Finding $\lim\limits_{n\to\infty}\frac1{n^2}\int_0^{\frac{\pi}2}x\left(\frac{\sin nx}{\sin x}\right)^4\,\mathrm dx$

Question

$$\lim_{n\to\infty}\frac1{n^2}\int_0^{\frac{\pi}2}x\left(\frac{\sin nx}{\sin x}\right)^4\,\mathrm dx$$I don't know how to solve this limit problem. I'm roughly clear about using the pinch theorem, but I haven't done it yet. Could somebody please tell me? Thanks!

Answer 1

From $$\lim_{x\to 0}x^2\left(\frac1{\sin^4 x}-\frac1{x^4}\right)=\frac23,$$ for $x\in(0,\frac\pi 2)$, there is $C$ such that $$\left|\frac{x\sin^4 nx}{\sin^4 x}-\frac{\sin^4 nx}{x^3}\right|\le \frac{C\sin^4 nx}{x}\le Cn.$$

Thus, \begin{align*} \lim_{n\to\infty}\frac1{n^2}\int_0^{\frac\pi2}x\left(\frac{\sin nx}{\sin x}\right)^4\, \mathrm dx&=\lim_{n\to\infty}\frac1{n^2}\int_0^{\frac\pi2}\frac{\sin^4 nx}{x^3}\, \mathrm dx\\ &= \lim_{n\to\infty}\int_0^{\frac{n\pi}2}\frac{\sin^4 x}{x^3}\, \mathrm dx\\ &= \int_0^{\infty}\frac{\sin^4 x}{x^3}\, \mathrm dx\\ \text{(IBP)}\quad &= \int_0^{\infty}\frac{2\sin^3 x\cos x}{x^2}\, \mathrm dx\\ \text{(IBP)}\quad &= \int_0^{\infty}\frac{6\sin^2 x\cos^2 x-2\sin^4 x}{x}\, \mathrm dx\\ &= \int_0^{\infty}\frac{\cos 2x-\cos 4x}{x}\, \mathrm dx\\ \text{(Frullani)}\quad &=\ln 2. \end{align*}

Answer 2

Result

Let

$$f(n) = \int_0^{\frac{\pi }{2}} x \left(\frac{\sin (n x)}{\sin (x)}\right)^4 \, dx$$

then

$$\lim_{n\to \infty } \, \frac{f(n)}{n^2}=\log (2)\simeq 0.6931471805599453 ...$$

This result is in good agreement with the numerical estimate of the integral.

Derivation

This is more heuristic and not really strict. Maybe others can fill in the gaps.

We split the integral into equally spaced intervals, and observe $\sin(\pi x/n) \simeq \pi x/n$ for $n\to \infty$ and get

$$\lim_{n\to \infty }f(n)/n^2 =\lim_{n\to \infty } \sum_{k=1}^\infty a(k)\tag{1}$$

where

$$a(k) = \int_{k-1}^k \pi ^2 s \left(\frac{\sin (\pi s)}{\pi s}\right)^4 \, ds$$

The first two values are

$$a(1) = \text{Ci}(2 \pi )-\text{Ci}(4 \pi )+\log (2)$$ $$a(2) = -\text{Ci}(-8 \pi )+\text{Ci}(-4 \pi )-\text{Ci}(2 \pi )+\text{Ci}(4 \pi )$$

and for $k \ge 3$

$$a(k) = -\text{Ci}(2 (k-1) \pi )+\text{Ci}(4 (k-1) \pi )-\text{Ci}(-4 k \pi )+\text{Ci}(-2 k \pi )$$

Here $\text{Ci}(z) = -\int_{-\infty }^z \frac{\cos (t)}{t} \, dt$ is the cosine integral.

The sum in $(1)$ telescopes and we are left with the announced result.

Discussion

My statement "the sum telescopes" is fairly bold.

In fact, the cancellation is like this

$$\text{Ci}(-4 \pi )-\text{Ci}(4 \pi )=i \pi$$ $$\text{Ci}(8 \pi )-\text{Ci}(-8 \pi )=-i \pi$$

see symmetry relations in http://people.math.sfu.ca/~cbm/aands/abramowitz_and_stegun.pdf, 5.2.20, and the sum of these two lines is $0$.