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

Question

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?

Answer 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, $$\int_0^\infty\frac{\ln\left(\frac{1+x^a}{1+x^b}\right)}{\ln x}\frac{dx}{1+x^2}=\frac\pi4(a-b).\tag3$$

Answer 2

I do not know how to answer your question. However, in order you to challenge your challenger, I give you a few amazing results (obtained using a CAS) for $$f(n)=\int_0^\infty\frac{\log\left(\frac{1+x^a}{1+x^b}\right)}{\left(1+x^2\right)\log x}dx$$ in which $a=2n+\sqrt{4 n^2-1}$ and $b=n+\sqrt{n^2-1}$. $$f(1)=\frac{1}{4} \left(1+\sqrt{3}\right) \pi$$ $$f(2)=\frac{1}{4} \left(2-\sqrt{3}+\sqrt{15}\right) \pi$$ $$f(3)=\frac{1}{4} \left(3-2 \sqrt{2}+\sqrt{35}\right) \pi$$ $$f(4)=\frac{1}{4} \left(4+3 \sqrt{7}-\sqrt{15}\right) \pi$$ $$f(5)=\frac{1}{4} \left(5-2 \sqrt{6}+3 \sqrt{11}\right) \pi$$ $$f(6)=\frac{1}{4} \left(6-\sqrt{35}+\sqrt{143}\right) \pi$$ $$f(7)=\frac{1}{4} \left(7-4 \sqrt{3}+\sqrt{195}\right) \pi$$ $$f(8)=\frac{1}{4} \left(8-3 \sqrt{7}+\sqrt{255}\right) \pi$$ $$f(9)=\frac{1}{4} \left(9-4 \sqrt{5}+\sqrt{323}\right) \pi$$ $$f(10)=\frac{1}{4} \left(10-3 \sqrt{11}+\sqrt{399}\right) \pi$$ which are exactly what Pranav Arora answered (what I missed) $$f(n)=\frac \pi 4(a-b)$$

Answer 3

Though this isn't an answer, this is interesting enough to me but too large for a comment. Based on Vladimir's solution, if we know $$f(a) = \int_0^\infty \frac{\ln[(1+x^a)/2]}{\ln x} \frac{1}{1+x^2}dx = a\frac{\pi}{4}$$ then we should have $$f'(a) = \int_0^\infty \frac{\ln x e^{a\ln x}}{\ln x (1+e^{a \ln x})} \frac{1}{1+x^2} dx = \pi/4$$ or $$f'(a) = \int_0^\infty \frac{x^a}{1+x^a} \frac{1}{1+x^2}dx = \int_0^\infty \left(1-\frac{1}{1+x^a} \right) \frac{1}{1+x^2}dx = \pi/4.$$ This is quite interesting because I would not expect the integral to be constant as a function of $a$. Furthermore, we should expect

$$f''(a) = \int_0^\infty \frac{\ln x \cdot x^a}{(1+x^a)^2}\frac{1}{1+x^2} dx = 0.$$ I didn't bother to check carefully for convergence issues (passing the derivative through), but I think everything cis okay. Does anyone know how to compute the above integrals without referring to Vladimir's answer?