text: # Definite integral with logarithm and arctangent inside of arctangent
  
  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?
  
  • 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 – user190080 Jul 1 '15 at 16:45
  • What makes you think that it might have a simple answer? – Alex M. Jul 1 '15 at 18:41
  • @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 – 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.
  
  • Oh, what has become of ArXiv! :( – Alex M. Jul 1 '15 at 19:25
explanation: The document contains a math question involving a definite integral with a complex integrand that includes logarithmic and arctangent functions. The question asks for a proof of the given integral's value. The document also includes comments and a detailed transformation of the integrand, leading to an equivalent form of the integral, which is referenced to an arXiv paper for further evaluation.
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distilabel_metadata.content: # Definite integral with logarithm and arctangent inside of arctangent
  
  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?
  
  • 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 – user190080 Jul 1 '15 at 16:45
  • What makes you think that it might have a simple answer? – Alex M. Jul 1 '15 at 18:41
  • @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 – 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.
  
  • Oh, what has become of ArXiv! :( – Alex M. Jul 1 '15 at 19:25
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question involving a definite integral with a complex integrand that includes logarithmic and arctangent functions. The question asks for a proof of the given integral's value. The document also includes comments and a detailed transformation of the integrand, leading to an equivalent form of the integral, which is referenced to an arXiv paper for further evaluation.
  
  Label: question-and-answer