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

10log(1+x2+3)1+xdx=0.094561677526995723016

It seems that the answer is
π212(13)+log(2)log(1+3)

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:
10log(1+x4+15)1+xdx=π212(215)+log(1+52)log(2+3)+log(2)log(3+5)10log(1+x6+35)1+xdx=π212(335)+log(1+52)log(8+37)+log(2)log(5+7)

Please take a look here.

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.

Attribution
Source : Link , Question Author : Shobhit Bhatnagar , Answer Author : Community

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