# 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

• Usually, if mathematica doesn't give a close form, it mean that you can't calculate it by hand. – idm Aug 28 '14 at 9:09
• @idm, if mathematica doesn't give a close form, it does not mean that you can't calculate it by hand. – xpaul Aug 28 '14 at 13:57
• @xpaul, precisely: residuetheorem.com/2014/03/14/… – Dmoreno Aug 28 '14 at 18:26
• @Dmoreno, please see this example math.stackexchange.com/questions/874431/…. Mathematica doesn't give a close form, but it has a closed form. – xpaul Aug 28 '14 at 20:52

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 $I=1.001...$

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