# Evaluate ∫10log(1+x2+√3)1+xdx\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx

I am trying to find a closed form for

It seems that the answer is

Mathematica is unable to give a closed form for the indefinite integral.

EDIT

Apart from this result, the following equalities are also known to exist:

$\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}$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

We set $m = k+2n$, in order to diagonalize the quadratic form:

It turns out to be more elegant to work with

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

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

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

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:

We set

Note that $\sigma_2(a)$ is $1$ if and only if $1+\sqrt{3}$ divides $a$.
So we have the Euler factorization

where $\pi$ runs over prime ideals of $R$.

So

From the class number formula, this last limit is

So

We now run the same trick with $V$. Again, we start with the Euler product:

So

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

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

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,

The original integral is

as desired.