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Prove: $$ {\Large\int_{0}^{\ln^{2}\left(2\right) \over4}}\, \frac{\arccos\left(\vphantom{\huge A} {\exp\left(\vphantom{\large A}\sqrt{x\,}\right) \over \sqrt{\vphantom{\large A}2\,}}\right)} {1-\exp\left(\sqrt{4x\,}\,\right)} \,{\rm d}x = -\,\frac{\,\,\pi^{3}}{192} $$

I haven't solved it yet.

Felix Marin
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Vladimir Reshetnikov
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2 Answers2

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With the substitution $x \mapsto \frac{1}{4}\log^{2}\left(\frac{2}{x^{2}+1}\right)$, it follows that

\begin{align*} \int_{0}^{\frac{\log^{2}2}{4}} \frac{\arccos\left( \frac{\exp\sqrt{x}}{\sqrt{2}} \right)}{1 - \exp\sqrt{4x}} \, dx &= \int_{0}^{1} \frac{x \arctan x}{1 - x^{2}} \log\left(\frac{1+x^{2}}{2}\right) \, dx \\ &= -\frac{1}{2} \int_{-1}^{1} \frac{\arctan x}{1 + x} \log\left(\frac{1+x^{2}}{2}\right) \, dx. \end{align*}

Now you can refer to this solution.

Actually, I obtained this integral representation by applying the following chain of much human-friendly substitutions:

$$ \exp\sqrt{4x} = t, \qquad t = 2\cos^{2}u, \qquad x = \tan u $$

Community
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Sangchul Lee
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    Have you already evaluated all possible integrals? ;) – Vladimir Reshetnikov Oct 12 '13 at 18:06
  • @VladimirReshetnikov, Sure I haven't :) I was just so lucky. You can prove my claim by giving me some integrals involving special functions (such as hypergeometric, Bessel, Airy, ...) and enjoying that I even don't know their definitions. – Sangchul Lee Oct 12 '13 at 18:16
  • @sos440 I think you're being excessively modest. I was going to ask you a question about your evaluation in the other post, but the answer came to me as I was typing the question. – Random Variable Oct 12 '13 at 20:36
  • Well, maybe I do have a question. Why does $$I_{\epsilon} = J_{\epsilon} + K_{\epsilon}$$

    necessarily imply that $$ \int_{-1}^{1} \frac{\arctan x}{1 + x} \log\left(\frac{1+x^{2}}{2}\right) , dx= \lim_{ \epsilon \to 0} J_{\epsilon} +\lim_{\epsilon \to 0} K_{\epsilon} \ ?$$

    Is is simply because $ \displaystyle\int_{-1}^{1} \frac{\arctan x}{1 + x} \log\left(\frac{1+x^{2}}{2}\right) , dx$ converges?

     – Random Variable Oct 12 '13 at 21:56
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    @RandomVariable, That's because the integrand not only converges, but also is holomoprhic at $x = -1$ (and in particular, continuous.) – Sangchul Lee Oct 12 '13 at 22:46
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use the residue theorem !

we first assume that :

$\exp\sqrt{x\,}=z$

then your formula turn out to be :

$ {\Large\int_{1}^{\sqrt{2}}}\, \frac{\arccos\left(\vphantom{\huge A} {z \over \sqrt{\vphantom{\large A}2\,}}\right)} {1-z^2} \,{\rm d}x = -\,\frac{\,\,\pi^{3}}{192} $

the residue theorem gives ,

$ {\Large\int_{1}^{\sqrt{2}}}\, \frac{\arccos\left(\vphantom{\ huge A} {z \over \sqrt{\vphantom{\large A}2\,}}\right)} {1-z^2} \,{\rm d}x =\frac{b\pi}{2}\cdot\frac{\arccos(1/\sqrt{2})}{2}\cdot\frac{\arccos(-1/\sqrt{2})}{-2}$$=-\frac{b\pi}{2}\cdot(\frac{\pi}{8})^2$

the last step is to calculate the radius :

$\frac{1}{b}=\frac{z}{2(Inz)^{'}}=$$z^{2}=z_{1}^{2}+z_{2}^{2}=\frac{3}{2}$$\Longrightarrow$$b=\frac{2}{3}$

then, your solution holds !

Karl Marx
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