I was playing with integrals, and came up with

$$L=\lim_{n\to\infty}\int_0^1 \frac{1}{(1+|\sin (nx)|)^2}dx.$$

Conjecture: $L=\dfrac{4}{3\pi}$

Is my conjecture true?

Remarks on numerical investigation:

Desmos and Wolfram don't do a good job with numerical investigation of this limit, but we can consider the series $f(n)=\dfrac{1}{n}\sum\limits_{k=1}^n \dfrac{1}{(1+|\sin{k}|)^2}$.




This suggests that $\lim\limits_{n\to\infty}f(n)=\frac{4}{3\pi}$.

Using Riemann sums, we have $\lim\limits_{n\to\infty}f(n)=L$.

My attempt:

I tried to use $\,\sin nx = \dfrac{1}{2i}(e^{nxi}-e^{-nxi})\,$ in $\;\displaystyle\int_0^1 \dfrac{1}{(1+|\sin (nx)|)^2}\,\mathrm dx\;,\;\;$ to no avail.

I also tried to use complex numbers in the series $f(n)$, as in answers to a question about $\sum_{n=1}^{\infty} \frac{\cos (n)}{n}$, to no avail.

  • 9
    $\begingroup$ For the periodic function $g(x)$ with the period $p$: $$\lim_{n\to\infty}\int_a^b f(x)g(nx)dx=\frac1p\int_a^bf(t)dt\int_0^pg(x)dx$$ $$\lim_{n\to\infty}\int_0^1 \frac{1}{(1+|\sin (nx)|)^2}dx=\frac1\pi\int_0^\pi\frac{dx}{(1+\sin x)^2}\overset{\tan \frac x2=t}{=}\frac2\pi\int_0^\infty\frac{1+t^2}{(1+t)^4}dt$$ $\endgroup$
    – Svyatoslav
    Jul 16, 2023 at 5:15
  • 4
    $\begingroup$ @Svyatoslav. Simple and very nice (as usual). Cheers :-) $\endgroup$ Jul 16, 2023 at 5:44
  • $\begingroup$ @Svyatoslav Does this identity have a name? Looks pretty powerful $\endgroup$
    – Imago
    Jul 16, 2023 at 21:39
  • 1
    $\begingroup$ @Imago, I'm not sure about the name, but there is a lot of proofs and applications of this lemma on the sites - please, see the answer of f@Mittens below (see also approach0.xyz/search/… ) $\endgroup$
    – Svyatoslav
    Jul 16, 2023 at 22:08

3 Answers 3


Use the substitution $y = nx$:

$$I = \lim_{n\to\infty}\frac{1}{n}\int_0^n\frac{dy}{(1+|\sin y|)^2} = \lim_{n\to\infty}\frac{1}{n}\int_0^{\pi\lfloor\frac{n}{\pi}\rfloor}\frac{dy}{(1+|\sin y|)^2}+\frac{1}{n}\int_{\pi\lfloor\frac{n}{\pi}\rfloor}^n\frac{dy}{(1+|\sin y|)^2}$$

Since the integrand is $\pi$-periodic, the second piece is bounded by $\frac{\pi}{n}$ and goes to zero by squeeze theorem. The first piece can be further divided into

$$I = \lim_{n\to\infty}\frac{1}{\pi}\int_0^{\pi}\frac{dy}{(1+|\sin y|)^2} - \frac{\left\{\frac{n}{\pi}\right\}}{n}\int_0^{\pi}\frac{dy}{(1+|\sin y|)^2}$$

where again the second piece is bounded by $\frac{\pi}{n}$ and goes to zero by squeeze theorem. To evaluate the integral, from symmetry considerations we have the equivalent

$$I = \frac{2}{\pi}\int_0^{\frac{\pi}{2}}\frac{dy}{(1+\cos y)^2} = \frac{1}{\pi}\int_0^{\frac{\pi}{2}}\frac{1}{2}\sec^2\left(\frac{y}{2}\right)\Bigr(1+\tan^2\left(\frac{y}{2}\right)\Bigr)\:dy$$

$$= \frac{1}{\pi}\Biggr[\tan\left(\frac{y}{2}\right)+\frac{1}{3}\tan^3\left(\frac{y}{2}\right)\Biggr]_0^{\frac{\pi}{2}} = \boxed{\frac{4}{3\pi}}$$


Just to provide an alternitive solution: \begin{align*} & \lim_{n\rightarrow \infty} \int_0^1{\left( \frac{1}{1+\left| \sin \left( nx \right) \right|} \right) ^2\mathrm{d}x} \\& =\lim_{n \rightarrow \infty} \left( \lim_{r\rightarrow 1^+} \int_0^1{\left( \frac{1}{r+\left| \sin \left( n x \right) \right|} \right) ^2\mathrm{d}x} \right) \\& =\lim_{n \rightarrow \infty} \left( \lim_{r\rightarrow 1^+} \int_0^1{\frac{\partial}{\partial r}\left( \frac{-1}{r+\left| \sin \left( n x \right) \right|} \right) \mathrm{d}x} \right) \\& \stackrel{\mathrm{(1)}}{=}\lim_{n \rightarrow \infty} \left( \lim_{r\rightarrow 1^+} \left(\frac{\mathrm{d}}{\mathrm{d}r}\int_0^1{\frac{-1}{r+\left| \sin \left( n x \right) \right|}\mathrm{d}x}\ \right) \right) \\& \stackrel{\mathrm{(2)}}{=} \lim_{r\rightarrow 1^+} \left( \lim_{n \rightarrow \infty} \left( \frac{\mathrm{d}}{\mathrm{d}r}\int_0^1{\frac{-1}{r+\left| \sin \left( n x \right) \right|}\mathrm{d}x} \right) \right) \\& \stackrel{\mathrm{(3)}}{=} \lim_{r\rightarrow 1^+} \left( \frac{\mathrm{d}}{\mathrm{d}r}\left( \lim_{n \rightarrow \infty} \int_0^1{\frac{-1}{r+\left| \sin \left( nx \right) \right|}\mathrm{d}x} \right) \right) \\& \stackrel{\mathrm{(4)}}{=}\lim_{r\rightarrow 1^+} \left( \frac{\mathrm{d}}{\mathrm{d}r}\left( \lim_{n\rightarrow \infty} \int_0^1{\frac{-1}{r+\left| \sin \left( n\pi x \right) \right|}\mathrm{d}x}\right) \right) \\& \stackrel{\mathrm{(5)}}{=}\lim_{r\rightarrow 1^+} \left( \frac{\mathrm{d}}{\mathrm{d}r}\left(\lim_{n\rightarrow \infty} \int_0^{n\pi}{\frac{-1/n\pi}{r+\left| \sin \left( t \right) \right|}\mathrm{d}t} \right) \right) \\& \stackrel{\mathrm{(6)}}{=}\lim_{r\rightarrow 1^+} \left( \frac{\mathrm{d}}{\mathrm{d}r}\left( \lim_{n\rightarrow \infty} \int_{\left( n-1 \right) \pi}^{n\pi}{\frac{-1/\pi}{r+\left| \sin \left( t \right) \right|}\mathrm{d}t}\right) \right) \\& \stackrel{\mathrm{(7)}}{=} \lim_{r\rightarrow 1^+} \left( \frac{\mathrm{d}}{\mathrm{d}r}\left( \lim_{n\rightarrow \infty} \int_0^{\pi}{\frac{-1/\pi}{r+\left| \sin \left( t \right) \right|}\mathrm{d}t} \right) \right) \\& =\lim_{r\rightarrow 1^+} \left( \frac{\mathrm{d}}{\mathrm{d}r}\left(\int_0^{\pi}{\frac{-1/\pi}{r+\left| \sin \left( t \right) \right|}\mathrm{d}t}\right) \right) \\& =\lim_{r\rightarrow 1^+} \left( \frac{\mathrm{d}}{\mathrm{d}r}\left( -\frac{2}{\pi}\left( \frac{\mathrm{arctan} \left( \begin{array}{c}\sqrt{r^2-1}\end{array} \right)}{\begin{array}{c}\sqrt{r^2-1}\end{array}} \right) \right) \right) \\& =\frac{2}{\pi}\lim_{r\rightarrow 1^+} \left( \frac{r^2\mathrm{arctan} \left( \begin{array}{c}\sqrt{r^2-1}\end{array} \right)-\begin{array}{c}\sqrt{r^2-1}\end{array}}{r\left( \begin{array}{c}\sqrt{r^2-1}\end{array} \right) ^3} \right) \\& \stackrel{\mathrm{(8)}}{=}\frac{2}{\pi}\lim_{r\rightarrow 0^+} \left( \frac{\left( r^2+1 \right) \mathrm{arctan} \left( r \right)-r}{\sqrt{r^2+1}r^3} \right) \\& =\frac{2}{\pi}\lim_{r\rightarrow 0^+} \left( \frac{\left( r^2+1 \right) \left( r-r^3/3+o\left( r^5 \right) \right)-r}{r^3} \right) \\& =\frac{2}{\pi}\lim_{r\rightarrow 0^+} \left( \frac{2}{3}+o\left( r \right) \right) =\frac{4}{3\pi} \end{align*} (1) Since the integrand is bounded, Lebesgue Dominated Convergence theorem could be utilized here to drag the $d$ out.

(2) By Moore-Osgood theorem, we can interchange the limits.

(3) Same as (2) here.

(4) The limit won't change if I put a extra $\pi$ in there.

(5) Let $n\pi x=t$ here.

(6) Use Stolz–Cesàro theorem here.

(7) By the periodicity of the integrand, we can drag the domain of integration back to $[0,\pi]$

(8) Let $\sqrt{r^2-1}\rightsquigarrow r$ here.


This can also be seen as an application of Fejer's theorem which is a Generalization of Riemann-Lebesgue's Lemma, also discussed here

Theorem: Suppose $g$ is a $T$-periodic function and that $g\in L^{loc}_q(\mathbb{R},m)$ for some $1<q\leq \infty$. For any $\infty<a< b<\infty$, and $f\in L_p([a,b])$, $\frac1p+\frac1q=1$, the limit \begin{align} I(f; g,[a,b]):=\lim_{\lambda\rightarrow\infty}\int^b_a f(t) g(\lambda t)\,dt\tag{1}\label{one} \end{align} exists and \begin{align} I(f; g, [a,b])=\Big(\frac1T\int^T_0 g(t)\,dt\Big)\Big(\int^b_a f(t)\,dt\Big)\tag{2}\label{two} \end{align}

(See Kahane, C. S., Generalizations of the Riemann-Lebesgue and Cantor-Lebesgue lemmas, Czechoslovak Mathematical Journal, Vol 30 (1980), No. 1, 108-117 for results of these type).

In the OP's case, $f(t)=\frac{1}{(1+|\sin x|)^2}$ is a $\pi$-periodic function and $f\in L_p([0,\pi])$ for all $1\leq p<\infty$. Let $g(t)=\mathbb{1}_{[0,1]}(t)$, $0\leq t\leq \pi$. Then $$\lim_n\int^1_0\frac{1}{(1+|\sin nt|)^2}\,dt=\Big(\frac{1}{\pi}\int^\pi_0\frac{1}{(1+|\sin t|)^2}\,dt\Big)\Big(\int^\pi_0\mathbb{1}_{[0,1]}(t)\,dt\Big)=\frac{4}{3\pi} $$

  • $\begingroup$ Bravo! What a concise way! $\endgroup$
    – oO_ƲRF_Oo
    Jul 16, 2023 at 22:39

You must log in to answer this question.

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