# Evaluate $\int_0^1 \ln^2{\left(x^4+x^2+1\right)} \, \mathrm{d}x$

Evaluate $$\int_0^1 \ln^2{\left(x^4+x^2+1\right)} \, \mathrm{d}x$$

First thing I saw is $$x^4+x^2+1=(x^2+x+1)(x^2-x+1)$$ so the integral is the same as:

$$\begin{gather*} \int_0^1\ln^2{\left(x^2+x+1\right)} \, \mathrm{d}x + \int_0^1\ln^2{\left(x^2-x+1\right)} \, \mathrm{d}x \\ + 2\int_0^1\ln{\left(x^2+x+1\right)}\ln{\left(x^2-x+1\right)} \, \mathrm{d}x \end{gather*}$$

I dont know if this helps though. The last integral above seems to be the hardest, but still I don't even know how to evaluate first 2 integrals (perhaps Feynman's method)?

• @PeterForeman I saw it asked on a website and the other integrals have closed forms
– user801111
Aug 2, 2020 at 18:32
• Wolfram Alpha only gives a numeric value Aug 2, 2020 at 18:33
• Perhaps after doing integration by parts you can partial fraction the resulting expression with that factoring, or you could use geometric series $$1+x^2+x^4 = \frac{x^6-1}{x^2-1}$$ Aug 2, 2020 at 18:41
• Using Henry Lee's substitution and simplifying a bit, I managed to rewrite the integral as $$\ln\left(\frac{4}{3}\right)\left(-2\ln\left(6\right)-\pi\sqrt{3}+8\right)+\frac{1}{\sqrt{2}}\int_{0}^{\sqrt{2}}\ln^{2}\left(1+\frac{\left(x^{2}+1\right)^{2}}{3}\right)dx$$ Aug 3, 2020 at 3:07
• @Zophikel I didn't note down the exact steps, but I simplified $\frac{3}{4} \left( \tan^2(x)+1 \right) = \frac{3}{4} \sec^2(x)$, split the $\ln^2$ using this (which made $3$ integrals), and then solved the $3$ integrals (I think using the substitution $u = \sec(x)$). Aug 17, 2020 at 0:06

\begin{align}\boxed{I=\int_0^1 \ln^2(1+x^2+x^4)dx \\ = 32-12\ln 3-4\ln 2\ln 3-4 \pi \sqrt 3 +3\ln^2 3+2\pi \sqrt 3\ln 2+\frac{7\pi \ln 3}{\sqrt 3}\\ -8\sqrt 3 \operatorname{Ti}_2\left(\sqrt 3\right)+3\operatorname{Li}_2\left(-\frac13\right)+\operatorname{Li}_2\left(-3\right)-\operatorname{Li}_2\left(-8\right)}\end{align}

Where $$\operatorname{Ti}_{2}(x)$$ is the inverse tangent integral and $$\operatorname{Li}_2(x)$$ is the Dilogarithm.

To show this result we will start by using $$(a+b)^2=2a^2+2b^2-(a-b)^2$$ in order to rewrite the integral as: $$I=2\int_0^1 \ln^2(1+x+x^2)dx+2\int_0^1 \ln^2(1-x+x^2)dx -\int_0^1\ln^2\left(\frac{1+x+x^2}{1-x+x^2}\right)dx$$ $$\overset{x\to \frac{1-x}{1+x}}=\color{blue}{4\int_0^1 \frac{\ln^2\left(\frac{3+x^2}{(1+x)^2}\right)}{(1+x)^2}dx+4\int_0^1 \frac{\ln^2\left(\frac{1+3x^2}{(1+x)^2}\right)}{(1+x)^2}dx}\color{red}{-2\int_0^1 \frac{\ln^2\left(\frac{3+x^2}{1+3x^2}\right)}{(1+x)^2}dx}$$ $$\overset{x\to \frac{1}{x}}=\color{blue}{4\int_0^\infty \frac{\ln^2\left(\frac{3+x^2}{(1+x)^2}\right)}{(1+x)^2}dx}\color{red}{-\int_0^\infty \frac{\ln^2\left(\frac{3+x^2}{1+3x^2}\right)}{(1+x)^2}dx}$$ Now we will integrate by parts, in the same time we'll also simplify things using partial fractions. \begin{align}\Rightarrow I=3\ln^2 3+12\underbrace{\int_0^\infty \frac{\ln(3+x^2)}{3+x^2}dx}_{=I_1}+4\underbrace{\int_0^\infty \frac{\ln(1+3x^2)}{3+x^2}dx}_{=I_2}-32\underbrace{\int_0^\infty \frac{\ln(1+x)}{3+x^2}dx}_{=I_3(1)}\\ -16\underbrace{\int_0^\infty \frac{\ln(3+x^2)}{(1+x)^2}dx}_{=I_4}+32\underbrace{\int_0^\infty \frac{\ln(1+x)}{(1+x)^2}dx}_{=I_5} -24\underbrace{\int_0^\infty \frac{\ln(3+x^2)}{(3+x^2)(1+x)}dx}_{=I_6(1)}\\+8\underbrace{\int_0^\infty \frac{\ln(1+3x^2)}{(3+x^2)(1+x)}dx}_{=I_7(3)}+32\underbrace{\int_0^\infty \frac{\ln(1+x)}{(3+x^2)(1+x)}dx}_{=I_8(1)}\end{align} And all that's left to do is to evaluate each integral in order to obtain the closed form.

We obtain immediately, using $$\int_0^\infty \frac{\ln(a+x^2)}{b+x^2}dx=\frac{\pi}{b}\ln(a+b)$$ that: $$I_1=\int_0^\infty \frac{\ln(3+x^2)}{(3+x^2)}dx=\frac{\pi \ln 2}{\sqrt 3}+\frac{\pi \ln 3}{2\sqrt 3}$$ $$I_2=\int_0^\infty \frac{\ln(1+3x^2)}{(3+x^2)}dx=\frac{2\pi \ln 2}{\sqrt 3}$$

$$I_3(t)=\int_0^\infty \frac{\ln(1+tx)}{3+x^2}dx\Rightarrow I_3'(t)=\int_0^\infty\frac{x}{(1+tx)(3+x^2)} dx$$ $$=\frac{3\pi}{2\sqrt 3}\frac{t}{1+3t^2}-\frac{\ln 3}{2}\frac{1}{1+3t^2}-\frac{\ln t}{1+3t^2}$$ $$I_3(1)=\int_0^1 I_3'(t)dt=\frac{\pi \ln 2}{2\sqrt 3}-\frac{\pi \ln 3}{6\sqrt 3}+\frac{\operatorname{Ti}_2(\sqrt 3)}{\sqrt 3}$$ Where the inverse tangent integral appears after integrating by parts the last term as: $$\small \int_0^1 \frac{\ln t}{1+3t^2}dt\overset{IBP}=-\frac{1}{\sqrt 3}\int_0^1\frac{\arctan(\sqrt 3t)}{t}dt\overset{\sqrt 3 t=x}=-\frac{1}{\sqrt 3}\int_0^\sqrt 3\frac{\arctan x}{x}dx=-\frac{\operatorname{Ti}_2(\sqrt 3)}{\sqrt 3}$$

$$I_4=\int_0^\infty \frac{\ln(3+x^2)}{(1+x)^2}dx\overset{IBP}=\frac{3\pi}{4\sqrt 3}+\frac34\ln 3$$

$$I_5=\int_0^\infty\frac{\ln(1+x)}{(1+x)^2}dx\overset{IBP}=1$$

$$I_6(t)=\int_0^\infty \frac{\ln(3+tx^2)}{(3+x^2)(1+x)}dx\Rightarrow I_6'(t)=\int_0^\infty \frac{x^2}{(3+tx^2)(3+x^2)(1+x)}dx$$ $$=\frac{1}{8}\frac{\ln \left(\frac3t\right)}{3+t}+\frac{\pi}{8\sqrt 3}\frac{\sqrt t}{3+t}-\frac{1}{8}\frac{\ln t}{1- t}-\frac{\pi}{8\sqrt 3}\frac{1}{1+\sqrt t}$$ $$\small I_6(1)=\int_0^1I_6'(t)dt+\underbrace{\frac{\pi\ln 3}{8\sqrt 3}+\frac{\ln^2 3}{8}}_{=I_6(0)}=\frac{\ln 2 \ln 3}{4}-\frac18\operatorname{Li}_2\left(-\frac13\right)-\frac{\pi^2}{48}+\frac{\pi\ln 2}{4\sqrt 3}+\frac{\pi\ln 3}{8\sqrt 3}$$ Also that Dilogarithm comes from the first term, since: $$\int_0^1 \frac{\ln\left(\frac{3}{t}\right)}{3+t}dt\overset{\frac{t}{3}=x}=-\int_0^\frac13 \frac{\ln t}{1+t}dt\overset{IBP}=2\ln 2\ln 3-\ln^2 3-\operatorname{Li}_2\left(-\frac13\right)$$

$$I_7(t)=\int_0^\infty \frac{\ln(1+tx^2)}{(3+x^2)(1+x)}dx\Rightarrow I_7'(t)=\int_0^\infty \frac{x^2}{(1+tx^2)(3+x^2)(1+x)}dx$$ $$=\frac{\pi}{8}\frac{\sqrt t}{1+t}+\frac{3\pi \sqrt t -\pi\sqrt 3}{8}\frac{1}{1-3t}-\frac18\frac{\ln t}{1+t}-\frac38\frac{\ln (3t)}{1-3t}$$ $$I_7(3)=\int_0^3I_7'(t)dt=\frac{\pi \ln 2}{2\sqrt 3}-\frac{\pi^2}{16}-\frac{\ln 2\ln 3}{4}-\frac{\operatorname{Li}_2(-3)}{8}-\frac18\operatorname{Li}_2(-8)$$ It's perhaps worth to mention here the last integral: $$\small \int_0^1 \frac{\ln(3t)}{1-3t}dt\overset{3t=x}=\frac13\int_0^9\frac{\ln x}{1-x}dx=\frac13\operatorname{Li}_2(1-x)\bigg|_0^9=\frac13\operatorname{Li}_2(-8)-\frac{\pi^2}{18}$$

$$I_8(t)=\int_0^\infty \frac{\ln(1+tx)}{(3+x^2)(1+x)}dx\Rightarrow I_8'(t)=\int_0^\infty\frac{x}{(1+tx)(3+x^2)(1+x)} dx$$ $$=\frac14\frac{\ln t}{1-t}+\frac{3\pi}{8\sqrt 3}\frac{1+t}{1+3x^2}+\frac{\ln 3}{8}\frac{3t-1}{1+3t^2}-\frac14\frac{\ln t}{1+3t^2}+\frac34\frac{t\ln t}{1+3t^2}$$ $$I_8(1)=\int_0^1 I_8'(t)dt=\frac{\pi\ln 2}{8\sqrt 3}+\frac{\ln 2\ln 3}{8}-\frac{\pi \ln 3}{24\sqrt 3}+\frac{\operatorname{Ti}_2(\sqrt 3)}{4\sqrt 3}+\frac{\operatorname{Li}_2(-3)}{16}$$

• You can use \begin{aligned} and \end{aligned} to make long formulas fall in a single TeX block. Aug 3, 2020 at 13:18
• That is some dedication to the problem! Hats off.
– zwim
Aug 3, 2020 at 19:32
• More than impressive work ! Aug 4, 2020 at 1:29
• Thanks a lot for the compliments! Aug 4, 2020 at 10:14

I would maybe try: $$x^4+x^2+1=(x^2+\frac12)^2+\frac34$$ and so: $$\int_0^1\ln^2(x^4+x^2+1)dx=\int_0^1\ln^2\left[(x^2+\frac12)^2+\frac{\sqrt{3}}{2}^2\right]dx$$ Now we know that: $$\tan^2u+1=\sec^2u$$ so by letting: $$\left[\frac{2}{\sqrt{3}}\left(x^2+\frac12\right)\right]=\tan(u)$$ $$\frac{4}{\sqrt{3}}xdx=\sec^2udu$$ we get: $$I=\frac{\sqrt{3}}4\int\limits_{\frac\pi6}^{\frac\pi3}\ln^2\left[\frac32\tan^2u+\frac32\right]\left(\frac{\sqrt{3}\tan(u)-1}{2}\right)^{-1/2}\sec^2(u)du$$ and whilst the inside of the natural log simplifies nicely the rest is still rather ugly so I'm unsure if I can get a nice result from it but its worth a try :)

• Numerically, the two integrals don't seem to be equal. It should equal $\approx 0.247$, but your integral is $\approx 1.238$. Aug 2, 2020 at 21:33
• I think it should be $$\ln^2\left[\frac{3}{\color{red}{4}}\tan^2u+\frac{3}{\color{red}{4}}\right]$$ in the integrand. Aug 2, 2020 at 21:44

If you enjoy very, bery long formulae, try another CAS to see the antiderivative (it would take pages to type).

The problem is that the symbolic evaluation at the bounds almost killed my computer. So, numerical evaluation of the symbolic results.

At the upper bound $$23.67702048724287969803516653795923942977580171907339873883701501010155$$ At the lower bound $$23.43008216952775709246803606740565726521761645296916931243549426722402$$

Then, for the definite integral $$0.2469383177151226055671304705535821645581852661042294264015207428775327$$ which is not recognized by inverse symbolic calculators.

I seriously wonder if the square applies to the logarithm and not to its arguments; if this was the case, the problem would be quite simple.

• My guess is so, as in $\sin^2(\theta) = (\sin(\theta))^2$ Aug 3, 2020 at 5:33