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?

  • $\begingroup$ 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

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.