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: $$\int\limits_0^{+\infty}\frac{\sin x}{x}\,\mathrm{d}x= \lim_{n \to +\infty}b_n=\lim_{n \to +\infty}a_n=\frac{\pi}{2}.$$

  • 3
    $\begingroup$ Why don't you apply the Riemann-Lebesgue lemma? en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma $\endgroup$
    – Botond
    Nov 24, 2019 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, 2019 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, 2019 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, 2019 at 2:52
  • $\begingroup$ @zhw of course, just edited. $\endgroup$ Nov 24, 2019 at 11:29

1 Answer 1


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 $C$.

  • $\begingroup$ Imagine if you were not tired. Totally fine indeed. Thank you! $\endgroup$ Nov 24, 2019 at 3:26

You must log in to answer this question.

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