On the integral $\int_0^1 \frac{\ln(2-x)}{1+x^2} \, \mathrm{d}x$

Question

When I found closed form of integrals $1$ and $2$ \begin{align} \int_0^1 \frac{\ln(1+x)}{1+x^2} \, \mathrm{d}x \tag{1}\\ \int_0^1 \frac{\ln(1-x)}{1+x^2} \, \mathrm{d}x \tag{2}\\ \end{align}

I thought why not generalize this for any $ b \in (-\infty, 1]$ like this $$ \mathcal{I}(b) = \int_0^1 \frac{\ln(1-bx)}{1+x^2} \, \mathrm{d}x \tag{3}\\$$

Using differentiation under integral sign I got that,

$$ \mathcal{I}'(b) = \frac{1}{b^2+1} \bigg[ -\ln(1-b) + \frac{1}{2}\ln(2) - b\frac{\pi}{4} \bigg] \tag{4}$$ Now after integrating [$4$] from $0$ to $y$ we get,

$$ \mathcal{I}(y)= -\int_0^y \frac{\ln(1-x)}{x^2+1} \, \mathrm{d}x + \frac{1}{2} \ln(2) \arctan(y) - \frac{\pi}{8}\ln(y^2+1) \tag{5} $$

I wanted to find closed form of [3] at $b=1/2$,

$$ \mathcal{I}(1/2) = \int_0^1 \frac{\ln(2-x) - \ln(2)}{1+x^2} \, \mathrm{d}x $$

from [5] I got,

$$ \mathcal{I}(1/2)= -\underbrace{\int_0^{1/2} \frac{\ln(1-x)}{x^2+1} \, \mathrm{d}x}_{\mathcal{J}} + \frac{1}{2} \ln(2) \arctan\left( \frac{1}{2}\right) - \frac{\pi}{8}\ln\left( \frac{5}{4}\right) \tag{6} $$

Now I have been struggling with $\mathcal{J}$. I have no idea how to find its closed form. I tried by substituting $x \to x/2$ then I got, $$ \mathcal{J} = \int_0^1 \frac{2\ln(2-x)}{x^2+4} \, \mathrm{d}x $$ Now I am totally stuck at this one.... Series expansion is making it more difficult to evaluate.

I search on Approach Zero but couldn't find which can help me.

Thank you very much!

Answer 1

You might find helpful the following beautiful results from More (Almost) Impossible Integrals, Sums, and Series: A New Collection of Fiendish Problems and Surprising Solutions (2023), page $7$, which is the sequel of (Almost) Impossible Integrals, Sums, and Series (2019): $$\int_0^a \frac{\log(1+x)}{1+x^2}\textrm{d}x$$ $$=\arctan(a)\log(1+a)-\frac{\pi}{4}\operatorname{arctanh}(a)+\frac{1}{2}G-\frac{1}{2}\operatorname{Ti}_2(a)$$ $$-\frac{1}{2}\operatorname{Ti}_2\left(\frac{1-a}{1+a}\right)+\frac{1}{4}\operatorname{Ti}_2\left(\frac{2a}{1-a^2}\right), \ |a|<1, \ a\in \mathbb{R},$$ where $G$ represents the Catalan's constant and $\displaystyle \operatorname{Ti}_2(x)=\int_0^x\frac{\arctan(t)}{t}\textrm{d}t$ denotes the Inverse tangent integral.

Following the pathway style you started with, if you have the case $\displaystyle \int_0^b \frac{\log(1+x)}{1+x^2}\textrm{d}x$, where $b >1$, you can let the variable change $x\mapsto 1/x$, use that $1/b=a$, and reduce again all to the initial integral together with trivial integrals. That simple! (in the blink of an eye)

Also, the following forms are given in the second book mentioned above $$\ \int_0^1 \frac{\log(1-a x)}{1+x^2}\textrm{d}x$$ $$=\arctan(a)\log(1-a)-\frac{1}{2}\log(2)\arctan(a)+\frac{\pi}{8}\log(1+a^2)+\frac{\pi}{4}\operatorname{arctanh}(a)$$ $$-\frac{1}{2}G-\frac{1}{2}\operatorname{Ti}_2(a)+\frac{1}{2}\operatorname{Ti}_2\left(\frac{1-a}{1+a}\right)+\frac{1}{4}\operatorname{Ti}_2\left(\frac{2a}{1-a^2}\right), \ |a|<1, \ a\in \mathbb{R}.$$ $$ \int_0^1 \frac{\log(1+ax)}{1+x^2}\textrm{d}x$$ $$=\frac{1}{2}\log(2)\arctan(a)-\arctan(a)\log(1+a)+\frac{\pi}{8}\log(1+a^2)+\frac{\pi}{4}\operatorname{arctanh}(a)$$ $$-\frac{1}{2}G+\frac{1}{2}\operatorname{Ti}_2(a)+\frac{1}{2}\operatorname{Ti}_2\left(\frac{1-a}{1+a}\right)-\frac{1}{4}\operatorname{Ti}_2\left(\frac{2a}{1-a^2}\right), \ |a|<1, \ a\in \mathbb{R}.$$

End of story (everything can be properly adjusted with a tiny bit of creativity)

Answer 2

Using first $$\frac{1}{1+x^2}=\frac{1}{(x+i)(x-i)}=\frac i 2\left(\frac{1}{x+i}-\frac{1}{x-i}\right)$$ Assuming $a>b>0$ $$\int_0^1 \frac{\log(a-b x)}{x+i}\,dx=-\text{Li}_2\left(\frac{a}{a+i b}\right)+\text{Li}_2\left(\frac{a-b}{a+i b}\right)-\log (a) \log \left(\frac{i b}{a+i b}\right)+$$ $$\log (a-b) \log \left(\frac{(1+i) b}{a+i b}\right)$$ Doing the smae for the second and recombining all pieces $$\int_0^1 \frac{\log(a-b x)}{1+x^2}\,dx=\log (a) \tan ^{-1}\left(\frac{a}{b}\right)-\log (a-b) \tan ^{-1}\left(\frac{a-b}{a+b}\right)+$$ $$\frac i 2\left(\text{Li}_2\left(\frac{a}{a-i b}\right)-\text{Li}_2\left(\frac{a-b}{a-i b}\right)-\text{Li}_2\left(\frac{a}{a+i b}\right)+\text{Li}_2\left(\frac{a-b}{a+i b}\right)\right)$$

For $a=2$ and $b=1$

$$\int_0^1 \frac{\log(2-x)}{1+x^2} \, \mathrm{d}x=\log (2) \tan ^{-1}(2)+$$ $$\frac i 2 \left(\text{Li}_2\left(\frac{4+2i}{5} \right)-\text{Li}_2\left(\frac{4-2i}{5} \right)+\text{Li}_2\left(\frac{2-i}{5}\right)-\text{Li}_2\left (\frac{2+i}{5}\right)\right)$$

Inverse symbolic calculators do not find what is $$-0.43232421462851709114401308429334124864499857979115\cdots$$

Answer 3

Preliminary integral $$\newcommand{\Li}{\operatorname{Li}}\newcommand{\Im}{\operatorname{Im}} \begin{align} \int_0^1\frac{\log(1-ax)}{x+b}\,\mathrm{d}x &=\int_b^{1+b}\frac{\log(1+ab-ax)}x\,\mathrm{d}x\tag{1a}\\ &=\log(1+ab)\log\left(\frac{1+b}b\right)+\int_b^{1+b}\frac{\log\left(1-\frac{ax}{1+ab}\right)}x\,\mathrm{d}x\tag{1b}\\ &=\log(1+ab)\log\left(\frac{1+b}b\right)+\int_{\frac{ab}{1+ab}}^{\frac{ab+a}{1+ab}}\frac{\log(1-x)}x\,\mathrm{d}x\tag{1c}\\ &=\log(1+ab)\log\left(\frac{1+b}b\right)-\Li_2\left(\frac{ab+a}{1+ab}\right)+\Li_2\left(\frac{ab}{1+ab}\right)\tag{1d} \end{align} $$ Explanation:
$\text{(1a):}$ substitute $x\mapsto x-b$
$\text{(1b):}$ apply $\int_b^{1+b}\frac{\log(1+ab)}x\,\mathrm{d}x=\log(1+ab)\log\left(\frac{1+b}b\right)$
$\text{(1c):}$ substitute $x\mapsto\frac{1+ab}ax$
$\text{(1d):}$ $\int\frac{\log(1-x)}x\,\mathrm{d}x=-\Li_2(x)+C$

---

Integral in Question $$ \begin{align} &\int_0^1\frac{\log(2-x)}{1+x^2}\,\mathrm{d}x\\ &=\frac\pi4\log(2)+\int_0^1\frac{\log\left(1-\frac12x\right)}{1+x^2}\,\mathrm{d}x\tag{2a}\\ &=\frac\pi4\log(2)-\frac1{2i}\int_0^1\frac{\log\left(1-\frac12x\right)}{x+i}\,\mathrm{d}x+\frac1{2i}\int_0^1\frac{\log\left(1-\frac12x\right)}{x-i}\,\mathrm{d}x\tag{2b}\\ &=\frac\pi4\log(2)\\ &-\frac1{2i}\left(\log\left(1+\frac12i\right)\log(1-i)-\Li_2\left(\frac{3+i}5\right)+\Li_2\left(\frac{1+2i}5\right)\right)\\ &+\frac1{2i}\left(\log\left(1-\frac12i\right)\log(1+i)-\Li_2\left(\frac{3-i}5\right)+\Li_2\left(\frac{1-2i}5\right)\right)\tag{2c}\\ &=\frac\pi4\log(2)+\Im\left(\log\left(1-\frac12i\right)\log(1+i)-\Li_2\left(\frac{3-i}5\right)+\Li_2\left(\frac{1-2i}5\right)\right)\tag{2d}\\ &=\frac\pi8\log\left(\frac54\right)+\frac12\log(2)\tan^{-1}(2)-\Im\left(\Li_2\left(\frac{3-i}5\right)-\Li_2\left(\frac{1-2i}5\right)\right)\tag{2e}\\[5pt] &\approx0.33509279757101529304\tag{2f} \end{align} $$ Explanation:
$\text{(2a):}$ pull $\int_0^1\frac{\log(2)}{1+x^2}\,\mathrm{d}x=\frac\pi4\log(2)$ out of the integral
$\text{(2b):}$ $\frac1{1+x^2}=-\frac1{2i}\frac1{x+i}+\frac1{2i}\frac1{x-i}$
$\text{(2c):}$ apply $(1)$
$\text{(2d):}$ $\frac1{2i}z-\frac1{2i}\bar z=\Im(z)$
$\text{(2e):}$ $\Im\left(\log\left(1-\frac12i\right)\log(1+i)\right)$
$\phantom{\text{(2e):}}$ $=\frac\pi8\log\left(\frac54\right)-\frac\pi4\log(2)+\frac12\log(2)\tan^{-1}(2)$
$\text{(2f):}$ evaluate