$$\lim_{n\to\infty}\frac1{n^2}\int_0^{\frac{\pi}2}x\left(\frac{\sin nx}{\sin x}\right)^4\,\mathrm dx$$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!
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$\begingroup$ (pinch theorem) $\endgroup$ – mathworker21 Mar 27 '19 at 7:08
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$\begingroup$ there's some formula like $\sum_{|n| \le N} e(n\theta) = \frac{\sin(N\theta)}{\sin(\theta)}$. Something like that. I think that should be useful $\endgroup$ – mathworker21 Mar 27 '19 at 7:26
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$\begingroup$ I checked through W|A for values of $n$ as high as $n\sim 20000$ and it seems that the limit does exist and is equal to $\ln 2$. But at present I don't have a rigorous proof for the same. $\endgroup$ – Rohan Shinde Mar 27 '19 at 8:00
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$\begingroup$ After a lengthy semi-heuristic study I suspect the simple result $\lim = \log(2) \simeq 0.6931471805599453..$ $\endgroup$ – Dr. Wolfgang Hintze Mar 27 '19 at 8:09
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2$\begingroup$ Another way to reach the integral in Tianlalu's answer is as follows: Substituting $nx = u$ and writing $\operatorname{sinc}(x) = \frac{\sin x}{x}$, we have $$ \frac{1}{n^2} \int_{0}^{\frac{\pi}{2}} x \frac{\sin^4(nx)}{\sin^4 x} \, \mathrm{d}x = \int_{0}^{\frac{n \pi}{2}} \frac{\sin^4 u}{u^3 \operatorname{sinc}^4(u/n)} \, \mathrm{d} u \ \xrightarrow[n\to\infty]{\text{DCT}} \ \int_{0}^{\infty} \frac{\sin^4 u}{u^3} \, \mathrm{d} u, $$ where the convergence follows from the dominated convergence together with the integrable dominating function $Cu^{-3}\sin^4 u$ for some constant $C > 0$. $\endgroup$ – Sangchul Lee Mar 27 '19 at 11:48
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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*}
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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 $0$.
answered Mar 27 '19 at 8:34
