Evaluating $\int_0^1 \frac{\ln(1+x)}{(1+x)(1+x^2)} \mathrm dx$

Question

I have an integration problem in which I tried to use king's property but failed. $$\int_0^1\frac{\ln(1+x)}{(1+x)(1+x^2)}\mathrm dx$$ First thought was $x=\tan\theta$ but it didn't help me much. $$\int_0^{\pi/4}\frac{\ln(1+\tan\theta)}{1+\tan\theta}\mathrm d\theta$$

Answer 1

Partial fraction decomposition of the integral would lead to,$$\begin{align}\int_0^1 \frac{\ln(1+x)}{(1+x)(1+x^2)} \, dx& = \frac{1}{2}\int_0^1\frac{\ln(1+x)}{1+x}\ dx\ + \frac12 \int_0^1 \frac{\ln(1 + x)(1-x)}{1 + x^2} dx\\&= \frac{1}{2}\int_0^1\frac{\ln(1+x)}{1+x}\ dx + \frac{1}{2} \int_0^1 \frac{\ln(1 + x)}{1 + x^2} dx - \frac{1}{2} \int_0^1 \frac{x\ln(1 + x)}{1 + x^2} dx\end{align}$$

The first integral evaluates to $\frac{\log^2(2)}{4}$ and the second integral can be continued with a substitution $x = \tan(\theta)$ and the third integral is here.

Answer 2

Another one $$I=\int \frac{\log(1+x)}{(1+x)(1+x^2)} \, dx$$ Write $$\frac{1}{(1+x)(1+x^2)}=\frac{1}{2 (x+1)}-\frac{1-i}{4(x+i)}-\frac{1+i}{4(x-i)}$$ to face three integrals $$J(a)=\int \frac{\log(1+x)}{x+a}\,dx=\text{Li}_2\left(-\frac{x+1}{a-1}\right)+\log (x+1) \log \left(\frac{a+x}{a-1}\right)$$ So, before simplifications $$\int_0^1 \frac{\log(1+x)}{(1+x)(1+x^2)} \, dx=\frac 14 \log^2(2)-$$ $$\frac {1-i}{384}\Big(\pi ^2+12 \log ^2(2)-12 i \pi \log (2) \Big)-\frac {1+i}{384}\Big(\pi ^2+12 \log ^2(2)+12 i \pi \log (2)\Big)$$

$$\int_0^1 \frac{\log(1+x)}{(1+x)(1+x^2)} \, dx=-\frac{\pi ^2}{192}+\frac{3 \log^2(2)}{16}+\frac{\pi \log(2)}{16} $$

Answer 3

King's property indeed simplifies the integral:

$$\begin{align}\int_0^1\frac{\ln(1+x)}{(1+x)(1+x^2)}\mathrm dx&=\int_0^\frac\pi4\frac{\ln(1+\tan x)}{1+\tan x}\mathrm dx\\&=\frac12\int_0^\frac\pi4\left(\ln2-\ln(1+\tan x)\right)(1+\tan x)\mathrm dx\\&=\frac{\ln2}2\left(\frac{\ln2}2+\frac\pi4\right)-\frac12\left(\frac\pi8\ln2\right)-\frac12\int_0^\frac\pi4\tan x\ln(1+\tan x)\mathrm dx&\left(\because\int_0^\frac\pi4\ln(1+\tan x)\mathrm dx=\frac\pi8\ln2\right)\\&=\frac{\ln^22}4+\frac{\pi\ln2}{16}-\frac12\underbrace{\int_0^1\frac{x\ln(1+x)}{1+x^2}\mathrm dx}_{\mathcal I}\end{align}$$

I will evaluate $\mathcal I$ using a series expansion approach:

$$\begin{align}\mathcal I&=\int_0^1x^2\left(1-\frac{x}2+\frac{x^2}3\cdots\right)\left(1-x^2+x^4\cdots\right)\mathrm dx\\&=\int_0^1x^2-\frac{x^3}2-\left(1-\frac13\right)x^4+\left(\frac12-\frac14\right)x^5+\left(1-\frac13-\frac15\right)x^6-\left(\frac12-\frac14+\frac16\right)x^7\cdots\mathrm dx\\&=\underbrace{\left(\frac1{1\cdot3}-\frac1{1\cdot5}+\frac1{3\cdot5}+\frac1{1\cdot7}-\frac1{3\cdot7}+\frac1{5\cdot7}\cdots\right)}_{S_1}+\underbrace{\left(-\frac1{2\cdot4}+\frac1{2\cdot6}-\frac1{4\cdot6}-\frac1{2\cdot8}+\frac1{4\cdot8}-\frac1{6\cdot8}\cdots\right)}_{S_2}\end{align}$$

Using the expansion of $\frac\pi4=1-\frac13+\frac15-\frac17\cdots$ and $\frac1{1^2}+\frac1{3^2}+\frac1{5^2}\cdots=\frac34\left(\frac1{1^2}+\frac1{2^2}+\frac1{3^2}\cdots\right)=\frac{\pi^2}8,$

$$\begin{align}S_1&=1\left(\frac13-\frac15+\frac17\cdots\right)-\frac13\left(-\frac15+\frac17-\frac19\cdots\right)+\frac15\left(\frac17-\frac19+\frac1{11}\cdots\right)\cdots\\&=1\left(-\frac\pi4+1\right)-\frac13\left(-\frac\pi4+1-\frac13\right)+\frac15\left(-\frac\pi4+1-\frac13+\frac15\right)\cdots\\&=-\frac\pi4\left(1-\frac13+\frac15\cdots\right)+\left(\frac1{1^2}+\frac1{3^2}+\frac1{5^2}\cdots\right)-S_1\\&=\frac{\pi^2}{32}\end{align}$$

Using the expansion of $\ln\sqrt2=\frac12-\frac14+\frac16-\frac18\cdots$ and $\frac1{2^2}+\frac1{4^2}+\frac1{6^2}\cdots=\frac14\left(\frac1{1^2}+\frac1{2^2}+\frac1{3^2}\cdots\right)=\frac{\pi^2}{24},$

$$\begin{align}S_2&=\frac12\left(-\frac14+\frac16-\frac18\cdots\right)-\frac14\left(\frac16-\frac18+\frac1{10}\cdots\right)+\frac16\left(-\frac18+\frac1{10}-\frac1{12}\cdots\right)\cdots\\&=\frac12\left(\ln\sqrt2-\frac12\right)-\frac14\left(\ln\sqrt2-\frac12+\frac14\right)+\frac16\left(\ln\sqrt2-\frac12+\frac14-\frac16\right)\cdots\\&=\ln\sqrt2\left(\frac12-\frac14+\frac16\cdots\right)-\left(\frac1{2^2}+\frac1{4^2}+\frac1{6^2}\cdots\right)-S_2\\&=\frac{\ln^22}8-\frac{\pi^2}{48}\end{align}$$

So, $$\mathcal I=\frac{\ln^22}8+\frac{\pi^2}{96}$$ $$\therefore\boxed{\int_0^1\frac{\ln(1+x)}{(1+x)(1+x^2)}\mathrm dx=\frac{3\ln^22}{16}+\frac{\pi\ln2}{16}-\frac{\pi^2}{192}}$$

Answer 4

One can evaluate this integral after the change of variable $x=\tan\theta$. Under $\theta=\frac\pi4-\phi$,

$$\int_0^{\pi/4}\frac{\log(1+\tan\theta)}{1+\tan\theta}d\theta=\frac12\int_0^{\pi/4}(1+\tan\phi)\left[\log2-\log(1+\tan\phi)\right]d\phi\\\implies2I=\frac{\log2}{2}\int_0^{\pi/4}A(\theta)d\theta+\int_0^{\pi/4}J(\theta)\left[\frac{1}{A(\theta)}-\frac{A(\theta)}{2}\right]d\theta$$ Where $A(\theta)=1+\tan\theta$ and $J(\theta)=\log(1+\tan\theta)$.

$$J:=\int_0^{\pi/4}J(\theta)\left[\frac{1}{A(\theta)}-\frac{A(\theta)}{2}\right]d\theta\stackrel{t=\tan\theta}{=}\int_0^1\log(1+t)\left[\frac{1-2t-t^2}{2(1+t)}\right]\frac{dt}{1+t^2}\\=-\frac{\log^22}{8}-\frac18\left[\operatorname{Li_2}\left(\frac1{\sqrt2}\right)-\operatorname{Li_2}\left(-\frac1{\sqrt2}\right)\right]. $$

Answer 5

With\begin{gather*} I=\int_{0}^{1}\dfrac{\ln(1+x)}{(1+x)(1+x^2)}\, \mathrm{d}x\\[2ex] I_1 = \dfrac{1}{2}\int_{0}^{1}\dfrac{\ln(1+x)}{1+x^2}\, \mathrm{d}x\\[2ex] f(s)=\dfrac{1}{2}\int_{0}^{1}\dfrac{x\ln(1+sx)}{1+x^2}\, \mathrm{d}x \end{gather*} we get (cf @Utkarsh) \begin{equation} I=\dfrac{1}{2}\int_{0}^{1}\dfrac{\ln(1+x)}{1+x}\, \mathrm{d}x+I_1-f(1)= \dfrac{1}{4}\ln^2(2)+I_1-f(1). \end{equation} If we in $I_1$ change $x$ to $\dfrac{1-x}{1+x}$ we get \begin{equation*} I_1 = \dfrac{1}{2}\int_{0}^{1}\dfrac{\ln(1+\frac{1-x}{1+x})}{1+x^2}\, \mathrm{d}x =\dfrac{1}{2}\int_{0}^{1}\dfrac{\ln(2)}{1+x^2}\, \mathrm{d}x -I_1 = \dfrac{\pi}{16}\ln(2). \end{equation*} Futhermore \begin{equation} f(1) = f(1)-f(0) = \int_{0}^{1}f'(s)\, \mathrm{d}s \end{equation} where \begin{gather*} f'(s) = \dfrac{1}{2}\int_{0}^{1}\dfrac{x^2}{(1+x^2)(1+sx)}\, \mathrm{d}x =\\[2ex] \dfrac{1}{2(1+s^2)}\int_{0}^{1}\left(\dfrac{sx}{1+x^2}-\dfrac{1}{1+x^2}+\dfrac{1}{1+sx}\right)\, \mathrm{d}x =\\[2ex] \dfrac{1}{4}\dfrac{s\ln(2)}{1+s^2}-\dfrac{\pi}{8(1+s^2)}+\dfrac{\ln(1+s)}{2s(1+s^2)}=\\[2ex] \dfrac{1}{4}\dfrac{s\ln(2)}{1+s^2}-\dfrac{\pi}{8(1+s^2)}+ \dfrac{1}{2}\dfrac{\ln(1+s)}{s}-\dfrac{1}{2}\dfrac{s\ln(1+s)}{1+s^2}. \end{gather*} If we return to $(2)$ we get \begin{gather*} f(1)= \dfrac{\ln^2(2)}{8}-\dfrac{\pi^{2}}{32}+\dfrac{1}{2}\int_{0}^{1}\dfrac{\ln(1+s)}{s}\, \mathrm{d}s -f(1) =\\[2ex] \dfrac{\ln^2(2)}{16}-\dfrac{\pi^{2}}{64} +\dfrac{\pi^{2}}{48} = \dfrac{\ln^2(2)}{16}+\dfrac{\pi^{2}}{192} \end{gather*} where we have used the Maclaurin expansion of $\ln(1+s)$. Finally we sum up in $(1)$. \begin{equation*} I=\dfrac{1}{4}\ln^2(2)+\dfrac{\pi}{16}\ln(2)-\dfrac{\ln^2(2)}{16}-\dfrac{\pi^{2}}{192} = \dfrac{3}{4}\ln^2(2)+\dfrac{\pi}{16}\ln(2)-\dfrac{\pi^{2}}{192}. \end{equation*}

Answer 6

Another solution using Feyman's trick. $$I(a)=\int_0^1 \frac{\log(1+ax)}{(1+x)(1+x^2)} \, dx$$ $$I'(a)=\int_0^1 \frac{x}{(x+1) \left(x^2+1\right) (a x+1)}\,dx$$ Partial fraction decomposition leads to $$I'(a)=\frac{\log (2)}{2 (a-1)}+\frac{\pi +2\log (2)}{8\left(a^2+1\right)}+\frac{a (\pi -2 \log (2))}{8 \left(a^2+1\right)}-$$ $$\frac{\log(a+1)}{2 (a-1)}+\frac{a \log (a+1)}{2 \left(a^2+1\right)}-\frac{\log(a+1)}{2 \left(a^2+1\right)}$$ For the last two, use again $$\frac{a}{a^2+1}=\frac 12\left(\frac{1}{a+i}+\frac{1}{a-i}\right)$$ $$\frac{1}{a^2+1}=\frac 12\left(\frac{i}{a+i}-\frac{i}{a-i}\right)$$

Recombine everything before using the bounds and obtain the result.