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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.

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    is the sine in the denominator considered under product, i.e is it effectively $$\frac{\prod_{k=1}^n\sin kx}{\left(\sin x\right)^n}$$?2012-06-09
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    @Valentin It seems so.2012-06-09
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    @Valentin yes, you are right, the same you wrote.2012-06-09
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    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](http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html). Now $$T_{k}\left(x\right)=2^{k-1}\prod_{k=1}^{n}\left\{ x-\cos\left[\frac{\left(2k-1\right)\pi}{2n}\right]\right\}$$. etc2012-06-09
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    @Valentin Do you mean to wrap the integral around circles of radius 2n like for Chebyshev Polynomials by Michael Trott you mentioned at Mathworld?2012-06-10

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