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How to prove $$\int_0^1 \left[ \frac{2}{\pi }\arctan \left(\frac 2 \pi \arctan \frac{1}{x} + \frac{1}{\pi }\ln \frac{1 + x}{1 - x}\right) - \frac{1}{2} \right]\frac{\mathrm{d}x} x = \frac{1}{2} \ln \left( \frac \pi {2\sqrt 2 } \right).$$

I have tried let $t=\frac1x$, but it seems no use! Could you help me to solve it?

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  • $\begingroup$ tough... I just gave it a try with mathematica which used some numerical scheme, the result seems to be correct. I can't see an easy way to substitute, may be taylor can be of some use $\endgroup$ – user190080 Jul 1 '15 at 16:45
  • $\begingroup$ What makes you think that it might have a simple answer? $\endgroup$ – Alex M. Jul 1 '15 at 18:41
  • $\begingroup$ @AlexM. that's quite a meta question :) On the other hand, why it shouldn't have an easy access through a substitution of one term or another? When it comes to such integrals I have already often seen people just using a "simple" trick and tada...the leftovers were easily tractable . So my answer would then be this: my experience $\endgroup$ – user190080 Jul 1 '15 at 19:22
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One may show that this integral is equivalent to the integral in this problem as follows:

First, rewrite

$$\frac12 \log{\left ( \frac{1+x}{1-x} \right )} = \tanh^{-1}{x}$$

Then note that

$$\tan^{-1}{\left ( \frac1{x} \right )} = \frac{\pi}{2} - \tan^{-1}{x} $$

The integrand is then equal to

$$\left [\frac{2}{\pi} \tan^{-1}{\left (\frac{2}{\pi} \left (\frac{\pi}{2} - \tan^{-1}{x} + \tanh^{-1}{x} \right ) \right )}-\frac12 \right ] \frac1{x}$$

which is equal to

$$\left [\frac{2}{\pi} \tan^{-1}{\left (1+\frac{2}{\pi} \left ( \tanh^{-1}{x} - \tan^{-1}{x} \right ) \right )}-\frac12 \right ] \frac1{x}$$

Now, let

$$\tan^{-1}{(1+y)} = \frac{\pi}{4} + \tan^{-1}{w} $$

Then it is straight forward to show that

$$y = \frac{1+w}{1-w} - 1 = \frac{2 w}{1-w} \implies w=\frac{y}{y+2} $$

With $y=(2/\pi) (\tanh^{-1}{x} - \tan^{-1}{x} )$, we have

$$\frac{2}{\pi} \tan^{-1}{\left (1+\frac{2}{\pi} \left ( \tanh^{-1}{x} - \tan^{-1}{x} \right ) \right )}-\frac12 = \frac{2}{\pi} \tan^{-1}{\left (\frac{\tanh^{-1}{x} - \tan^{-1}{x} }{\tanh^{-1}{x} - \tan^{-1}{x} +\pi} \right )} $$

Thus, the integral in question is equal to

$$\frac{2}{\pi} \int_0^1 \frac{dx}{x} \tan^{-1}{\left (\frac{\tanh^{-1}{x} - \tan^{-1}{x} }{\tanh^{-1}{x} - \tan^{-1}{x} +\pi} \right )} $$

The evaluation of this integral is detailed in this arXiv paper.

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  • 1
    $\begingroup$ Oh, what has become of ArXiv! :( $\endgroup$ – Alex M. Jul 1 '15 at 19:25

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