# Calculation of Integrals with reciproce Logarithm, Euler's constant $\gamma=0.577…$

Evaluate the improper integral $$\int\limits_0^1\left(\frac1{\log x} + \frac1{1-x}\right)^2 dx = \log2\pi - \frac12 = 0.33787...$$

With integration by parts we get from $$\int\limits_0^1\left(\frac1{\log x} + \frac1{1-x}\right) dx = \gamma$$

the similar integral $$\int\limits_0^1\left(\frac1{\log^2 x} - \frac{x}{(1-x)^2}\right)dx = \gamma-\frac12$$

But we need $$\int\limits_0^1\left(\frac2{(1-x)\log x} + \frac{1+x}{(1-x)^2}\right)dx = 0.260661401507813...$$

to get the integral in question. In question series from one of Coffey's papers involving digamma, $\gamma$, and binomial there is a hint of connection to Stieltjes constants.

Most of my ideas on this come from using $$(\ln \ln(1-x))' = \frac{-1}{(1-x)\ln(1-x)}$$.
$$\frac{2}{x\ln(1-x)} = \frac{1-x}{x}\frac{2}{(1-x)\ln(1-x)} = \frac{2}{(1-x)\ln(1-x)} - \frac{2x}{(1-x)\ln(1-x)}$$.