Integral $\int_0^1\frac{\ln x}{\left(1+x\right)\left(1+x^{-\left(2+\sqrt3\right)}\right)}dx$

Question

There is a curious known integral: $$\int_0^1\frac{\ln\left(1+x^{2+\sqrt{3\vphantom{\large3}}}\right)}{1+x}dx=\frac{\pi^2}{12}\left(1-\sqrt{3\vphantom{\large3}}\right)+\ln \left(1+\sqrt{3\vphantom{\large3}}\right)\ln2.$$ If we consider $\alpha=2+\sqrt{3\vphantom{\large3}}$ as a parameter and take a derivative w.r.t. $\alpha$ at this point, we get the following: $$I=\int_0^1\frac{\ln x}{\left(1+x\right)\left(1+x^{-\left(2+\sqrt{3\vphantom{\large3}}\right)}\right)}dx.$$ Is it possible to express the integral $I$ in a closed form?

Answer 1

Here is a partial progress report. I am basically repeating Jim Belk's analysis from the previous answer.

Set $F(a) = \int_{x=0}^1 \frac{\log(1+x^a)}{1+x} dx$. Then $$F(a) = \int_{x=0}^1 \int_{y=0}^{x^a} \frac{dx dy}{(1+x)(1+y)} = \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)}$$ so $$F(a) + F(a^{-1}) = \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)} + \int_{0 \leq y \leq x^{1/a} \leq 1} \frac{dx dy}{(1+x)(1+y)}$$ $$= \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)} + \int_{0 \leq y^a \leq x \leq 1} \frac{dx dy}{(1+x)(1+y)} = \int_{0 \leq x,y \leq 1} \frac{dx dy}{(1+x)(1+y)} = (\log 2)^2.$$ (In order to combine the integrals, first switch the names of $x$ and $y$ in the second one.)

So $$F'(a) - a^{-2} F'(a^{-1})=0.$$ This gives a linear relation between $F'(2 + \sqrt{3})$ and $F'(2-\sqrt{3})$. If we find a second one, we can solve the linear equations and be done.

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Notice that $$F'(a) = \int_{x=0}^1 \frac{x^a \log x dx}{(1+x)(1+x^a)} = \sum_{m,n=0}^{\infty} \int_{x=0}^1 (-1)^{m+n} x^{m+(n+1) a} \log x dx.$$ Integrating by parts, $\int_{x=0}^1 x^b \log x dx = \frac{-1}{(b+1)^2}$. So, ignoring issues of convergence, we should have $$F'(a) = \sum_{m,n=0}^{\infty} \frac{(-1)^{m+n+1}}{(m+(n+1) a + 1)^2} = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n+1}}{(m+n a)^2}$$ In the last step, we turned $m+1$ and $n+1$ into $m$ and $n$ to make things pretty. My guess is that the convergence issues can be dealt with for any $a>0$, but I haven't thought much about it.

So $$F'(a) + F'(a^{-1}) = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \left( \frac{(-1)^{m+n+1}}{(m+n a)^2} +\frac{(-1)^{m+n+1}}{(m+n a^{-1})^2} \right).$$ Putting $a=2 + \sqrt{3}$, this is $$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} (-1)^{m+n+1} \frac{2 (m^2+4mn+7n^2)}{(m^2+4mn+n^2)^2} $$ $$= 2 \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n+1}}{m^2+4mn+n^2} +12 \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n+1} n^2}{(m^2+4mn+n^2)^2}.$$

Here is where I run out of ideas. The first sum is basically the one at the end of Jim Belk's post, but I have no ideas for the second one.

Answer 2

Jim Belk's analysis is very impressive. But I'm afraid there is an assumption that $\alpha\ge 0$ being implicitly made in his analysis. Here presents the other half of the answer. \begin{eqnarray} F(\alpha)&=&\int_0^1 {\frac{\ln (1+x^\alpha)}{1+x}dx} \\ I(\alpha)&=&\frac{dF}{d\alpha}=\int_0^1 {\frac{\ln x}{(1+x)(1+x^{-\alpha})}dx} \\ &=&\int_0^1 {\frac{\ln x}{1+x}\frac{x^\alpha}{1+x^\alpha}dx} \\ &=&\int_0^1 {\frac{\ln x}{1+x} \left( 1 - \frac{1}{1+x^\alpha} \right) dx } \\ &=&\int_0^1 {\frac{\ln x}{1+x}dx}-\int_0^1{ \frac{1}{1+x^\alpha} \frac{\ln x}{1+x}dx } \\ &=&-\frac{\pi^2}{12}-I(-\alpha) \\ \end{eqnarray}