A conjectured closed form of ∞∫0x−1√2x−1 ln(2x−1)dx\int\limits_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx

Consider the following integral:
I=0x12x1 ln(2x1)dx.

I tried to evaluate I in a closed form (both manually and using Mathematica), but without success.

However, if WolframAlpha is provided with a numerical approximation I3.2694067500684…, it returns a possible closed form:
I?=π2ln22.
Further numeric caclulations show that this value is correct up to at least 103 decimal digits. So, I conjecture that this is the exact value of I.

Question: Is this conjecture correct?

Answer

Sub u=log(2x1). Then x=log(1+eu)/log2, dx=(1/log2)(du/(1+eu). The integral then becomes

1log2du1+eueu/2log(1+eu)log21u=12log22ducosh(u/2)log(1+eu)log2u=12log220ducosh(u/2)log(1+eu)log2uuu+12log220ducosh(u/2)log(1+eu)log2u=12log220ducosh(u/2)log(1+eu)log2ulog(1+eu)=log(1+eu)u+12log220ducosh(u/2)log(1+eu)log2u

The nasty pieces of the integral cancel, and we are left with

12log220ducosh(u/2)=π2log22

as correctly conjectured.

Attribution
Source : Link , Question Author : Vladimir Reshetnikov , Answer Author : Ron Gordon

Leave a Comment