How to find $\int_0^\infty\frac{\log\left(\frac{1+x^{4+\sqrt{15}}}{1+x^{2+\sqrt{3}}}\right)}{\left(1+x^2\right)\log x}\mathrm dx$

I was challenged to prove this identity $$\int_0^\infty\frac{\log\left(\frac{1+x^{4+\sqrt{15\vphantom{\large A}}}}{1+x^{2+\sqrt{3\vphantom{\large A}}}}\right)}{\left(1+x^2\right)\log x}\mathrm dx=\frac{\pi}{4}\left(2+\sqrt{6}\sqrt{3-\sqrt{5}}\right).$$ I was not successful, so I want to ask for your help. Can it be somehow related to integrals listed in that question?

Solution 1:

This integral can be evaluated in a closed form for arbitrary real exponents, and does not seem to be related to Herglotz-like integrals.

Assume $a,b\in\mathbb{R}$. Note that $$\int_0^\infty\frac{\ln\left(\frac{1+x^a}{1+x^b}\right)}{\ln x}\frac{dx}{1+x^2}=\int_0^\infty\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}-\int_0^\infty\frac{\ln\left(\frac{1+x^b}2\right)}{\ln x}\frac{dx}{1+x^2}.\tag1$$ Both integrals on the right-hand side have the same shape, so we only need to evaluate one of them: $$\begin{align}&\phantom=\underbrace{\int_0^\infty\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}}_\text{split the region}\\&=\int_0^1\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}+\underbrace{\int_1^\infty\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}}_{\text{change variable}\ y=1/x}\\&=\int_0^1\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}+\underbrace{\int_1^0\frac{\ln\left(\frac{1+y^{-a}}2\right)}{\ln\left(y^{-1}\right)}\frac1{1+y^{-2}}\left(-\frac1{y^2}\right)dy}_\text{flip the bounds and simplify}\\&=\int_0^1\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}-\underbrace{\int_0^1\frac{\ln\left(\frac{1+y^{-a}}2\right)}{\ln y}\frac{dy}{1+y^2}}_{\text{rename}\ y\ \text{to}\ x}\\&=\underbrace{\int_0^1\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}-\int_0^1\frac{\ln\left(\frac{1+x^{-a}}2\right)}{\ln x}\frac{dx}{1+x^2}}_\text{combine logarithms}\\&=\int_0^1\frac{\ln\left(\frac{1+x^a}{1+x^{-a}}\right)}{\ln x}\frac{dx}{1+x^2}=\underbrace{\int_0^1\frac{\ln\left(\frac{x^a\left(x^{-a}+1\right)}{1+x^{-a}}\right)}{\ln x}\frac{dx}{1+x^2}}_{\text{cancel}\ \ 1+x^{-a}}\\&=\int_0^1\frac{\ln\left(x^a\right)}{\ln x}\frac{dx}{1+x^2}=a\int_0^1\frac{dx}{1+x^2}=a\,\Big(\arctan1-\arctan0\Big)\\&=\vphantom{\Bigg|^0}\frac{\pi\,a}4.\end{align}\tag2$$ So, finally, $$\int_0^\infty\frac{\ln\left(\frac{1+x^a}{1+x^b}\right)}{\ln x}\frac{dx}{1+x^2}=\frac\pi4(a-b).\tag3$$

Solution 2:

Let the considered integral be $I$. Just to make it easier to write, let $4+\sqrt{15}=a$ and $2+\sqrt{3}=b$. Use the substitution $x=\tan\theta$ to get: $$I=\int_0^{\pi/2} \frac{\ln\left(\dfrac{1+(\tan\theta)^a}{1+(\tan\theta)^b}\right)}{\ln\tan\theta}\,d\theta$$ Next, use the substitution $\theta=\pi/2-t$ to obtain: $$I=\int_0^{\pi/2} \frac{\ln\left(\dfrac{1+(\tan t)^a}{(1+(\tan t)^b)(\tan t)^{a-b}}\right)}{\ln\cot t}\,dt=\int_0^{\pi/2} \frac{-\ln\left(\dfrac{1+(\tan\theta)^a}{1+(\tan\theta)^b}\right)+(a-b)\ln\tan\theta}{\ln\tan\theta}\,d\theta$$ where I used $\ln(\tan(\pi/2-\theta))=\ln(\cot\theta)=-\ln(\tan\theta)$.

Add the two expressions for I and notice that you are left with: $$2I=\int_0^{\pi/2} \frac{(a-b)\ln \tan\theta}{\ln \tan\theta}\,d\theta=\frac{\pi}{2}(a-b)$$ $$I=\frac{\pi}{4}(a-b)$$ Therefore, $$\boxed{I=\dfrac{\pi}{4}(2+\sqrt{15}-\sqrt{3})}$$

Solution 3:

I have a short way to solve this problem. Let $x=\frac{1}{u}, a=4+\sqrt{15},b=2+\sqrt{3}$. Then \begin{eqnarray*} I&=&-\int_\infty^0\frac{\ln\frac{1+u^{-a}}{1+u^{-b}}}{(1+u^{-2})\ln(u^{-1})}u^{-2}du\\ &=&-\int_0^\infty\frac{\ln\frac{1+u^{-a}}{1+u^{-b}}}{(1+u^{2})\ln u}du\\ &=&-\int_0^\infty\frac{\ln \left(u^{b-a}\frac{1+u^{a}}{1+u^{b}}\right)}{(1+u^{2})\ln u}du\\ &=&-\int_0^\infty\frac{\ln \left(\frac{1+u^{a}}{1+u^{b}}\right)}{(1+u^{2})\ln u}du-\int_0^\infty\frac{(b-a)\ln u}{(1+u^{2})\ln u}du \end{eqnarray*} and hence $$ 2I = -\int_0^\infty\frac{(b-a)}{1+u^{2}}du=(a-b)\frac{\pi}{2}. $$ So $$ I = (a-b)\frac{\pi}{4}. $$