Need help with a definite integral $\int_0^{\infty} \frac{x-1}{\sqrt{2^x-1}\ln(2^x-1)}\,dx$

Question

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!

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. — Pranav Arora, May 20 '14 at 11:56
@PranavArora Just sub $t\mapsto \dfrac{1}{t}$ to regenerate the integral and get the desired result. — MathGod, Jan 03 '17 at 10:13

score 4 · Accepted Answer · answered May 20 '14 at 11:30

4

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.

answered May 20 '14 at 11:30

evil999man

6,018

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

Need help with a definite integral $\int_0^{\infty} \frac{x-1}{\sqrt{2^x-1}\ln(2^x-1)}\,dx$

1 Answers1