Evaluate: $$\int_0^{\infty} \frac{x-1}{\sqrt{2^x-1}\ln(2^x-1)}\,dx$$

I am not sure where to start or what should be the best approach towards this problem. I tried the substitution $2^x-1=t^2$ but that seems to make things more worse. Using this substitution, I got: $$\int_0^{\infty} \frac{1}{\ln^2 2}\frac{\ln\left(\frac{1+t^2}{2}\right)}{(1+t^2)\ln t}\,dt$$ I don't see how to proceed after this. :(

Any help is appreciated. Thanks!


marked as duplicate by Najib Idrissi, user127.0.0.1, M Turgeon, Hakim, Davide Giraudo May 20 '14 at 13:12

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  • 3
    $\begingroup$ Have you tried $t=\tan\theta$ ? $\endgroup$ – evil999man May 20 '14 at 11:22
  • $\begingroup$ Ah, how could I not see it, thanks a lot Awesome! :) $\endgroup$ – Pranav Arora May 20 '14 at 11:23
  • $\begingroup$ Thanks but those approaches are a little complicated. I like Awesome's approach and Awesome's method is quite a standard trick, I have been using it all the time but this time, I missed it. $\endgroup$ – Pranav Arora May 20 '14 at 11:56
  • $\begingroup$ @PranavArora Just sub $t\mapsto \dfrac{1}{t}$ to regenerate the integral and get the desired result. $\endgroup$ – MathGod Jan 3 '17 at 10:13

Try $t=\tan\theta$

Simplify everything in terms of $\sin$ and $\cos$. Substitute $\pi/2-\theta=\phi$ and add and do what you do with other definite questions. Be sure to completely simplify your numerator to break into two integrals.

  • $\begingroup$ I am not sure what you meant by break into two integrals but I got the answer though. $\endgroup$ – Pranav Arora May 20 '14 at 11:32
  • $\begingroup$ @PranavArora It is irrelevant now. $\endgroup$ – evil999man May 20 '14 at 11:35

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