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.


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.

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    $\begingroup$ How did you manage to come up with the proposed exact answer? $\endgroup$ Jun 21, 2013 at 17:04
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    $\begingroup$ @N3buchadnezzar: After differentiating with respect to $\alpha$ we get $$\int_0^1 \frac{x^\alpha \log x}{(1+x)(1+x^\alpha)}dx$$ How can we evaluate it? $\endgroup$ Jun 22, 2013 at 5:22
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    $\begingroup$ Numerically, $\displaystyle\int_0^1\frac{\log\bigl(1+x^{12+\sqrt{143}}\bigr)}{x}dx$ seems to be equal to $\displaystyle \frac{\pi^2}{12}(6-\sqrt{143})-\log\left(\frac{3+\sqrt{13}}{2}\right)\log(10+ 3\sqrt{11} ) +\log(2) \log( \sqrt{11} + \sqrt{13} )$. $\endgroup$
    – Jim Belk
    Jun 24, 2013 at 17:17
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    $\begingroup$ Looks like if $n$ is such that $2n-1$ and $2n+1$ are twin primes, $\int_{0}^{1} \frac{\log(1+x^{2n+\sqrt{4n^{2}-1}})}{1+x}dx$ should be $$\frac{\pi^{2}}{12}\left(n-\sqrt{4n^{2}-1}\right)+\log(a+b\sqrt{2n-1})\log(c+d \sqrt{2n+1})+\log(2)\log(\sqrt{2n-1}+\sqrt{2n+1})$$ for some $a,b,c,d$. $\endgroup$ Jun 25, 2013 at 9:23
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    $\begingroup$ If $n=1,2,3,6$, $s_{\pm}=\sqrt{2n\pm 1}$, and $\eta_{\pm} \in \mathbb{Z}[s_{\pm}]$ are fundamental units, respectively, then the integral $$I_{n}=\frac{\pi^{2}}{12}\left(n-s_{+}s_{-}\right)+\log(\eta_{+})\log(\eta_{-})+\log(2)\log(s_{+}+s_{-}).$$ Numerically only works for those $n$. $\endgroup$ Jun 25, 2013 at 18:54

4 Answers 4


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

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    $\begingroup$ Very nice! Presumably this works for the other integrals as well? $\endgroup$
    – Jim Belk
    Jun 25, 2013 at 2:11
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    $\begingroup$ Unless I am misunderstanding something, $T=-0.738367$, but $\frac12U = -\frac{\log 2\log(2+\sqrt{3})}{2\sqrt{3}} = -0.263516$. $\endgroup$
    – Kirill
    Jun 25, 2013 at 5:38
  • $\begingroup$ $$T=\frac{F(2+\sqrt3)-F(2-\sqrt3)}{-2\sqrt{3}}-\frac{\pi^2}{12} = -\frac{\pi^2}{12\sqrt3}+\frac{\log2\log(\sqrt3-1)}{2\sqrt{3}}-\frac{\log2\log(1+\sqrt3)}{2\sqrt3}. $$ $\endgroup$
    – Kirill
    Jun 25, 2013 at 6:00
  • $\begingroup$ @JimBelk Large parts of it work, but the other rings aren't PID's. I have ideas about how to deal with this, but not a complete solution. $\endgroup$ Jun 25, 2013 at 10:00
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    $\begingroup$ @ David Speyer, is it possible to use your methods to prove that for $B=\sqrt{2}$: $$\sum_{k,n=0}^\infty (-1)^{k+n} \frac{4k+1-(2n+1)B}{(4k+1+(2n+1)B)^2}=-\sum_{k,n=0}^\infty (-1)^{k+n} \frac{4k+3-(2n+1)B}{(4k+3+(2n+1)B)^2}$$? See math.stackexchange.com/q/1928329/269624 $\endgroup$
    – Yuriy S
    Jan 16, 2017 at 17:03

This isn't an answer to the question, but I thought I should post some of my work here.

Consider the function $$ F(a) \;=\; \int_0^1 \frac{\log(1+x^a)}{1+x}dx. $$ The question asks us to prove that $F(2+\sqrt{3}) = \dfrac{\pi^2}{12}(1+\sqrt{3}) + \log(2)\log(1+\sqrt{3})$.

  1. Mathematica isn't able to compute a closed form for $F(2+\sqrt{3})$, but it can compute $F(a)$ for certain other values of $a$:

    • $F(0) \;=\; (\log 2)^2.$ (Very easy.)

    • $F(1) \;=\; \dfrac{1}{2}(\log 2)^2.$ (Also easy to do by hand.)

    • $F(2) \;=\; -\dfrac{\pi^2}{48} \,+\, \dfrac{3}{4}(\log 2)^2$

    • $F(3) \;=\; \dfrac{\pi^2}{9}+\dfrac{1}{2}(\log 2)^2 \,+\, \dfrac{1}{2}\mathrm{Li}_2\left(-\dfrac13\right) - \mathrm{Li}_2\left(\dfrac{1+i\sqrt{3}}6\right) - \mathrm{Li}_2\left(\dfrac{1-i\sqrt{3}}6\right)$

    It can also compute $F(1/3)$, $F(1/2)$, $F(-2)$, $F(-1)$, $F(-1/2)$, and $F(-1/3)$, which follow from the symmetry properties below, as well as very long expressions for $F(4)$, $F(-4)$, $F(1/4)$, and $F(-1/4)$.

    In the formula for $F(3)$, the function $\mathrm{Li}_2$ is the dilogarithm, which is defined by the integral $$ \mathrm{Li}_2(a) \;=\; -\int_0^1 \frac{\ln(1-ax)}{x}dx. $$ Interestingly, the only values of $a$ for which $\mathrm{Li}_2(a)$ is known to have a closed form are $-1$, $0$, $1/2$, $1$, $2$, and various expressions involving $\sqrt{5}$. (See here.)

  2. The function $F(a)$ defined above has a few symmetry properties. It is easy to show that $$ F(-a) \;=\; F(a) - a\int_0^1 \frac{\log x}{1+x}dx. $$ which gives $$ F(-a) \;=\; F(a) - \frac{\pi^2 a}{12}. $$ In addition, integration by parts followed by a substitution can be used to show that $$ F(a) \,+\, F(1/a) \;=\; (\log 2)^2 $$ for any positive value of $a$. Since $(2+\sqrt{3})^{-1} = 2-\sqrt{3}$, it follows that the given question is equivalent to the equation $$ F(2-\sqrt{3}) \;=\; \frac{\pi^2}{12}(\sqrt{3}-1)\,+\,\log(2)\log(\sqrt{3}-1). $$

  3. It's not too hard to find a series for $F(\alpha)$. We have $$ \begin{align*} \frac{\log(1+x^\alpha)}{1+x} \;&=\; \left(\frac{1}{1+x}\right)\log(1+x^\alpha) \\[6pt] &=\; \left(\sum_{k=1}^\infty (-1)^{k+1}x^{k-1}\right)\sum_{n=1}^\infty (-1)^{n+1}\frac{x^{\alpha n}}{n} \\[6pt] &=\; \sum_{n,k=1}^\infty (-1)^{n+k}\frac{x^{\alpha n+k-1}}{n} \end{align*} $$ We can now (by Abel's Theorem) integrate this series term-by-term to get $$ F(\alpha) \;=\; \sum_{n,k=1}^\infty \frac{(-1)^{n+k}}{n(\alpha n+k)} $$ In particular $$ F(2+\sqrt{3}) \;=\; \sum_{n,k=1}^\infty \frac{(-1)^{n+k}}{n\bigl((2+\sqrt{3})n+k\bigr)} $$ The $2+\sqrt{3}$ in the denominator of this series is unfortunate. However, we already know the value of the sum $F(2+\sqrt3)+F(2-\sqrt3)$, so all we really want is the value of the difference $F(2+\sqrt3)-F(2-\sqrt3)$. In general: $$ \begin{align*} F(\alpha)-F(\alpha^{-1}) \;&=\; \sum_{n,k=1}^\infty (-1)^{n+k}\left(\frac{1}{n(\alpha n+k)}-\frac{1}{n(\alpha^{-1} n+k)}\right) \\[6pt] &=\; \sum_{n,k=1}^\infty (-1)^{n+k}\frac{(\alpha^{-1} n+k)-(\alpha n+k)}{n(\alpha n+k)(\alpha^{-1} n+k)} \\[6pt] &=\; \bigl(\alpha^{-1}-\alpha\bigr)\sum_{n,k=1}^\infty \frac{(-1)^{n+k}}{n^2 + (\alpha+\alpha^{-1})nk + k^2} \end{align*} $$ So: $$ F(2+\sqrt{3})-F(2-\sqrt{3}) \;=\; -2\sqrt{3} \sum_{n,k=1}^\infty \frac{(-1)^{n+k}}{n^2 + 4nk + k^2} $$ This series is more pleasant, but I still have no idea how to evaluate it.


I have found an interesting reference related to this integral. It may be of interest to many users:

A restricted Epstein zeta function and the evaluation of some definite integrals by Habib Muzaffar and Kenneth S. Williams


Unfortunately, the following generalization works for only positive integer $a$:

$$\int_0^1\frac{\ln(1+x^{2a})}{1+x}dx=\ln^2(2)-\frac{2a^2-1}{8a}\zeta(2)+\frac12\sum_{j=0}^{2a-1}\ln^2\left(2\sin\left(\frac{(2j+1)\pi}{4a}\right)\right)$$ $$-\frac12\sum_{j=0}^{a-1}\ln^2\left(2\sin\left(\frac{(2j+1)\pi}{2a}\right)\right).$$


$$\int_0^1\frac{\ln(1+x^{2a})}{1+x}dx=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\int_0^1\frac{x^{2an}}{1+x}dx$$

$$=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\left(\ln(2)+H_{an}-H_{2an}\right)$$ $$=\ln^2(2)+\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\left(H_{an}-H_{2an}\right).$$ Omram Kouba provided in his paper page (12-13) the following general result:

$$\sum_{n=1}^\infty(-1)^{n-1}\frac{H_{an}}{n} = \frac{a^2+1}{4a}\zeta(2) - \frac{1}{2} \sum_{j=0}^{a-1} \ln^2\left(2 \sin \frac{(2j+1)\pi}{2a} \right)$$

from which, the proof follows.

By the way, we can find the integral representation of $\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n-1}H_{an}}{n}$:

$$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_{an}}{n}=a\sum_{n=1}^\infty (-1)^{n-1}\left(\frac{H_{an}}{an}\right)$$

$$=a\sum_{n=1}^\infty (-1)^{n-1}\left(-\int_0^1 x^{an-1}\ln(1-x)dx\right)$$


$$=a\int_0^1\frac{\ln(1-x)}{x}\left(\frac{-x^a}{1+x^a}\right)dx$$ $$=a\int_0^1\frac{\ln(1-x)}{x}\left(-1+\frac{1}{1+x^a}\right)dx$$


Substitute the result of $\displaystyle\sum_{n=1}^\infty\frac{(-1)^nH_{an}}{n}$, we also get

$$\int_0^1\frac{\ln(1-x)}{x(1+x^a)}dx=\frac{1-3a^2}{4a^2}\zeta(2)- \frac{1}{2a} \sum_{j=0}^{a-1} \ln^2\left(2 \sin \frac{(2j+1)\pi}{2a} \right).$$


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