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**

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.

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.

**Attribution***Source : Link , Question Author : Vladimir Reshetnikov , Answer Author :
2 revs, 2 users 92%*