$I(x) = -\int_0^1 \frac{1}{z}\ln\left(\frac{1-x z + \sqrt{1-2 x z+ z^2}}{2}\right)\,dz$

Question

Is there a closed form integral for $$I(x) = -\int_0^1 \frac{1}{z}\ln\left(\frac{1-x z + \sqrt{1-2 x z+ z^2}}{2}\right)\,dz$$ for $-1 < x < 1$?

This integral is related to Legendre polynomials as $$\frac{-1}{z}\ln\left(\frac{1-x z + \sqrt{1-2 x z+ z^2}}{2}\right)=\sum_{n=1}^{\infty} \frac{P_{n}(x)}{n} z^{n-1}$$ $$I(x) = \sum_{n=1}^{\infty} \frac{P_{n}(x)}{n^2}$$ where $P_{n}(x)$ satisfies $\sum_{n=0}^{\infty} P_n(x) z^2 = (1-2 x z + z^2)^{- 1/2}$ for all $-1< x< 1$ and $-1< z< 1$. Integrating twice with respect to $z$ gives a relation \begin{align}\sum_{n=0}^{\infty} \frac{P_{n}(x)}{(n+1)(n+2)} &= \int_0^1 dz \int_0^z dz'\frac{1}{\sqrt{1 - 2 z' x +x^2}}\,. \\ &= 1 - \sqrt{2(1-x)} +(1-x) \ln\left(1+\sqrt{\frac{2}{1-x}}\right) \\&= 1-2s+2s^2 \ln\frac{1+s}{s} \end{align} where $s=\sin(\theta/2)$ and $x=\cos\theta$. The elements of this sum are asymptotically equal to the sum defining $I$, but I am not sure if this is useful to get a closed form for $I$. Another related relation is $$I(x) = \int_0^1 \frac{dz}{z} \int_0^z \frac{dz'}{z'} \left[\frac{1}{\sqrt{1 - 2 z' x +x^2}}-1\right]\,. $$

Answer 1

Here's an approach using differentiation under the integral sign:

$$\begin{align}I'(x) &= \int_{0}^{1}\frac{\frac{1}{\sqrt{-2xz+z^2+1}}+1}{\sqrt{-2xz+z^2+1}-xz+1}\, dz\\ &=\left.\frac{(x+1)\ln\left(\sqrt{z^2-2xz+1}+x-z\right)-2x\ln\left(\sqrt{z^2-2xz+1}-z+1\right)}{x^2-1}\right]_{z=0}^{z=1}\\ &=\frac{x\ln(2)-x\ln(1-x)-(x+1)\ln(x+1)+(x+1)\ln\left(x+\sqrt{2-2x}-1\right)}{x^2-1}\end{align}$$

Mathematica now gives:

$$\begin{align}I(x) &= \int \frac{x\ln(2)-x\ln(1-x)-(x+1)\ln(x+1)+(x+1)\ln\left(x+\sqrt{2-2x}-1\right)}{x^2-1}\,dx\\ &=\frac{1}{2}\left(4\operatorname{Li}_2\left(\frac{1}{2}\left(2-\sqrt{2-2x}\right)\right)+\operatorname{Li}_2\left(\frac{1}{2}-\frac{x}{2}\right)-\ln^2(1-x)+2\ln\left(2\left(x+\sqrt{2-2x}-1\right)\right)\ln(2-2x)-\ln\left(2-\sqrt{2-2x}\right)\ln(8-8x)+\ln\left(\sqrt{2-2x}+2\right)(\ln(2)-\ln(1-x))-3\ln^2(2)\right)+C\end{align}$$

Now using, as @Steven Clark in the comments said

$$I(0) = \frac{1}{4} \left(2 \operatorname{Li}_2 \left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)-\left(\operatorname{arsinh} (1)-\ln (2)\right)^2\right)$$ we can solve for $C$.

Upon doing so, one determines the rather horrible expression (which might be possible to simplify)

$$\boxed{I(x) = 2\operatorname{Li}_2\left(1-\frac{1}{2}\sqrt{2-2x}\right)+\frac{1}{2}\text{Li}_2\left(\frac{1}{2}-\frac{x}{2}\right)-2\operatorname{Li}_2\left(1-\frac{1}{\sqrt{2}}\right)+\frac{1}{2}\operatorname{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)-\frac{1}{2}\ln^2(2-2x)+\ln\left(2x+2\sqrt{2-2x}-2\right)\ln(2-2x)+\frac{1}{2}\ln(2)\ln\left(\sqrt{2-2x}+2\right)-\frac{1}{2}\ln\left(2-\sqrt{2-2x}\right)\ln(8-8x)-\frac{1}{2}\ln\left(\sqrt{2-2x}+2\right)\ln(1-x)-\frac{\pi^2}{24}+\frac{\ln^2(2)}{4}-\frac{1}{2}\ln(2)\ln\left(2+\sqrt{2}\right)-\frac{1}{4}\operatorname{arsinh}^2(1)}$$

EDIT

Through some dilogarithm identities, one can simplify this expression to $$I(x)=\frac{\pi^2}{6}-\operatorname{Li}_{2}{\left(p\right)}+\operatorname{Li}_{2}{\left(-p\right)}-\ln{\left(p\right)}\ln{\left(1-p^{2}\right)}$$ with $p=\sqrt{\frac{1-x}{2}}$ as mentioned in the comments below by @David H

Answer 2

Let $x=\cos2a$ for notational convenience, i.e.

$$I(a) = \int_0^1 \frac{1}{z}\ln\bigg(\frac{1-z\cos2a+ \sqrt{1-2z \cos2a+ z^2}}{2}\bigg) dz$$ and substitute $z-\cos2a =\sin2a\sinh t$ to remove the square-roots in evaluating $I’(a)$ below

\begin{align} I’(a) =& -\int_0^1 \frac{1+\frac1{\sqrt{1-2z \cos2a+ z^2}} }{1-z\cos2a+ \sqrt{1-2z \cos2a+ z^2}}\ dz\\ =& \ 2\int_{\ln (\tan a)}^{\ln(\tan a+\sec a)} \frac{1+\sin2a\cosh t}{\sin2a-\cos2a\sinh t+\cosh t}\ dt\\ =& \ 2\bigg[t\tan a +\left(\cot a-\tan a\right)\ln(1+e^t\tan a) \bigg] _{\ln (\tan a)}^{\ln(\tan a+\sec a)}\\ = & \ {\cot a}\ln\frac{1+\sin a}{1-\sin a}+2\left( \frac{\ln\cos a}{\tan a}- \frac{\ln\sin a}{\cot a}\right) \end{align} Then, with $I(0)=\int_0^1 \frac{\ln(1-z)}zdz=-\frac{\pi^2}6$ \begin{align} I(a)=& \ I(0) +\int_0^a I’(t)\ dt \\ =& \ -\frac{\pi^2}6 +\int_0^a {\cot t}\ln\frac{1+\sin t}{1-\sin t}+2\left( \frac{\ln\cos t}{\tan t}- \frac{\ln\sin t}{\cot t}\right) dt\\ =& \ -\frac{\pi^2}6+\bigg[\text{Li}_2(\sin t)-\text{Li}_2(-\sin t) +2\ln(\sin t)\ln(\cos t)\bigg]_0^a\\ =&\ -\frac{\pi^2}6+\text{Li}_2(\sin a)-\text{Li}_2(-\sin a) +2\ln(\sin a)\ln(\cos a) \end{align} where $\int_0^a \cot t \ln(1\pm \sin t)dt=-\text{Li}_2(\mp \sin a) $.