Integral $ \int_{0}^1 \sqrt{\frac{\ln{x}}{x^2-1}} dx$ Please help evaluating this integral $$ \large\int_{0}^1 \sqrt{\frac{\ln{x}}{x^2-1}}  dx$$
Mathematica could not evaluate it in a closed form. Numerically it is about $$I\approx1.060837861412045137097...,$$
ISC+ did not return any closed form for it.
It is related to a previous post
 A: Let's make a variable change $x=e^{-t}$:  
$$I=\int_0^{\infty}\frac{\sqrt{t}e^{-t}}{\sqrt{1-e^{-2t}}}dt$$  
Now, let's expand the integrand into Taylor series: 
$$\frac{1}{\sqrt{1-e^{-2t}}}=\sum_{n=0}^{\infty}\frac{(2n)!}{(2^nn!)^2}e^{-2nt}$$
Thus  
$$I=\sum_{n=0}^{\infty}\frac{(2n)!}{(2^nn!)^2}\int_0^{\infty}\sqrt{t}e^{-(2n+1)t}dt=$$  
$$=\frac{\sqrt{\pi}}{2}\sum_{n=0}^{\infty}\frac{(2n)!}{(2^nn!)^2}\frac{1}{(2n+1)^{\frac{3}{2}}}$$  
Now, by hand, if we take only the three first terms of the sum we get $I=1.001...$  
Not bad!
A: The Lerch transcendent, initially defined by
$$\Phi(z,s,a):=\sum_{k=0}^\infty\frac{z^k}{(a+k)^s}, \quad a>0,\Re s>1,|z|<1,$$
admits the following integral representation
$$
\Phi(z,s,a)=\int_0^{\infty}\frac{x^{s-1}e^{-ax}}{1-ze^{-x}}{\rm d}x.
$$ By differentiation
$$
\partial_z^r\Phi(z,s,a)=(-1)^r\int_0^{\infty}\frac{x^{s-1}e^{-(a+r)x}}{(1-ze^{-x})^{r+1}}{\rm d}x,
$$ then, by extension, your integral $I$ may be formally rewritten as

$$
I=\int_{0}^1 \sqrt{\frac{\ln{x}}{x^2-1}}  dx=-i\sqrt{2\pi}\:\partial_z^{\! -\frac12}\Phi\left(1,\frac32,1\right)
$$ 

This is to show the level of complexity of this integral: fractional calculus.
