Closed form for ∫10ln2x√1−x+x2dx{\large\int}_0^1\frac{\ln^2x}{\sqrt{1-x+x^2}}dx

I want to find a closed form for this integral:
I=10ln2xx2x+1dx
Mathematica and Maple cannot evaluate it directly, and I was not able to find it in tables. A numeric approximation for it is
I2.100290124838430655413586565140170651784798511276914224…
(click here to see more digits).

Mathematica is able to find a closed form for a parameterized integral in terms of the Appell hypergeometric function:
I(a)=10xax2x+1dx=1a+1F1(a+1;12,12;a+2;(1)1/3,(1)2/3).
I suspect this expression could be rewritten in a simpler form, but I could not find it yet.

It’s easy to see that I=I but it’s unclear how to find a closed-form derivative of the Appell hypergeometric function with respect to its parameters.

Could you help me to find a closed form for I?


Update: Numerical calculations suggest that for all complex z with \Re(z)>0 the following functional equation holds:
z\,I(z-1)-\!\left(z+\tfrac12\right)\,I(z)+(z+1)\,I(z+1)=1.\tag4

Answer

Edited for a more concise derivation:

Define \mathcal{I} to be the value of the definite integral,

\mathcal{I}:=\int_{0}^{1}\frac{\ln^{2}{\left(x\right)}}{\sqrt{x^{2}-x+1}}\,\mathrm{d}x.\tag{1}

The definite integral \mathcal{I} is found to have as an approximate numerical value

\mathcal{I}\approx2.10029.\tag{2}


We begin by transforming the integral via an Euler substitution of the first kind:

\sqrt{x^{2}-x+1}=x+t\implies x=\frac{1-t^{2}}{1+2t},\tag{3}

\implies\mathrm{d}x=\frac{\left(-2\right)\left(1+t+t^{2}\right)}{\left(1+2t\right)^{2}}\,\mathrm{d}t,

\implies\sqrt{x^{2}-x+1}=\frac{1-t^{2}}{1+2t}+t=\frac{1+t+t^{2}}{1+2t}.

Under the transformation (3), the integral \mathcal{I} becomes

\begin{align}
\mathcal{I}
&=\int_{0}^{1}\frac{\ln^{2}{\left(x\right)}}{\sqrt{x^{2}-x+1}}\,\mathrm{d}x\\
&=2\int_{0}^{1}\frac{\ln^{2}{\left(\frac{1-t^{2}}{1+2t}\right)}}{1+2t}\,\mathrm{d}t;~~~\small{\left[\sqrt{x^{2}-x+1}=x+t\right]}\\
&=2\int_{0}^{1}\frac{\left[\ln{\left(1-t\right)}+\ln{\left(1+t\right)}-\ln{\left(1+2t\right)}\right]^{2}}{1+2t}\,\mathrm{d}t.\tag{4}\\
\end{align}

Next, using the algebraic identity

\left(a+b-c\right)^{2}=a^{2}+2b^{2}-\left(a-b\right)^{2}+\left(a-c\right)^{2}-2bc,

with a=\ln{\left(1-t\right)}\land b=\ln{\left(1+t\right)}\land c=\ln{\left(1+2t\right)}, we may expand the integral \mathcal{I} as a sum of five simpler logarithmic integrals:

\begin{align}
\mathcal{I}
&=2\int_{0}^{1}\frac{\left[\ln{\left(1-t\right)}+\ln{\left(1+t\right)}-\ln{\left(1+2t\right)}\right]^{2}}{1+2t}\,\mathrm{d}t\\
&=2\int_{0}^{1}\frac{\ln^{2}{\left(1-t\right)}}{1+2t}\,\mathrm{d}t+4\int_{0}^{1}\frac{\ln^{2}{\left(1+t\right)}}{1+2t}\,\mathrm{d}t\\
&~~~~~-2\int_{0}^{1}\frac{\left[\ln{\left(1-t\right)}-\ln{\left(1+t\right)}\right]^{2}}{1+2t}\,\mathrm{d}t\\
&~~~~~+2\int_{0}^{1}\frac{\left[\ln{\left(1-t\right)}-\ln{\left(1+2t\right)}\right]^{2}}{1+2t}\,\mathrm{d}t\\
&~~~~~-4\int_{0}^{1}\frac{\ln{\left(1+t\right)}\ln{\left(1+2t\right)}}{1+2t}\,\mathrm{d}t\\
&=2\int_{0}^{1}\frac{\ln^{2}{\left(1-t\right)}}{1+2t}\,\mathrm{d}t+4\int_{0}^{1}\frac{\ln^{2}{\left(1+t\right)}}{1+2t}\,\mathrm{d}t\\
&~~~~~-2\int_{0}^{1}\frac{\ln^{2}{\left(\frac{1-t}{1+t}\right)}}{1+2t}\,\mathrm{d}t+2\int_{0}^{1}\frac{\ln^{2}{\left(\frac{1-t}{1+2t}\right)}}{1+2t}\,\mathrm{d}t\\
&~~~~~-4\int_{0}^{1}\frac{\ln{\left(1+t\right)}\ln{\left(1+2t\right)}}{1+2t}\,\mathrm{d}t.\tag{5}\\
\end{align}

Now, we’ll find it convenient to introduce the following two auxiliary functions:

J{\left(z\right)}:=\int_{0}^{1}\frac{\ln^{2}{\left(y\right)}}{1+zy}\,\mathrm{d}y;~~~\small{z\ge-1},\tag{6a}

and

H{\left(a,c\right)}:=\int_{0}^{1}\frac{\ln^{2}{\left(1+ay\right)}}{1+cy}\,\mathrm{d}y;~~~\small{a\ge-1\land c>-1}.\tag{6b}

As we shall see, the integral \mathcal{I} can be expressed entirely in terms of the auxiliary functions J and H, and hence, entirely in terms of elementary functions and the standard polylogarithms:

\begin{align}
\mathcal{I}
&=2\int_{0}^{1}\frac{\ln^{2}{\left(1-t\right)}}{1+2t}\,\mathrm{d}t+4\int_{0}^{1}\frac{\ln^{2}{\left(1+t\right)}}{1+2t}\,\mathrm{d}t\\
&~~~~~-2\int_{0}^{1}\frac{\ln^{2}{\left(\frac{1-t}{1+t}\right)}}{1+2t}\,\mathrm{d}t+2\int_{0}^{1}\frac{\ln^{2}{\left(\frac{1-t}{1+2t}\right)}}{1+2t}\,\mathrm{d}t\\
&~~~~~-\int_{0}^{1}\frac{4\ln{\left(1+t\right)}\ln{\left(1+2t\right)}}{1+2t}\,\mathrm{d}t\\
&=2\int_{0}^{1}\frac{\ln^{2}{\left(u\right)}}{3-2u}\,\mathrm{d}u;~~~\small{\left[1-t=u\right]}\\
&~~~~~+4\,H{\left(1,2\right)}\\
&~~~~~-\int_{0}^{1}\frac{4\ln^{2}{\left(v\right)}}{\left(3-v\right)\left(1+v\right)}\,\mathrm{d}v;~~~\small{\left[\frac{1-t}{1+t}=v\right]}\\
&~~~~~+2\int_{0}^{1}\frac{\ln^{2}{\left(w\right)}}{1+2w}\,\mathrm{d}w;~~~\small{\left[\frac{1-t}{1+2t}=w\right]}\\
&~~~~~\small{-\left(\left[\ln{\left(1+t\right)}\ln^{2}{\left(1+2t\right)}\right]_{0}^{1}-\int_{0}^{1}\frac{\ln^{2}{\left(1+2t\right)}}{1+t}\,\mathrm{d}t\right)};~~~\small{I.B.P.s}\\
&=\frac23\int_{0}^{1}\frac{\ln^{2}{\left(u\right)}}{1-\frac23u}\,\mathrm{d}u+4\,H{\left(1,2\right)}\\
&~~~~~-\frac13\int_{0}^{1}\frac{\ln^{2}{\left(v\right)}}{1-\frac13v}\,\mathrm{d}v-\int_{0}^{1}\frac{\ln^{2}{\left(v\right)}}{1+v}\,\mathrm{d}v+2\,J{\left(2\right)}\\
&~~~~~-\ln{(2)}\ln^{2}{(3)}+H{\left(2,1\right)}\\
&=\frac23\,J{\left(-\frac23\right)}+4\,H{\left(1,2\right)}-\frac13\,J{\left(-\frac13\right)}\\
&~~~~~-J{\left(1\right)}+2\,J{\left(2\right)}-\ln{(2)}\ln^{2}{(3)}+H{\left(2,1\right)}.\tag{7}\\
\end{align}

The function H can be expressed in terms of J, dilogarithms, and elementary functions. Define \gamma:=\frac{a-c}{c}. For 0<a\land0<c, we have -1<\gamma and

\begin{align}
H{\left(a,c\right)}
&=\int_{0}^{1}\frac{\ln^{2}{\left(1+ay\right)}}{1+cy}\,\mathrm{d}y\\
&=\int_{\frac{1}{1+a}}^{1}\frac{\ln^{2}{\left(x\right)}}{x\left[c+\left(a-c\right)x\right]}\,\mathrm{d}x;~~~\small{\left[\frac{1}{1+ay}=x\right]}\\
&=\frac{1}{c}\int_{\frac{1}{1+a}}^{1}\frac{\ln^{2}{\left(x\right)}}{x}\,\mathrm{d}x-\frac{a-c}{c}\int_{\frac{1}{1+a}}^{1}\frac{\ln^{2}{\left(x\right)}}{c+\left(a-c\right)x}\,\mathrm{d}x\\
&=\frac{1}{3c}\ln^{3}{\left(1+a\right)}-\frac{a-c}{c^{2}}\int_{\frac{1}{1+a}}^{1}\frac{\ln^{2}{\left(x\right)}}{1+\left(\frac{a-c}{c}\right)x}\,\mathrm{d}x\\
&=\frac{1}{3c}\ln^{3}{\left(1+a\right)}-\frac{\gamma}{c}\int_{\frac{1}{1+a}}^{1}\frac{\ln^{2}{\left(x\right)}}{1+\gamma x}\,\mathrm{d}x\\
&=\frac{1}{3c}\ln^{3}{\left(1+a\right)}-\frac{\gamma}{c}\int_{0}^{1}\frac{\ln^{2}{\left(x\right)}}{1+\gamma x}\,\mathrm{d}x\\
&~~~~~+\frac{\gamma}{c}\int_{0}^{\frac{1}{1+a}}\frac{\ln^{2}{\left(x\right)}}{1+\gamma x}\,\mathrm{d}x\\
&=\frac{1}{3c}\ln^{3}{\left(1+a\right)}-\frac{\gamma}{c}J{\left(\gamma\right)}\\
&~~~~~+\frac{\gamma}{c\left(1+a\right)}\int_{0}^{1}\frac{\ln^{2}{\left(\frac{w}{1+a}\right)}}{1+\left(\frac{\gamma}{1+a}\right)w}\,\mathrm{d}w;~~~\small{\left[\left(1+a\right)x=w\right]}\\
&=\frac{1}{3c}\ln^{3}{\left(1+a\right)}-\left(\frac{a-c}{c^{2}}\right)J{\left(\frac{a-c}{c}\right)}\\
&~~~~~\small{+\frac{\gamma}{c\left(1+a\right)}\int_{0}^{1}\frac{\ln^{2}{\left(w\right)}-2\ln{\left(w\right)}\ln{\left(1+a\right)}+\ln^{2}{\left(1+a\right)}}{1+\left(\frac{\gamma}{1+a}\right)w}\,\mathrm{d}w}\\
&=\frac{1}{3c}\ln^{3}{\left(1+a\right)}-\left(\frac{a-c}{c^{2}}\right)J{\left(\frac{a-c}{c}\right)}\\
&~~~~~+\frac{\gamma}{c\left(1+a\right)}\int_{0}^{1}\frac{\ln^{2}{\left(w\right)}}{1+\left(\frac{\gamma}{1+a}\right)w}\,\mathrm{d}w\\
&~~~~~-\frac{2}{c}\ln{\left(1+a\right)}\int_{0}^{1}\frac{\left(\frac{\gamma}{1+a}\right)\ln{\left(w\right)}}{1+\left(\frac{\gamma}{1+a}\right)w}\,\mathrm{d}w\\
&~~~~~+\frac{1}{c}\ln^{2}{\left(1+a\right)}\int_{0}^{1}\frac{\left(\frac{\gamma}{1+a}\right)}{1+\left(\frac{\gamma}{1+a}\right)w}\,\mathrm{d}w\\
&=\frac{1}{3c}\ln^{3}{\left(1+a\right)}-\left(\frac{a-c}{c^{2}}\right)J{\left(\frac{a-c}{c}\right)}\\
&~~~~~+\frac{\gamma}{c\left(1+a\right)}\,J{\left(\frac{\gamma}{1+a}\right)}\\
&~~~~~+\frac{2}{c}\ln{\left(1+a\right)}\int_{0}^{1}\frac{\ln{\left(1+\left(\frac{\gamma}{1+a}\right)w\right)}}{w}\,\mathrm{d}w\\
&~~~~~+\frac{1}{c}\ln^{2}{\left(1+a\right)}\ln{\left(1+\left(\frac{\gamma}{1+a}\right)\right)}\\
&=\frac{1}{3c}\ln^{3}{\left(1+a\right)}-\left(\frac{a-c}{c^{2}}\right)J{\left(\frac{a-c}{c}\right)}\\
&~~~~~+\frac{a-c}{c^{2}\left(1+a\right)}\,J{\left(\frac{a-c}{c\left(1+a\right)}\right)}\\
&~~~~~-\frac{2}{c}\ln{\left(1+a\right)}\operatorname{Li}_{2}{\left(-\left(\frac{\gamma}{1+a}\right)\right)}\\
&~~~~~+\frac{1}{c}\ln^{2}{\left(1+a\right)}\ln{\left(\frac{a\left(1+c\right)}{\left(1+a\right)c}\right)}\\
&=\frac{a-c}{c^{2}\left(1+a\right)}\,J{\left(\frac{a-c}{c\left(1+a\right)}\right)}-\frac{a-c}{c^{2}}\,J{\left(\frac{a-c}{c}\right)}\\
&~~~~~-\frac{2}{c}\ln{\left(1+a\right)}\operatorname{Li}_{2}{\left(\frac{c-a}{c\left(1+a\right)}\right)}\\
&~~~~~+\frac{1}{3c}\ln^{3}{\left(1+a\right)}+\frac{1}{c}\ln^{2}{\left(1+a\right)}\ln{\left(\frac{a\left(1+c\right)}{\left(1+a\right)c}\right)}.\tag{8}\\
\end{align}

The function J reduces to the trilogarithm. For -1\le z\land z\neq0,

\begin{align}
J{\left(z\right)}
&=\int_{0}^{1}\frac{\ln^{2}{\left(y\right)}}{1+zy}\,\mathrm{d}y\\
&=-\frac{2}{z}\operatorname{Li}_{3}{\left(-z\right)}.\tag{9}\\
\end{align}

Thus, continuing from where we left off at the last line of (7),

\begin{align}
\mathcal{I}
&=\frac23\,J{\left(-\frac23\right)}-\frac13\,J{\left(-\frac13\right)}-J{\left(1\right)}+2\,J{\left(2\right)}\\
&~~~~~+4\,H{\left(1,2\right)}+H{\left(2,1\right)}-\ln{(2)}\ln^{2}{(3)}\\
&=2\operatorname{Li}_{3}{\left(\frac23\right)}-2\operatorname{Li}_{3}{\left(\frac13\right)}+2\operatorname{Li}_{3}{\left(-1\right)}-2\operatorname{Li}_{3}{\left(-2\right)}\\
&~~~~~\small{+4\left[-\operatorname{Li}_{3}{\left(\frac14\right)}+\operatorname{Li}_{3}{\left(\frac12\right)}-\ln{(2)}\operatorname{Li}_{2}{\left(\frac14\right)}-\frac56\ln^{3}{(2)}+\frac12\ln^{2}{(2)}\ln{(3)}\right]}\\
&~~~~~\small{+\left[2\operatorname{Li}_{3}{\left(-1\right)}-2\operatorname{Li}_{3}{\left(-\frac13\right)}-\ln{(9)}\operatorname{Li}_{2}{\left(-\frac13\right)}-\frac23\ln^{3}{(3)}+\ln{(4)}\ln^{2}{(3)}\right]}\\
&~~~~~-\ln{(2)}\ln^{2}{(3)}\\
&=2\operatorname{Li}_{3}{\left(\frac23\right)}-2\operatorname{Li}_{3}{\left(\frac13\right)}-2\operatorname{Li}_{3}{\left(-\frac13\right)}-2\operatorname{Li}_{3}{\left(-2\right)}\\
&~~~~~-4\operatorname{Li}_{3}{\left(\frac14\right)}+4\operatorname{Li}_{3}{\left(\frac12\right)}+4\operatorname{Li}_{3}{\left(-1\right)}\\
&~~~~~-4\ln{(2)}\operatorname{Li}_{2}{\left(\frac14\right)}-2\ln{(3)}\operatorname{Li}_{2}{\left(-\frac13\right)}\\
&~~~~~-\frac{10}{3}\ln^{3}{(2)}+2\ln^{2}{(2)}\ln{(3)}+\ln{(2)}\ln^{2}{(3)}-\frac23\ln^{3}{(3)}.\blacksquare\\
\end{align}


For z\notin[1,\infty),

\small{\operatorname{Li}_{3}{\left(z\right)}=-\operatorname{Li}_{3}{\left(\frac{z}{z-1}\right)}-\operatorname{Li}_{3}{\left(1-z\right)}+\frac{\ln^3{\left(1-z\right)}}{6}-\frac{\ln{\left(z\right)}\ln^2{\left(1-z\right)}}{2}+\frac{\pi^2}{6}\ln{\left(1-z\right)}+\zeta{(3)}}.

For 0<z<1,

\small{\operatorname{Li}_{3}{\left(z\right)}+\operatorname{Li}_{3}{\left(1-z\right)}+\operatorname{Li}_{3}{\left(\frac{z}{z-1}\right)}=\frac{\ln^3{\left(1-z\right)}}{6}-\frac{\ln{\left(z\right)}\ln^2{\left(1-z\right)}}{2}+\frac{\pi^2}{6}\ln{\left(1-z\right)}+\zeta{(3)}}.

Setting z=\frac13,

\small{\operatorname{Li}_{3}{\left(\frac13\right)}+\operatorname{Li}_{3}{\left(\frac23\right)}+\operatorname{Li}_{3}{\left(-\frac12\right)}=\frac{\ln^3{\left(\frac23\right)}}{6}-\frac{\ln{\left(\frac13\right)}\ln^2{\left(\frac23\right)}}{2}+\zeta{(2)}\ln{\left(\frac23\right)}+\zeta{(3)}}.

\small{\operatorname{Li}_{3}{\left(\frac13\right)}+\operatorname{Li}_{3}{\left(\frac23\right)}+\operatorname{Li}_{3}{\left(-\frac12\right)}=\frac{\ln^{3}{(2)}-3\ln{(2)}\ln^{2}{(3)}+2\ln^{3}{(3)}}{6}+\zeta{(2)}\ln{\left(\frac23\right)}+\zeta{(3)}}

\small{\operatorname{Li}_{3}{\left(\frac13\right)}+\operatorname{Li}_{3}{\left(\frac23\right)}+\operatorname{Li}_{3}{\left(-\frac12\right)}=\frac{\ln^{2}{\left(\frac32\right)}\ln{\left(18\right)}}{6}-\zeta{(2)}\ln{\left(\frac32\right)}+\zeta{(3)}}


Attribution
Source : Link , Question Author : Vladimir Reshetnikov , Answer Author : David H

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