Integral of $$\int_0^{\frac{\pi 2^n}{2}}\sin^{2+d}(u)\prod_{i=1}^{n}\cos^{2-d}(u/2^i)du$$

I've tried a number of ways to re-write this in a way that makes sense, but no luck thus far. The integrand looks a bit like the beta function, but i'm not sure if that's leading anywhere.

Any help?

  • $\begingroup$ You sure about this integral, I think you made a couple of typos. The index $k$ doesn't by any chance run from $k=1$ to $n$? So the last cosine in the product is actually $\cos^{2-d}(u/2^{2^n})$? $\endgroup$ – Hamed Nov 10 '15 at 1:06
  • $\begingroup$ Yup, you're right. Sorry, made the change. $\endgroup$ – measure Nov 10 '15 at 1:16
  • 1
    $\begingroup$ A first step might be to use the product formula $$\prod_{i=1}^n \cos(u/2^i)=2^{-n}\frac{\sin u}{\sin(2^{-n}u)}.$$ But then it gets ugly... $\endgroup$ – mickep Nov 10 '15 at 5:31

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