2
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.