Evaluating $\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx$

Question

How can i evaluate this integral, maybe differentiation under the integral sign? i started expressing the integral as the following, $$\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx=\int _0^1\frac{\ln \left(x^2+x+1\right)}{x}\:dx-\int _0^1\frac{\ln \left(x^2+x+1\right)}{x+1}\:dx\:$$ But i dont know how to keep going, ill appreciate any solutions or hints.

Answer 1

I dont think Feynman's trick would work best here, following your path: $$\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx=\int _0^1\frac{\ln \left(x^2+x+1\right)}{x}\:dx-\underbrace{\int _0^1\frac{\ln \left(x^2+x+1\right)}{x+1}\:dx}_{x=\frac{1-t}{1+t}}\:$$ $$=\int _0^1\frac{\ln \left(x^3-1\right)}{x}\:dx-\int _0^1\frac{\ln \left(x-1\right)}{x}\:dx-\int _0^1\frac{\ln \left(x^2+3\right)}{x+1}\:dx+2\int _0^1\frac{\ln \left(x+1\right)}{x+1}\:dx$$ $$-\sum _{k=1}^{\infty }\frac{1}{k}\int _0^1x^{3k-1}\:dx\:+\sum _{k=1}^{\infty }\frac{1}{k}\:\int _0^1x^{k-1}\:dx-\int _0^1\frac{\ln \left(x^2+3\right)}{x+1}\:dx+\ln ^2\left(2\right)$$ To solve that remaining integral you can use the identity i derived here

So, $$=\frac{2\zeta \left(2\right)}{3}-(-\frac{\ln ^2\left(3\right)}{4}-\frac{\text{Li}_2\left(-\frac{1}{3}\right)}{2}-\frac{\ln ^2\left(4\right)}{4}+\frac{\ln \left(3\right)\ln \left(4\right)}{2}-\arctan ^2\left(\sqrt{\frac{1}{3}}\right)+\ln \left(2\right)\ln \left(4\right))+\ln ^2\left(2\right)$$ $$\frac{\pi ^2}{9}+\frac{\ln ^2\left(3\right)}{4}+\frac{\text{Li}_2\left(-\frac{1}{3}\right)}{2}+\ln ^2\left(2\right)-\ln \left(3\right)\ln \left(2\right)+\frac{\pi ^2}{36}-2\ln ^2\left(2\right)+\ln ^2\left(2\right)$$ So your integral's solution is, $$\boxed{\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx=\frac{5\pi ^2}{36}+\frac{\ln ^2\left(3\right)}{4}+\frac{\text{Li}_2\left(-\frac{1}{3}\right)}{2}-\ln \left(3\right)\ln \left(2\right)}$$

Answer 2

Here is a less clever but almost self-contained solution:

1. Let $a, b \in \mathbb{C} \setminus(-\infty, 0)$ and define

$$ F(a, b) := \int_{0}^{1} \frac{\log(1-ax)}{x+b} \, \mathrm{d}x, $$

where $\log$ is the complex logarithm with the branch cut along $(-\infty, 0]$. Then assuming for a moment that both $|a|$ and $|b|$ are sufficiently small (so as to not worry about the effect of branch cut of the complex logarithm),

\begin{align*} F(a, b) &= \int_{b}^{b+1} \frac{\log(1+ab-ax)}{x} \, \mathrm{d}x \\ &= \int_{b}^{b+1} \frac{\log(1+ab) + \log(1-\frac{a}{1+ab}x)}{x} \, \mathrm{d}x \\ &= \log(1+ab)\log\left(1+\frac{1}{b}\right) - \left[ \operatorname{Li}_2\left( \frac{ax}{1+ab} \right) \right]_{x=b}^{x=1+b} \\ &= \log(1+ab)\log\left(1+\frac{1}{b}\right) + \operatorname{Li}_2\left(\frac{ab}{1+ab} \right) - \operatorname{Li}_2\left(\frac{a(1+b)}{1+ab}\right), \tag{1} \end{align*}

where

$$ \operatorname{Li}_2(z) := -\int_{0}^{z} \frac{\log (1-t)}{t} \, \mathrm{d}t$$

is the dilogarithm function. The branch cut of $\log$ induces that of $\operatorname{Li}_2$ as $[1,\infty)$. So the above identity continues to hold as soon as $(a, b)$ can be connected to $(0, 0)$ via a path avoiding the set

$$ B := \biggl\{ (z, w) \in \mathbb{C}^2 : z < 0 \text{ or } w < 0 \text{ or } \frac{zw}{1+zw} > 1 \text{ or } \frac{z(1+w)}{1+zw} > 1 \biggr\}. $$

Also, when $b = 0$ we regard $\log(1+ab)\log\left(1+\frac{1}{b}\right) = 0$ by appealing to the continuity.

2. Write $\omega = e^{2\pi i/3} = \frac{-1+i\sqrt{3}}{2}$. Denoting by $I$ the integral, we have

\begin{align*} I &= \int_{0}^{1} \frac{\log(1-\omega x) + \log(1-\bar{\omega}x)}{x} \, \mathrm{d}x - \int_{0}^{1} \frac{\log(1-\omega x) + \log(1-\bar{\omega}x)}{1+x} \, \mathrm{d}x \\ &= F(\omega, 0) + F(\bar{\omega}, 0) - F(\omega, 1) + F(\bar{\omega}, 1). \end{align*}

Plugging the formula $\text{(1)}$,

$$ \begin{aligned} I &= \operatorname{Li}_2(1+i\sqrt{3}) + \operatorname{Li}_2(1-i\sqrt{3})\\ &\quad - \operatorname{Li}_2(e^{2\pi i/3}) - \operatorname{Li}_2(-e^{2\pi i/3}) - \operatorname{Li}_2(e^{-2\pi i/3}) - \operatorname{Li}_2(-e^{-2\pi i/3}). \end{aligned} \tag{2} $$

3. $\text{(2)}$ can be further simplified by using the dilogarithm identities:

\begin{align*} \operatorname{Li}_2(z) + \operatorname{Li}_2(-z) &= \tfrac{1}{2}\operatorname{Li}_2(z^2) \tag{DI1} \\ \operatorname{Li}_2(z) + \operatorname{Li}_2(1-z) &= \zeta(2) - \log z \log(1-z) \tag{DI2} \\ \operatorname{Li}_2(z) + \operatorname{Li}_2(1/z) &= -\zeta(2) - \tfrac{1}{2}\log^2(-z) \tag{DI3} \end{align*} Here, $\text{(DI1)}$ and $\text{(DI2)}$ hold for $z \notin (-\infty, 0]\cup[1,\infty)$, and $\text{(DI3)}$ holds for $z \notin [0, 1)$.

• From $\text{(DI1)}$, $$ \operatorname{Li}_2(e^{\pm2\pi i/3}) + \operatorname{Li}_2(-e^{\pm2\pi i/3}) = \frac{1}{2}\operatorname{Li}_2(e^{\pm4\pi i/3}) = \frac{1}{2}\operatorname{Li}_2(e^{\mp2\pi i/3}). $$ Summing these up for the choices of signs $\pm$ and simplifying, $$ \operatorname{Li}_2(e^{2\pi i/3}) + \operatorname{Li}_2(e^{-2\pi i/3}) = - 2\Bigl( \operatorname{Li}_2(-e^{2\pi i/3}) + \operatorname{Li}_2(-e^{-2\pi i/3}) \Bigr). $$ Then by $\text{(DI2)}$, $$ \operatorname{Li}_2(-e^{2\pi i/3}) + \operatorname{Li}_2(-e^{-2\pi i/3}) = \zeta(2) - \log (e^{\pi i/3}) \log (e^{-\pi i/3}) = \frac{\pi^2}{6} - \frac{\pi^2}{9} = \frac{\pi^2}{18}. $$ Therefore we get $$ I = \operatorname{Li}_2(1+i\sqrt{3}) + \operatorname{Li}_2(1-i\sqrt{3}) + \frac{\pi^2}{18}. \tag{3} $$

• By $\text{(DI2)}$ again, \begin{align*} \operatorname{Li}_2(1\pm i\sqrt{3}) &= \zeta(2) - \log(1\pm i\sqrt{3})\log(\mp i\sqrt{3}) - \operatorname{Li}_2(\mp i\sqrt{3}) \\ &= -\frac{(\log 2)(\log 3)}{2} \pm \frac{i\pi \log(8/3)}{6} - \operatorname{Li}_2(\mp i\sqrt{3}). \end{align*} Summing these up for the choices of signs $\pm$, \begin{align*} &\operatorname{Li}_2(1+i\sqrt{3}) + \operatorname{Li}_2(1-i\sqrt{3}) \\ &= -(\log 2)(\log 3) - \operatorname{Li}_2(i\sqrt{3}) - \operatorname{Li}_2(-i\sqrt{3}) \\ &= -(\log 2)(\log 3) - \frac{1}{2}\operatorname{Li}_2(-3) \tag*{by (DI1)} \\ &= -(\log 2)(\log 3) + \frac{\pi^2}{12} + \frac{1}{4} \log^2 3 + \frac{1}{2}\operatorname{Li}_2(-\tfrac{1}{3}) \tag*{by (DI3)} \end{align*} Plugging this back to $\text{(3)}$ proves that $$ I = \boxed{ \frac{5\pi^2}{36} -(\log 2)(\log 3) + \frac{1}{4} \log^2 3 + \frac{1}{2}\operatorname{Li}_2(-\tfrac{1}{3}) } $$ This is identical to @Dennis Orton's answer.

Answer 3

Solution using harmonic series

$$\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx=\int _0^1\frac{\ln \left(x^2+x+1\right)}{x}\:dx-\int _0^1\frac{\ln \left(x^2+x+1\right)}{x+1}\:dx\:$$

---

$$\int _0^1\frac{\ln \left(x^2+x+1\right)}{x}\:dx=\underbrace{\int _0^1\frac{\ln \left(1-x^3\right)}{x}\:dx}_{x^3\to x}-\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx$$

$$=-\frac23\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx=\frac23\zeta(2)$$

---

$$\int _0^1\frac{\ln \left(1+x+x^2\right)}{1+x}\:dx\overset{IBP}{=}\ln(2)\ln(3)-\int_0^1\frac{(2x+1)\ln(1+x)}{1+x+x^2}dx$$

For the latter integral, set $a=\frac{2\pi}{3}$ in the identity

$$\sum_{n=1}^{\infty}x^{n-1} \cos(na)=\frac{\cos(a)-x}{1-2x\cos(a)+x^2}, \ |x|<1$$

we have

$$-2\sum_{n=1}^{\infty}x^{n-1} \cos(n\frac{2\pi}{3})=\frac{2x+1}{1+x+x^2}$$

$$\Longrightarrow \int_0^1\frac{(2x+1)\ln(1+x)}{1+x+x^2}dx=-2\sum_{n=1}^\infty \cos(n\frac{2\pi}{3})\int_0^1 x^{n-1}\ln(1+x)dx$$

$$=-2\sum_{n=1}^\infty \cos(n\frac{2\pi}{3})\left(\frac{H_n-H_{n/2}}{n}\right)$$

$$=-2\Re\sum_{n=1}^\infty \left(e^{i\frac{2\pi}{3}}\right)^n\left(\frac{H_n-H_{n/2}}{n}\right)$$

And finally we use the generating functions

$$\sum_{n=1}^\infty x^n\frac{H_n}{n}=\frac12\ln^2(1-x)+\text{Li}_2(x)$$

$$\sum_{n=1}^\infty x^n\frac{H_{n/2}}{n}=i\pi\frac{\ln(1-x^2)-\ln(-x^2)}{x^2}$$

$$+\frac{\ln(x-1)\ln(-x^2)-\ln(x-1)\ln(1-x^2)}{x^2}$$

$$+\frac{\text{Li}_2\left(\frac{1-x}{1+x}\right)-\text{Li}_2\left(\frac{1}{1+x}\right)-\text{Li}_2\left(\frac{1}{1-x}\right)}{x^2}$$

I found the second generating function with help of Mathematica after I converted it to integral;

$$\sum_{n=1}^\infty x^n\frac{H_{n/2}}{n}=-\int_0^1\frac{xy^2\ln(1-y^2)}{1-xy}dy$$