Consider the following integral:

I=∞∫0x−1√2x−1 ln(2x−1)dx.I tried to evaluate I in a closed form (both manually and using

Mathematica), but without success.However, if

WolframAlphais provided with a numerical approximation I≈3.2694067500684…, 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(2x−1). Then x=log(1+eu)/log2, dx=(1/log2)(du/(1+e−u). The integral then becomes

1log2∫∞−∞du1+e−ue−u/2log(1+eu)log2−1u=12log22∫∞−∞ducosh(u/2)log(1+eu)−log2u=12log22∫0−∞ducosh(u/2)log(1+eu)−log2u⏟u→−u+12log22∫∞0ducosh(u/2)log(1+eu)−log2u=−12log22∫∞0ducosh(u/2)log(1+e−u)−log2u⏟log(1+e−u)=log(1+eu)−u+12log22∫∞0ducosh(u/2)log(1+eu)−log2u

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

12log22∫∞0ducosh(u/2)=π2log22

as correctly conjectured.

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