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$\begingroup$

I ran across a curious integral that seems to be rather tough that some on the site may enjoy.

Show that $\displaystyle \int_{0}^{1}\frac{\sqrt{1-x^{2}}}{1-x^{2}\sin^{2}(x)}dx = \frac{5\sqrt[5]{{\pi}^{8}}}{32\sqrt[5]{{\zeta(5)}^{9}}}$

How in the world can $\zeta(5)$ be incorporated into this?. I tried series and several methods, but made no real progress. Any ideas?. Thanks very much.

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    I computed both sides to 20 places in Maple. It concluded they agree only to 8 places, and differ by about $2 \times 10^{-9}$.2011-11-24

1 Answers 1

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The purported identity is false, as GEdgar already indicated in the comment, but remarkably accurate: $ \begin{eqnarray} \int_0^1 \frac{\sqrt{1-x^2}}{1-x^2 \sin^2(x)} \mathrm{d} x &\approx& \color{red}{ 0.91392913}60302011781728596 \\ \frac{5 \pi^{8/5}}{32 \zeta(5)^{9/5}} &\approx& \color{red}{0.91392913}77247633495515212 \end{eqnarray} $

Here is the Mathematica code used:

In[19]:= N[  NIntegrate[Sqrt[1 - x^2]/(1 - x^2 Sin[x]^2), {x, 0, 1},    WorkingPrecision -> 60], 25]  Out[19]= 0.9139291360302011781728596  In[20]:= N[(5 Pi^(8/5))/(32 Zeta[5]^(9/5)), 25]  Out[20]= 0.9139291377247633495515212  In[21]:= % - %%  Out[21]= 1.694562171378662*10^-9 
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    @Cody We must always check the OP proposed answer $b$efore we try anything. Anyway, your question is still quite interesting.2014-08-28