Closed form for $\int_ 0^{1/2} {\frac {x} {3 +4 x^2}\ln\frac {\ln\left (1/2 +x \right)} {\ln\left (1/2 - x \right)}\mathrm {d} x} $

Question

Find the integral $$\int_ 0^{1/2} {\frac {x} {3 + 4 x^2}\ln\frac {\ln\left (1/2 + x \right)} {\ln\left (1/2 - x \right)}\mathrm {d} x} $$

Wolfram | Alpha tells me that the answer is $ - 0.0951875 ... $, and its possible closed form is $ - \frac18\ln2\ln3$. But I don't know how to do it.

Thanks in advance.

Answer 1

Here is an approach that largely relies on elementary techniques and manipulation of infinite sequences. To begin, first apply the integral identity below

$$\int\limits_a^b f(x)\,\mathrm dx=\int\limits_a^b f(a+b-x)\,\mathrm dx\tag{1}$$

Our integral simplifies a bit to

$$\mathfrak{I}\equiv\int\limits_0^{1/2}\frac x{3+4x^2}\log\left(\frac {\log\left(\frac 12+x\right)}{\log\left(\frac 12-x\right)}\right)\,\mathrm dx=\frac 18\int\limits_0^{1/2}\frac {1-2x}{1-x+x^2}\log\left(\frac {\log(1-x)}{\log x}\right)\,\mathrm dx\tag{2}$$

Our primary forcus now is the resulting integral, which will be denoted as $K$. To evaluate this integral, split up the natural logarithm and expand.

$$\begin{align*}K & =\int\limits_0^{1/2}\frac {1-2x}{1-x+x^2}\log\log(1-x)\,\mathrm dx-\int\limits_0^{1/2}\frac {1-2x}{1-x+x^2}\log\log x\,\mathrm dx\\ & =8\int\limits_0^{1/2}\frac {x}{4x^2+3}\log\log\left(x+\frac 12\right)\,\mathrm dx-\int\limits_0^{1/2}\frac {1-2x}{1-x+x^2}\log\log x\,\mathrm dx\tag{3}\\ & =-\int\limits_{1/2}^1\frac {1-2x}{1-x+x^2}\log\log x\,\mathrm dx-\int\limits_0^{1/2}\frac {1-2x}{1-x+x^2}\log\log x\,\mathrm dx\tag{4}\\ & =\int\limits_0^1\frac {2x-1}{1-x+x^2}\log\log x\,\mathrm dx\end{align*}$$

Where the first identity $(1)$ was applied to arrive at equation $(3)$ and the substitution $x\mapsto x+\tfrac 12$ was made to arrive at equation $(4)$. It can be shown through integration by parts that

$$\begin{align*}\int\limits_0^1\frac {2x-1}{1-x+x^2}\log\log x\,\mathrm dx & =\int\limits_0^1\frac {\log(1-x+x^2)}{x\log x}\,\mathrm dx\\ & =\int\limits_0^1\frac {2x-1}{1-x+x^2}\log\log\left(\frac 1x\right)\,\mathrm dx\end{align*}\tag{5}$$

Taking the last line of $(5)$, multiply the integrand by $1+x$ such that the denominator becomes $1+x^3$ and enforce the substitution $x\mapsto -\log x$ to arrive at

$$\begin{align*}K & =\int\limits_0^{+\infty}\frac {e^{-x}\left(2e^{-2x}+e^{-x}-1\right)}{1+e^{-3x}}\log x\,\mathrm dx\\ & =\sum\limits_{n=0}^{+\infty} (-1)^n\int\limits_0^{+\infty} e^{-x(3n+1)}\left(2e^{-2x}+e^{-x}-1\right)\log x\,\mathrm dx\tag{6}\end{align*}$$

After applying the formula

$$\int\limits_0^{+\infty} e^{-nx}\log x\,\mathrm dx=-\frac {\gamma+\log n}n$$

Equation $(6)$ becomes

$$\small K=2\sum\limits_{n=0}^{+\infty}(-1)^{n-1}\frac {\gamma+\log(3n+3)}{3n+3}+\sum\limits_{n=0}^{+\infty}(-1)^{n-1}\frac {\gamma+\log(3n+2)}{3n+2}+\sum\limits_{n=0}^{+\infty}(-1)^n\frac {\gamma+\log(3n+1)}{3n+1}$$

Fortunately for us, most of the terms readily cancel. The sums involving $\gamma$ cancel each other out, leaving behind only three sums of natural logarithms.

$$K=2\sum\limits_{n=1}^{+\infty}(-1)^{n}\frac {\log 3n}{3n}+\sum\limits_{n=0}^{+\infty}(-1)^n \frac {\log(3n+1)}{3n+1}+\sum\limits_{n=0}^{+\infty} (-1)^{n-1}\frac {\log(3n+2)}{3n+2}$$

Further simplification can be obtained by expanding each sum into its constituent terms and showing that only the $3n$ terms remain after simplification.

$$\begin{align*}K & =-\frac 23\log 2\log 3-\frac 13\sum\limits_{n=1}^{+\infty}(-1)^n\frac {\log n}n+\sum\limits_{n=1}^{+\infty}(-1)^n\frac {\log 3n}{3n}\\ & =-\log 2\log 3\end{align*}$$

Substituting our result for $K$ back into $(2)$ gives us our final answer

$$\int\limits_0^{1/2}\frac x{3+4x^2}\log\left(\frac {\log\left(\frac 12+x\right)}{\log\left(\frac 12-x\right)}\right)\,\mathrm dx\color{blue}{=-\frac 18\log 2\log 3}$$

Answer 2

Firstly, recognise the "symmetry" in the integral $$I = \int_{0}^{\frac{1}{2}} \frac{x}{3+4x^2} \ln \left(\frac{\ln(1/2+x)}{\ln(1/2-x)}\right)\, dx = \int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{x}{3+4x^2} \ln\left(-\ln\left(\frac{1}{2}+x\right)\right)\,dx$$

Now enforce the substitution $-\ln(1/2+x)=u$: $$\implies I = \frac{1}{8} \int_{0}^{\infty} \frac{\ln(u) (2-e^u)}{e^{2u}-e^u+1}\,du$$ Now consider firstly the Mellin transform $$\int_{0}^{\infty} \frac{x^{s-1}}{e^x-z}\,dx = \frac{1}{z} \Gamma (s) \operatorname{Li}_s (z)$$ which can be found here. Now, through partial fractions, we can determine that \begin{align}\int_{0}^{\infty} \frac{x^{s-1}}{(e^x+p-q)(e^x+p+q)}\,dx &= \int_{0}^{\infty} \frac{x^{s-1}}{e^{2x}+2pe^x+p^2-q^2}\,dx\\&=\frac{\Gamma(s)}{2q(p+q)} \operatorname{Li}_s (-p-q) - \frac{\Gamma(s)}{2q(p-q)} \operatorname{Li}_s (-p+q)\end{align} Thus setting $p=-\frac{1}{2}$ and $q=\frac{i \sqrt{3}}{2}$: $$\int_{0}^{\infty} \frac{x^{s-1}}{e^{2x}-e^x+1}\,dx=-\frac{\Gamma(s)}{2\sqrt{3}}\left(\left(\sqrt{3}-i\right)\operatorname{Li}_s\left(\frac{1}{2}-\frac{i\sqrt{3}}{2}\right)+\left(\sqrt{3}+i\right)\operatorname{Li}_s\left(\frac{1}{2}+\frac{i\sqrt{3}}{2}\right)\right)$$ Through a similar process, we can determine $$\int_{0}^{\infty} \frac{x^{s-1} e^x}{(e^x+p-q)(e^x+p+q)}\,dx = \frac{\Gamma(s)}{2q} \left(\operatorname{Li}_s (-p+q)-\operatorname{Li}_s (-p-q)\right)$$

In general: $$\int_{0}^{\infty} \frac{x^{s-1}}{a \cosh x+b\sinh x+c}\,dx=\frac{\Gamma(s)}{\sqrt{b^2+c^2-a^2}} \left(\operatorname{Li}_s \left(\frac{\sqrt{b^2+c^2-a^2}-c}{a+b}\right)-\operatorname{Li}_s \left(-\frac{\sqrt{b^2+c^2-a^2}+c}{a+b}\right)\right)$$ which I utilised here.

This gives $$\int_{0}^{\infty} \frac{x^{s-1} e^x}{e^{2x}-e^x+1} \,dx=-\frac{i\cdot\Gamma(s)}{\sqrt{3}}\left(\operatorname{Li}_s\left(\frac{1+i\sqrt{3}}{2}\right)-\operatorname{Li}_s\left(\frac{1-i\sqrt{3}}{2}\right)\right)$$

Thus $$\int_{0}^{\infty} \frac{x^{s-1}(2-e^x)}{e^{2x}-e^x+1}\,dx=-\Gamma(s) \left(\operatorname{Li}_s\left(\frac{1+i \sqrt{3}}{2}\right)+\operatorname{Li}_s\left(\frac{1-i \sqrt{3}}{2}\right)\right)$$

Giving finally $$I = -\frac{1}{8}\cdot\frac{d}{ds}\left(\Gamma(s) \left(\operatorname{Li}_s\left(\frac{1+i \sqrt{3}}{2}\right)+\operatorname{Li}_s\left(\frac{1-i \sqrt{3}}{2}\right)\right)\right)\Bigg|_{s=1} = -\frac{1}{8} \ln(2) \ln(3)$$ as desired. $\square$

Answer 3

After integration by parts we have \begin{gather*} I=\int_{0}^{1/2}\frac{x}{3+4x^2}\ln\dfrac{\ln(1/2+x)}{\ln(1/2-x)}\,\mathrm{d}x= \underbrace{\left[\dfrac{1}{8}\ln\left(\frac{3+4x^2}{4}\right)\ln\dfrac{\ln(1/2+x)}{\ln(1/2-x)}\right]_{0}^{1/2}}_{=0}-\\[2ex] \dfrac{1}{8}\int_{0}^{1/2}\ln\left(\frac{3+4x^2}{4}\right)\left(\dfrac{1}{\ln(1/2+x)(1/2+x)}+\dfrac{1}{\ln(1/2-x)(1/2-x)}\right)\,\mathrm{d}x= \\[2ex] -\dfrac{1}{8}\int_{-1/2}^{1/2}\ln\left(\frac{3+4x^2}{4}\right)\dfrac{1}{\ln(1/2+x)(1/2+x)}\,\mathrm{d}x. \end{gather*} Now substitute $1/2+x=y$. We get \begin{equation*} I= -\dfrac{1}{8}\int_{0}^{1}\dfrac{\ln(y^2-y+1)}{y\ln(y)}\,\mathrm{d}y. \end{equation*} Finally substitute $y=e^{-t}$. Then \begin{equation*} I=\dfrac{1}{8}\int_{0}^{\infty}\dfrac{\ln(e^{-2t}-e^{-t}+1)}{t}\,\mathrm{d}t=\dfrac{1}{8}\int_{0}^{\infty}\dfrac{\ln(e^{-3t}+1)-\ln(e^{-t}+1)}{t}\,\mathrm{d}t. \end{equation*} But this is a Frullani integral. See

https://en.wikipedia.org/wiki/Frullani_integral

Put $f(t)=\ln(e^{-t}+1)$. Then \begin{equation*} I=\dfrac{1}{8}\int_{0}^{\infty}\dfrac{f(3t)-f(t)}{t}\,\mathrm{d}t=\dfrac{1}{8}(f(\infty)-f(0))\ln\left(\dfrac{3}{1}\right) =-\dfrac{1}{8}\ln(2)\ln(3). \end{equation*}