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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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  • $\begingroup$ This one is really tough! $\endgroup$
    – Abramo
    Nov 23, 2011 at 23:22
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    $\begingroup$ Where did you see this relation? $\endgroup$
    – JavaMan
    Nov 23, 2011 at 23:24
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    $\begingroup$ I saw it on a site where someone had posted the solution but no method. At mymathforum.com. I checked this closed form against the numerical solution that Maple and Mathematica gave and it was exactly as posted. I do not know where the poster may have came up with it, but it appears to be correct. $\endgroup$
    – Cody
    Nov 23, 2011 at 23:44
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    $\begingroup$ I computed both sides to 20 places in Maple. It concluded they agree only to 8 places, and differ by about $2 \times 10^{-9}$. $\endgroup$
    – GEdgar
    Nov 24, 2011 at 0:46

1 Answer 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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    $\begingroup$ I appreciate you, for you saved my time from a deceptive non-identity! $\endgroup$
    – Sangchul Lee
    Nov 24, 2011 at 2:16
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    $\begingroup$ I am so sorry. I apologize for misleading. Apparently, I did not check far enough and took it on faith. I do not know from where the poster derived this. From now on I will make sure before I post. Again...sorry. Here is the link to the problem if you wan to look at it: mymathforum.com/… $\endgroup$
    – Cody
    Nov 24, 2011 at 9:10
  • $\begingroup$ @Cody We must always check the OP proposed answer before we try anything. Anyway, your question is still quite interesting. $\endgroup$
    – Felix Marin
    Aug 28, 2014 at 21:25

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