|\\ &=\left|-\int_0^{\pi/2}f'(x)\frac{\cos(2nx)}{2n}\mathrm{d}x-\frac{(-1)^n}{n \pi}\right|\\ &\leqslant \left|\int_0^{\pi/2}f'(x)\frac{\cos(2nx)}{2n}\mathrm{d}x\right|+\left|\frac{(-1)^n}{n\pi}\right| \\ &\leqslant \int_0^{\pi/2}\left|f'(x)\frac{\cos(2nx)}{2n}\right|\mathrm{d}x+\frac{1}{n\pi}\\ &\leqslant \int_0^{\pi/2}\left|C\frac{\cos(2nx)}{2n}\right|\mathrm{d}x+\frac{1}{n\pi}\\ &\leqslant \frac{C}{2n}\int_0^{\pi/2}\left|\cos(2nx)\right|\mathrm{d}x+\frac{1}{n\pi}\\ &\leqslant \frac{C}{2n}\int_0^{\pi/2}1\mathrm{d}x+\frac{1}{n\pi}\\ &\leqslant \frac{\pi C}{4n}+\frac{1}{n\pi} \to 0 \end{align} Since the absolute value of the derivative of $f$ on $[0, \pi/2]$ is bounded by a constant $C$.

$\endgroup$
1
  • $\begingroup$ Imagine if you were not tired. Totally fine indeed. Thank you! $\endgroup$ Nov 24 '19 at 3:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.