48
$\begingroup$

I want to find a closed form for this integral: $$I=\int_0^1\frac{\ln^2x}{\sqrt{x^2-x+1}}dx\tag1$$ Mathematica and Maple cannot evaluate it directly, and I was not able to find it in tables. A numeric approximation for it is $$I\approx2.100290124838430655413586565140170651784798511276914224...\tag2$$ (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: $$I(a)=\int_0^1\frac{x^a}{\sqrt{x^2-x+1}}dx\\=\frac1{a+1}F_1\left(a+1;\frac{1}{2},\frac{1}{2};a+2;(-1)^{\small1/3},-(-1)^{\small2/3}\right).\tag3$$ I suspect this expression could be rewritten in a simpler form, but I could not find it yet.

It's easy to see that $I=I''(0),$ 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: $$z\,I(z-1)-\!\left(z+\tfrac12\right)\,I(z)+(z+1)\,I(z+1)=1.\tag4$$

$\endgroup$
3
  • $\begingroup$ It can be rewritten as $~\displaystyle\int_{-\frac\pi6}^\frac\pi6\bigg(\ln\frac{\sqrt3~\tan x+1}2\bigg)^2\frac{dx}{\cos x}.~$ $\endgroup$
    – Lucian
    Commented Sep 4, 2014 at 6:04
  • $\begingroup$ Another equivalent form of the integral is $\displaystyle I=\frac8\pi\Re{\large\int}_0^1\frac{\operatorname{Li}_3\!\left(\frac12\left(1-x \sqrt{-3}\right)\right)}{\left(1-x\sqrt{-3}\right)\,\sqrt{1-x^2}}dx.$ $\endgroup$ Commented Sep 4, 2014 at 16:21
  • 9
    $\begingroup$ This looks promising or perhaps not: $$ 2\int_0^1\frac{\ln^2\left(\frac{1-t^2}{1+2t}\right)}{1+2t}\ dt. $$ $\endgroup$
    – Tunk-Fey
    Commented Sep 4, 2014 at 20:03

4 Answers 4

80
+50
$\begingroup$

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 $z=\frac13$,

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

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

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


$\endgroup$
10
  • 14
    $\begingroup$ Go David! +1! $\ddot\smile$ $\endgroup$
    – Tunk-Fey
    Commented Sep 6, 2014 at 6:58
  • 9
    $\begingroup$ I've to make a confession to you David, truth be told I've already had an idea to answer this OP since I posted my comment (I made my current answer two days ago) but I resisted to answer it and I plant to answer it when this is a bounty question. :P $\endgroup$
    – Tunk-Fey
    Commented Sep 6, 2014 at 7:21
  • 5
    $\begingroup$ David, if you allow me to assist in evaluating $I_4$, you may refer to my ADDENDUM in my edited answer. $\endgroup$
    – Tunk-Fey
    Commented Sep 7, 2014 at 12:06
  • 4
    $\begingroup$ @Tunk-Fey Your assistance is certainly appreciated, and I'm a little miffed I didn't notice that myself! I've been busy this week, but I finally got around to applying your suggestion and completing my answer. While it is numerically correct, I haven't yet verified it to be equivalent to Cleo's value. But this is probably just a matter of using the right polylog identities. TBC... $\endgroup$
    – David H
    Commented Sep 11, 2014 at 5:35
  • 4
    $\begingroup$ Like Tunk-Fey said in his answer, your calculation of $I_1,\cdots,I_6$ can be shortened by using his technique and equation $(5)$ $\endgroup$ Commented Oct 11, 2014 at 15:50
43
+100
$\begingroup$

$$I=12\operatorname{Li}_3\left(\frac13\right)+20\operatorname{Li}_3\left(\frac23\right)+\frac{32}3\operatorname{Li}_2\left(\frac23\right)\ln3+\left(4\ln2+\frac{10}{3}\ln3\right)\operatorname{Li}_2\left(\frac34\right)\\-\frac{163}6\zeta(3)+8\ln^32+\frac23\ln^22\cdot\ln3-\frac23\ln2\cdot\ln^23-\frac{7\pi^2}3\ln2+\frac{11\pi^2}{9}\ln3.$$

$\endgroup$
0
16
$\begingroup$

I have an answer, but it is somewhat too lengthy and messy so I use Wolfram Alpha to help me out. First we use Euler substitution by setting $t+x=\sqrt{x^2-x+1}$, then we have $$ x=\frac{1-t^2}{1+2t}\qquad\color{red}{\Rightarrow}\qquad dx=-\frac{2(1+t+t^2)}{(1+2t)^2} $$ and $$ \sqrt{x^2-x+1}=\frac{1+t+t^2}{1+2t}. $$ Hence \begin{align} I=\int_0^1\frac{\ln^2x}{\sqrt{x^2-x+1}}dx&=2\int_0^1\frac{\ln^2\left(\frac{1-t^2}{1+2t}\right)}{1+2t}\ dt\\ &=2\int_0^1\frac{\bigg[\ln(1-t)+\ln(1+t)-\ln(1+2t)\bigg]^2}{1+2t}\ dt.\tag1 \end{align} Set $u=1+2t\ \color{red}{\Rightarrow}\ t=\dfrac{1+u}{2}$ and expand the integrand, we have \begin{align} \frac{\bigg[\ln\left(\frac{3-u}{2}\right)+\ln\left(\frac{1+u}{2}\right)-\ln u\bigg]^2}u&=\frac{\bigg[\ln3-2\ln2+\ln\left(1-\frac{u}{3}\right)+\ln\left(1+u\right)-\ln u\bigg]^2}u \end{align} Let from this part Wolfram Alpha takes over since it is very cumbersome and complicated, the integrand turns out to be

enter image description here

It seems none of the integrand are difficult to integrate it term-wise, either by substitution, IBP multiple times, or series expansion, but of course it would be tedious so let Wolfram Alpha do it. Let the expanded form of integrand be $f(u)$, then $(1)$ becomes $$ I=\int_1^3 f(u)\ du\tag2 $$ and the indefinite integral of $(2)$ is

enter image description here

Unfortunately, Wolfram Alpha fails to give an answer like @Cleo's after substituting the limits of integration. Perhaps, Mathematica or Maple is able to give the closed-form. I hope David H can complete his answer and can prove @Cleo's claim. Wish him luck!!


ADDENDUM :

I hope David H allow me to assist in the evaluation of $\color{blue}{I_4}$, of course he will take all the credit for evaluating this problem since he has done the biggest part of the calculation.

Consider \begin{align} J=\int_0^1\frac{1}{1+2t}\ln^2\left(\frac{1-t}{1+t}\right)\ dt. \end{align} It's easy to see that \begin{align} J&=\int_0^1\frac{\ln^2(1-t)}{1+2t}\ dt+\int_0^1\frac{\ln^2(1+t)}{1+2t}\ dt-2\int_0^1\frac{\ln(1-t)\ln(1+t)}{1+2t}\ dt\\ &=\color{red}{I_1}+\color{green}{I_2}-2\color{blue}{I_4}.\tag3 \end{align} To evaluate $J$, let $u=\dfrac{1-t}{1+t}$ then we have $$ t=\frac{1-u}{1+u}\quad ,\ 1+2t=\frac{3-u}{1+u}\ ,\quad\text{and}\quad dt=-\frac{4}{(1+u)^2}\ du. $$ Hence \begin{align} J&=4\int_0^1\frac{\ln^2u}{(3-u)(1+u)}\ dt\\ &={\large\int_0^1}\left[\frac{\ln^2u}{u+1}-\frac{\ln^2u}{u-3}\right]\ du\tag4 \end{align} where it is easy to prove that $$ \int\frac{\ln^2x}{x+a}\ dx=2\operatorname{Li}_2\left(-\frac xa\right)\ln x-2\operatorname{Li}_3\left(-\frac xa\right)+\ln\left(\frac{x+a}a\right)\ln^2x+C.\tag5 $$ The rest part is just technical matter and I leave it to David. Note that, the similar technique can be used to evaluate $I_5$ and $I_6$ in his calculation.

$\endgroup$
2
  • 7
    $\begingroup$ Are you user Cleo? $\endgroup$
    – Venus
    Commented Sep 6, 2014 at 7:10
  • 3
    $\begingroup$ @Venus No. Why did you think so? $\endgroup$
    – Tunk-Fey
    Commented Sep 6, 2014 at 7:26
9
$\begingroup$

I have not been either to find a closed form expression for this integral; so forgive me if what I send you is of no interest.

What I did is to expand the denominator as a Taylor series at $x=0$, which gives $$\frac{1}{\sqrt{x^2-x+1}}=1+\frac{x}{2}-\frac{x^2}{8}-\frac{7 x^3}{16}-\frac{37 x^4}{128}+\frac{23 x^5}{256}+\frac{331 x^6}{1024}+\frac{457 x^7}{2048}-\frac{2413 x^8}{32768}-\frac{17557 x^9}{65536}-\frac{49343 x^{10}}{262144}+O\left(x^{11}\right)$$ and take into account the fact that $$\int x^n\log^2(x)dx=\frac{x^{n+1} \left((n+1)^2 \log ^2(x)-2 (n+1) \log (x)+2\right)}{(n+1)^3}$$ which implies $$\int_0^1 x^n\log^2(x)dx=\frac{2}{(n+1)^3}$$ Using the coefficients, the summation does not seem to converge very fast.

Added later

In a comment to my answer, Vladimir Reshetnikov pointed out that the coefficients of the Taylor series are just $P_n\left(\frac{1}{2}\right)$ (what I totally missed). So, as an infinite series, the result is $$I=\int_0^1\frac{\ln^2x}{\sqrt{x^2-x+1}}dx\tag1=2 \sum_{n=0}^\infty\frac{P_n\left(\frac{1}{2}\right) }{(n+1)^3}$$ for which I did not find any closed form and which has a very slow convergence (adding $10,000$ terms leads to $\simeq 2.100290124838414$).

$\endgroup$
4
  • 1
    $\begingroup$ The coefficients of the Taylor series are values of the Legendre polynomial $P_n\left(\tfrac12\right).$ $\endgroup$ Commented Sep 4, 2014 at 4:47
  • $\begingroup$ Sorry for the typo; I fixed it. So, do you then have the expression as an infinite series ? $\endgroup$ Commented Sep 4, 2014 at 5:17
  • $\begingroup$ Yes, I even have a polynomial recurrence relation for its partial sums. But it's still not clear how to find a closed form for the infinite series. $\endgroup$ Commented Sep 4, 2014 at 5:27
  • 1
    $\begingroup$ Note that ${1 \over \sqrt{1 - 2th + h^{2}}} = \sum_{\ell = 0}^{\infty}h^{\ell}{\rm P}_{\ell}\left(\,t\,\right)$. With $t = 1/2$ and $h = x$ you have a nice expansion in Legendre Polynomia. $\endgroup$ Commented Sep 11, 2014 at 6:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .