Does $\int_{0}^1\frac{\ln(1+x+x^2)}{x}\mathrm dx$ have a closed form?

Question

$$\int_0^1 \frac{\ln(1+x+x^2)}{x} \mathrm{d}x = 1.09662$$

I am trying to find a closed form of this integral.

I think one might exist, this integral looks like it might be related to $\pi$, but I don't know.

Answer 1

First $$1+x+x^2=\frac{1-x^3}{1-x}$$ So rewrite the integrand as $$\int_{0}^{1} \frac{\ln(1-x^3)}{x}-\frac{\ln(1-x)}{x} dx.$$ But using u-subtitution $u=x^3,du=3x^2dx$ $$\int_{0}^{1} \frac{\ln(1-x^3)}{x}dx=\int_{0}^{1} \frac{\ln(1-u)}{3u}du$$, so this means $$\int_{0}^{1} \frac{\ln(1-u)}{3u}du-\int_{0}^{1}\frac{\ln(1-x)}{x} dx=\frac{-2}{3}\int_{0}^{1}\frac{\ln(1-x)}{x} dx.$$

Now it is well known from the Basel Problem: $$\int_{0}^{1}\frac{\ln(1-x)}{x} dx=\int_{0}^{1}\int_{0}^{1}\frac{-1}{1-xy} dydx=-\zeta(2)=\frac{-\pi^2}{6}.$$ So $$\int_{0}^{1}\frac{\ln(1+x+x^2)}{x}dx=\frac{\pi^2}{9}$$

Answer 2

Note that $\ln(1+x+x^2)=\ln(1-x^3)-\ln(1-x)=-\sum_{k=1}^{\infty}(x^{3n}/n)+\sum_{k=1}^{\infty}(x^{n}/n)$. Change the order of integration and summation (it is valid, why?), and we have $$\begin{align}\int_0^1 \frac{\ln(1+x+x^2)}{x} \mathrm{d}x&=\int_0^1\left(-\sum_{k=1}^\infty\frac{x^{3n-1}}{n}+\sum_{k=1}^\infty\frac{x^{n-1}}{n}\right)\mathrm{d}x\\ &=-\sum_{k=1}^\infty\int_0^1\frac{x^{3n-1}}{n}\mathrm dx+\sum_{k=1}^\infty\int_0^1\frac{x^{n-1}}{n}\mathrm dx\\ &=-\sum_{k=1}^\infty\frac{1}{3n^2}+\sum_{k=1}^\infty\frac{1}{n^2}=\frac{\pi^2}9. \end{align}$$

Answer 3

Let $\omega=\frac{-1+\sqrt3i}2$. Then $$\begin{align}&\int_0^1 \frac{\ln(1+x+x^2)}{x} \mathrm{d}x\\ &=\int_0^1 \frac{\ln((1-\omega x)(1-\bar\omega x))}{x} \mathrm{d}x\\&=\int_0^1\left(-\sum_{n=1}^\infty\frac{\omega^nx^{n-1}}{n}-\sum_{n=1}^\infty\frac{\bar\omega^nx^{n-1}}{n}\right)\mathrm{d}x\\ &=-\sum_{n=1}^\infty\frac{\omega^n+\bar\omega^n}{n^2}\\ &=\sum_{n=1}^\infty\frac{1}{(3n-2)^2}+\sum_{n=1}^\infty\frac{1}{(3n-1)^2}-\sum_{n=1}^\infty\frac{2}{(3n)^2}\\ &=\sum_{n=1}^\infty\frac{1}{n^2}-\frac13\sum_{n=1}^\infty\frac{1}{n^2}\\ &=\frac{\pi^2}{9}. \end{align}$$

Answer 4

$$ \begin{aligned} \int_0^1 \frac{\ln \left(1+x+x^2\right)}{x} d x = & \int_0^1 \frac{\ln \left(1-x^3\right)-\ln (1-x)}{x} d x \\ = & \frac{1}{3} \int_0^1 \frac{\ln \left(1-x^3\right)}{x^3} d\left(x^3\right)-\int_0^1 \frac{\ln (1-x)}{x} d x \\ = & {\frac{1}{3}(-\operatorname{L i_2}(1))+ \operatorname{L i_2}(1)} \\ = & \frac{2}{3} \operatorname{L i_2}(1) \\ = & \frac{\pi^2}{9} \end{aligned} $$

Answer 5

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Since no one's yet posted a solution by differentiating under the integral, I thought I'd add it here, especially since it allows us to solve a broader class of integrals at once with not much additional effort.

Consider the function $G:\mathbb{R}\rightarrow\mathbb{R}$ defined by the definite integral

$$G{\left(\theta\right)}:=-\frac12\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\theta\right)}+x^{2}\right)}}{x}.$$

The integral from the question then corresponds to the following specific value $G$:

$$G{\left(\frac{2\pi}{3}\right)}=-\frac12\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x+x^{2}\right)}}{x}.$$

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Note that the value of $G$ at $\pi$ can be reduced to a particular value of the ordinary dilogarithm function:

$$\begin{align} G{\left(\pi\right)} &=-\frac12\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\pi\right)}+x^{2}\right)}}{x}\\ &=-\frac12\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+2x+x^{2}\right)}}{x}\\ &=-\frac12\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left((1+x)^{2}\right)}}{x}\\ &=-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}}{x}\\ &=\operatorname{Li}_{2}{\left(-1\right)}\\ &=-\frac{\pi^{2}}{12}.\\ \end{align}$$

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Calculate the derivative of $G$ using the Leibniz integral rule for differentiating under the integral sign:

$$\begin{align} G^{\prime}{\left(\vartheta\right)} &=\frac{d}{d\vartheta}\left[-\frac12\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-2x\cos{\left(\vartheta\right)}+x^{2}\right)}}{x}\right]\\ &=-\int_{0}^{1}\mathrm{d}x\,\frac{\partial}{\partial\vartheta}\left[\frac{\ln{\left(1-2x\cos{\left(\vartheta\right)}+x^{2}\right)}}{2x}\right]\\ &=-\int_{0}^{1}\mathrm{d}x\,\frac{\sin{\left(\vartheta\right)}}{1-2x\cos{\left(\vartheta\right)}+x^{2}}\\ &=-\int_{1}^{0}\mathrm{d}y\,\frac{(-2)}{\left(1+y\right)^{2}}\cdot\frac{\sin{\left(\vartheta\right)}}{1-2\left(\frac{1-y}{1+y}\right)\cos{\left(\vartheta\right)}+\left(\frac{1-y}{1+y}\right)^{2}};~~~\small{\left[x=\frac{1-y}{1+y}\right]}\\ &=-\int_{0}^{1}\mathrm{d}y\,\frac{2\sin{\left(\vartheta\right)}}{\left(1+y\right)^{2}-2\left(1-y^{2}\right)\cos{\left(\vartheta\right)}+\left(1-y\right)^{2}}\\ &=-\int_{0}^{1}\mathrm{d}y\,\frac{\sin{\left(\vartheta\right)}}{\left(1+y^{2}\right)-\left(1-y^{2}\right)\cos{\left(\vartheta\right)}}\\ &=-\int_{0}^{1}\mathrm{d}y\,\frac{\sin{\left(\vartheta\right)}}{\left[1-\cos{\left(\vartheta\right)}\right]+y^{2}\left[1+\cos{\left(\vartheta\right)}\right]}\\ &=-\int_{0}^{1}\mathrm{d}y\,\frac{2\sin{\left(\frac{\vartheta}{2}\right)}\cos{\left(\frac{\vartheta}{2}\right)}}{2\sin^{2}{\left(\frac{\vartheta}{2}\right)}+2y^{2}\cos^{2}{\left(\frac{\vartheta}{2}\right)}}\\ &=-\int_{0}^{1}\mathrm{d}y\,\frac{\cot{\left(\frac{\vartheta}{2}\right)}}{1+y^{2}\cot^{2}{\left(\frac{\vartheta}{2}\right)}}\\ &=-\int_{0}^{\cot{\left(\frac{\vartheta}{2}\right)}}\mathrm{d}t\,\frac{1}{1+t^{2}};~~~\small{\left[y\cot{\left(\frac{\vartheta}{2}\right)}=t\right]}\\ &=-\int_{0}^{\tan{\left(\frac{\pi}{2}-\frac{\vartheta}{2}\right)}}\mathrm{d}t\,\frac{1}{1+t^{2}}\\ &=-\arctan{\left(\tan{\left(\frac{\pi}{2}-\frac{\vartheta}{2}\right)}\right)}.\\ \end{align}$$

On the interval $(0,2\pi)$ the derivative reduces to simply

$$G^{\prime}{\left(\vartheta\right)}-\arctan{\left(\tan{\left(\frac{\pi}{2}-\frac{\vartheta}{2}\right)}\right)}=-\left(\frac{\pi}{2}-\frac{\vartheta}{2}\right)=\left(\frac{\vartheta}{2}-\frac{\pi}{2}\right);~~~\small{0<\vartheta<2\pi}.$$

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It then follows from the fundamental theorem of calculus that, for any $\theta\in[0,2\pi]$, we have

$$\begin{align} G{\left(\theta\right)} &=G{\left(\pi\right)}+\int_{\pi}^{\theta}\mathrm{d}\vartheta\,G^{\prime}{\left(\vartheta\right)}\\ &=-\frac{\pi^{2}}{12}+\int_{\pi}^{\theta}\mathrm{d}\vartheta\,\left(\frac{\vartheta}{2}-\frac{\pi}{2}\right)\\ &=-\frac{\pi^{2}}{12}+\left(\frac{\theta}{2}-\frac{\pi}{2}\right)^{2}.\\ \end{align}$$

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