I need an asymptotic expansion of J(n)

$J(n)=\frac {2} {\pi} \int_{0}^{\pi/n} \prod_{k=1}^n \frac {\sin kx} {\sin x} dx$, $n=2,3,4,\dots$

Can anybody help to find the asymptotic analytically or at least via numirical calculation please? Also, I wonder if there is a graphical interpretation of the result exists? Best to my knowledge the integral is not very simple to get an answer. Thank you for any help.

  • $\begingroup$ is the sine in the denominator considered under product, i.e is it effectively $$\frac{\prod_{k=1}^n\sin kx}{\left(\sin x\right)^n}$$? $\endgroup$ – Valentin Jun 9 '12 at 18:00
  • $\begingroup$ @Valentin It seems so. $\endgroup$ – Pedro Tamaroff Jun 9 '12 at 18:21
  • $\begingroup$ @Valentin yes, you are right, the same you wrote. $\endgroup$ – Mikhail G Jun 9 '12 at 18:21
  • $\begingroup$ Not fit for an answer, but maybe this argument will suggest the solution path. Consider a similar integral: $$I=\frac{2}{\pi} \int_{0}^{\frac{\pi}{n}} \prod_{k=1}^{n}\frac{\cos kx}{\cos x}dx$$ $$\cos x=t$$ $$dx=-\frac{dt}{\sqrt{1-t^{2}}}$$ $$I=\frac{2}{\pi} \int_{1}^{\cos\frac{\pi}{n}} \prod_{k=1}^{n}\frac{T_k(t)}{t\sqrt{1-t^2}}dt$$ Where $T_k$ are Chebyshev Polynomials. Now $$T_{k}\left(x\right)=2^{k-1}\prod_{k=1}^{n}\left\{ x-\cos\left[\frac{\left(2k-1\right)\pi}{2n}\right]\right\}$$. etc $\endgroup$ – Valentin Jun 9 '12 at 18:54
  • $\begingroup$ @Valentin Do you mean to wrap the integral around circles of radius 2n like for Chebyshev Polynomials by Michael Trott you mentioned at Mathworld? $\endgroup$ – Mikhail G Jun 10 '12 at 18:32

Here's what I've got:

For $x\in [0,\pi/n]$, we have $\sin(x)=x(1+O(1/n^2))$. Similarly, we have $\sin^n(x)=x^n(1+O(1/n))$. Let us write $f(x)=\frac{\sin(x)}{x}$. We have $|f(x)|\leq 1$, but for any $\epsilon>0$, $\exists \delta>0$ such that $f(x)\geq 1-\epsilon$ for $x\in [0,\delta]$

Then, for $x\in[0,\pi/n]$: $$ \prod_{k=1}^n\frac{\sin(kx)}{\sin(x)}=n!(1+O(1/n))\prod_{k=1}^n f(kx) $$

Now, we have \begin{eqnarray*} J(n)&=&\frac{2}{\pi}n!(1+O(1/n))\int_0^{\pi/n}\prod_{k=1}^n f(kx)\,dx\\ &\geq&n!(1+O(1/n))\int_0^{\delta/n}(1-\epsilon)^n\,dx\\ &\geq&\frac{2}{\pi}(n-1)!(\delta+O(1/n))(1-\epsilon)^n \end{eqnarray*}

Similarly, we have $$ J(n)\leq 2(n-1)!(1+O(1/n)) $$

We can combine these two estimates to get $$ J(n)=(n-1)!(1+o(1))^n $$

Of course, $(n-1)!=n!(1+o(1))^n$, so we can write $$ J(n)=n!(1+o(1))^n $$ It might be useful to use Stirling's formula to get $$ J(n)=\left(\frac{n}{e}(1+o(1))\right)^n $$

Using Mathematic, I computed $\left(\frac{J(500)}{499!}\right)^{1/500}=.99554\ldots$

  • $\begingroup$ Thank you, I have $J_1(n)=\frac {n!} {\sqrt{\pi A}}$, $A=n(n-1)(2n+5)/36$, so my calculation is slightly different: .98324... Generally I need an asymptotic expansion of the integral. It would be nice to represent it via special functions. $\endgroup$ – Mikhail G Jun 12 '12 at 13:01

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