I am trying to find a closed form for
$$\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx = 0.094561677526995723016 \cdots$$
It seems that the answer is
$$\frac{\pi^2}{12}\left( 1-\sqrt{3}\right)+\log(2) \log \left(1+\sqrt{3} \right)$$
Mathematica is unable to give a closed form for the indefinite integral.
How can we prove this result? Please help me.
EDIT
Apart from this result, the following equalities are also known to exist:
$$\begin{align*}
\int_0^1 \frac{\log \left( 1+x^{4+\sqrt{15}}\right)}{1+x}\mathrm dx &=\frac{\pi^2}{12} \left( 2-\sqrt{15}\right)+\log \left( \frac{1+\sqrt{5}}{2}\right)\log \left(2+\sqrt{3} \right) \\ &\quad +\log(2)\log\left( \sqrt{3}+\sqrt{5}\right)
\\ \int_0^1 \frac{\log \left( 1+x^{6+\sqrt{35}}\right)}{1+x}\mathrm dx &= \frac{\pi^2}{12} \left( 3-\sqrt{35}\right)+\log \left(\frac{1+\sqrt{5}}{2} \right)\log \left(8+3\sqrt{7} \right) \\
&\quad +\log(2) \log \left( \sqrt{5}+\sqrt{7}\right)
\end{align*}$$
Please take a look here.
Best Answer
$\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}$I will evaluate the sum at the end of Jim Belk's post.
There is clearly a lot more going on here; I hope a number theorist will come and explain what is really going on. I will be sloppy about convergence issues.
Set $$S:= \sum_{k,n=1}^{\infty} \frac{(-1)^{k+n}}{k^2+4kn+n^2}.$$ We set $m = k+2n$, in order to diagonalize the quadratic form: $$S = \sum_{0 < 2n < m} \frac{(-1)^{m-n}}{m^2 - 3 n^2}.$$ It turns out to be more elegant to work with $$T : = \sum_{0 \leq 2n < m} \frac{(-1)^{m-n}}{m^2 - 3 n^2} = S + \sum_{m=1}^{\infty} \frac{(-1)^m}{m^2} = S - \frac{\pi^2}{12}.$$
Let $R$ be the ring $\ZZ[\sqrt{3}]$. Let $D \subset R$ be $\{ m+n\sqrt{3} : 0 \leq 2n < m \}$. So, simply as a matter of formal rewriting $$T = \sum_{m+n \sqrt{3} \in D} \frac{(-1)^{m+n}}{m^2-3 n^2}.$$
We will now try to interpret each of these quantities in terms of the ring $R$. Recall that, for $m+n \sqrt{3} \in R$, the norm $N(m+n \sqrt{3})$ is $m^2 - 3 n^2$. Define $R_{+} = \{ a \in R : N(a) > 0\}$ and $R_{-} = \{ a \in R : N(a) < 0 \}$.
For $m+n \sqrt{3} \in R$, define $\sigma_2(m+n \sqrt{3}) = (-1)^{m+n}$.
Let $\Gamma$ denote the unit group of $R$. Explicitly, $\Gamma$ is $\{ \pm 1 \} \times (2+\sqrt{3})^{\ZZ}$.
Note that multiplication by $\Gamma$ takes $R_{+}$ and $R_{-}$ to themselves. Here is the first miracle: $D$ is a fundamental domain for the action of $\Gamma$ on $R_{+}$! For example, multiplication by $2+\sqrt{3}$ maps the ray $\RR_{\geq 0} (1+0 \sqrt{3})$ bounding one side of $D$ to the ray $\RR_{\geq 0} (2+\sqrt{3})$ on the other side. Moreover, multiplication by units leaves $N( \ )$ and $\sigma_2(\ )$ unchanged. So we can view the sum as $$T = \sum_{a \in R_{+}/\Gamma} \frac{\sigma_2(a)}{N(a)}$$ where the sum means to pick one representative for each orbit of the $\Gamma$ action.
I'd rather sum over $R$ than $R_{+}$. Define $\chi_{\infty}$ to be $\pm 1$ on $R_{\pm}$. So we can rewrite $$ T = \frac{1}{2} \sum_{a \in R_{\neq 0}/\Gamma} \frac{\sigma_2(a) \left( 1+\chi_{\infty}(a) \right)}{|N(a)|} = \frac{1}{2}\left( \sum_{a \in R_{\neq 0}/\Gamma} \frac{\sigma_2(a)}{|N(a)|} + \sum_{a \in R_{\neq 0}/\Gamma} \frac{\sigma_2(a) \chi_{\infty}(a)}{|N(a)|} \right) = : \frac{1}{2} (U+V).$$
Now, $a$ and $b$ in $R$ are in the same $\Gamma$ orbit if and only if the ideals $(a)$ and $(b)$ are equal. So the above sums are running over all principal ideals of $R$. Moreover, $R$ is a PID! And, finally, for a principal ideal $I = (a)$, we have $N(I) = |N(a)|$. So we can write: $$U = \sum_{I \subset R \ \mbox{an ideal}} \frac{\sigma_2(I)}{N(I)} \quad V = \sum_{I \subset R \ \mbox{an ideal}} \frac{\sigma_2(I) \chi_{\infty}(I)}{N(I)} .$$ We set $$U(s) = \sum_{I \subset R \ \mbox{an ideal}} \frac{\sigma_2(I)}{N(I)^s} \quad V(s) = \sum_{I \subset R \ \mbox{an ideal}} \frac{\sigma_2(I) \chi_{\infty}(I)}{N(I)^s} .$$
Note that $\sigma_2(a)$ is $1$ if and only if $1+\sqrt{3}$ divides $a$. So we have the Euler factorization $$U(s) = \left(-1 + 2^{-s} + 2^{-2s} + 2^{-3s} + \cdots \right) \prod_{\pi \neq (1+\sqrt{3})} \frac{1}{1-N(\pi)^{-s}} =$$ $$\frac{-1+2^{-s}+2^{-2s}+ 2^{-3s} + \cdots}{1+2^{-s}+2^{-2s}+2^{-3s}+\cdots} \prod_{\pi} \frac{1}{1-N(\pi)^{-s}} =(-1+2\cdot 2^{-s}) \zeta_R(s).$$ where $\pi$ runs over prime ideals of $R$.
So $$\lim_{s \to 1^{+}} U(s) = \lim_{s \to 1^{+}} \frac{-1+2 \cdot 2^{-s}}{s-1} \lim_{s \to 1^{+}} (s-1) \zeta_R(s) = - \log 2 \lim_{s \to 1^{+}} (s-1) \zeta_R(s).$$ From the class number formula, this last limit is $$\frac{2^2 \log (2+\sqrt{3})}{2 \cdot \sqrt{12}}.$$ So $$U(1) = - \frac{\log 2 \log(2+\sqrt{3})}{\sqrt{3}}.$$
We now run the same trick with $V$. Again, we start with the Euler product: $$V(s) = \left(-1 - 2^{-s} + 2^{-2s} - 2^{-3s} + 2^{-4s} - \cdots \right) \prod_{\pi \neq (1+\sqrt{3})} \frac{1}{1-\chi_{\infty}(\pi) N(\pi)^{-s}} =$$ $$\frac{-1 - 2^{-s} + 2^{-2s} - 2^{-3s} +2^{-4s} - \cdots}{1-2^{-s} + 2^{-2s} - 2^{-3s} + 2^{-4s} - \cdots} \prod_{\pi} \frac{1}{1-\chi_{\infty}(\pi) N(\pi)^{-s}} = \left( -1 - 2^{1-s} \right) L(s, \chi_{\infty}).$$
So $$V = V(1) = - 2 L(1, \chi_{\infty}).$$
We now must evaluate $L(1, \chi_{\infty})$. This $L$-function is defined by a sum over the ideals of $R$; we will rewrite it in terms of $L$-functions for $\ZZ$. Let $p \geq 5$ be prime. I claim that $p$ splits in $R$ if and only if $p \equiv \pm 1 \bmod 12$. For such a $p$, if we write $p = \pi \bar{\pi}$, I claim that $N(\pi) = N(\bar{\pi}) = p$ if $p \equiv 1 \bmod 12$ and $N(\pi) = N(\bar{\pi}) = -p$ if $p \equiv -1 \bmod 12$. Proof Sketch: The prime $p$ splits in $R$ if and only if $\left( \frac{3}{p} \right) = 1$, which is easily computed to be equivalent to $p \equiv \pm 1 \bmod 12$. Since $R$ is a PID, we can write such a $p$ as $\pi \bar{\pi}$ for $\pi$ and $\bar{\pi} \in R$. Then $N(\pi) = N(\bar{\pi}) = \pm p$. But, for $a=m+n \sqrt{3} \in R$, we have $N(a) = m^2-3 n^2 \not \equiv -1 \mod 3$. So only one of the two possibilities for $\pm p$ can occur. $\square$
So $L(1, \chi_{\infty})$ is $$\left(1+2^{-1} \right)^{-1} \left( 1+3^{-1} \right)^{-1} \prod_{p \equiv 1 \bmod 12} \left( 1- p^{-1} \right)^{-2} \prod_{p \equiv -1 \bmod 12} \left( 1+ p^{-1} \right)^{-2} \prod_{p \equiv \pm 5 \bmod 12} \left( 1- p^{-2} \right)^{-1}.$$
Define $\chi_4(n)$ to be $0$ if $n$ is even, $1$ if $n \equiv 1 \bmod 4$ and $-1$ is $n \equiv -1 \bmod 4$. Define $\chi_3(n)$ to be $1$, $-1$ or $0$ according to whether $n \equiv 1$, $2$ or $0 \bmod 3$. Then we have $$L(1, \chi_{\infty}) = \prod_p \left( 1- \chi_4(p) p^{-1} \right)^{-1} \prod_p \left( 1-\chi_3(p) p^{-1} \right)^{-1} = \left( \sum_{n=1}^{\infty} \frac{\chi_4(n)}{n} \right) \left( \sum_{n=1}^{\infty} \frac{\chi_3(n)}{n} \right) .$$ The sum $\sum \chi_4(n) = 1-1/3+1/5-1/7 + \cdots$ is well known to be $\pi/4$. The sum $\sum \chi_3(n)/n$ is only slightly less well known; it is $\pi/(3 \sqrt{3})$.
So I get $L(1,\chi_{\infty}) = \frac{\pi}{4} \frac{\pi}{3 \sqrt{3}} = \frac{\pi^2}{12 \sqrt{3}}$ and $V = - \frac{\pi^2}{6 \sqrt{3}}$.
Putting it all together, $$T = \frac{1}{2} \left( - \frac{\log 2 \log(2+\sqrt{3})}{\sqrt{3}} - \frac{\pi^2}{6 \sqrt{3}} \right) = - \frac{\log 2 \log(2+\sqrt{3})}{2 \sqrt{3}} - \frac{\pi^2}{12 \sqrt{3}} $$ $$S = \frac{\pi^2}{12} - \frac{\log 2 \log(2+\sqrt{3})}{2 \sqrt{3}} - \frac{\pi^2}{12 \sqrt{3}} .$$ The original integral is $$\frac{1}{2} \left( \log(2)^2 -2 \sqrt{3} S \right) = \frac{\log(2)^2}{2} - \frac{\pi^2 \sqrt{3}}{12} + \frac{\log 2 \log(2+\sqrt{3})}{2} + \frac{\pi^2}{12}$$ $$=\frac{\pi^2(1-\sqrt{3})}{12} + \log(2) \frac{\log(4 + 2 \sqrt{3})}{2} = \frac{\pi^2(1-\sqrt{3})}{12} + \log(2) \log(1+\sqrt{3})$$ as desired.