$$I=\int_0^1 \frac{\arctan(x^a)}{1+x^2}dx=\frac{1}{16}\ln(2)\ln(a)~~~~~~~a=3+\sqrt{8}$$

I try the substitution: $u=1/x$

$$I=\int_1^\infty \frac{\arctan(1/u^a)}{1+u^2}du=\int_1^\infty \frac{\frac{\pi}2-\arctan(u^a)}{1+u^2}du$$

But I cannot add them to seal the integral limit. Also, the given number $a=(\sqrt{2}+1)^2$ is very tricky. I am not sure if this complete square will play some role.

Or, if I do integration by part, I get:

$$I=\frac{\pi^2}{4}-a\int_0^1 \frac{\arctan(x)}{1+x^{2a}}x^{a-1}dx$$

The power index $a$ is irrational, so how should I proceed, any hint will be appreciated!

  • 1
    $\begingroup$ See: people.math.carleton.ca/~williams/papers/pdf/243.pdf $\endgroup$
    – Zacky
    Jul 27, 2022 at 1:38
  • $\begingroup$ @Zacky Thank you, I find in this paper, mentioned this integral was from this paper as early, seems this paper was written in German... G. Herglotz, Ub¨ er die Kroneckersche Grenzformel fur¨ reele, quadratische K¨orper. I., Ber. d. Sacks. Akad. d. Wiss. zu Leipzig 75 (1923), 3–14 $\endgroup$
    – MathFail
    Jul 27, 2022 at 2:44
  • $\begingroup$ Basically on the same level of difficulty as math.stackexchange.com/questions/426325. You need to at least understand solution there. $\endgroup$
    – pisco
    Jul 27, 2022 at 3:36
  • $\begingroup$ Thank you for the link, yes, they are the same level. For this link, it seems there is no elementary method? @pisco $\endgroup$
    – MathFail
    Jul 27, 2022 at 18:10
  • $\begingroup$ @MathFail Well, it would be too categorical to say an elementary method does not exist, but finding it is definitely as difficult as solving with non-elementary method. $\endgroup$
    – pisco
    Jul 27, 2022 at 19:25

1 Answer 1


This is a nontrivial result. I decided to write this answer to promote relevant techniques to general community. Some background in number theory is would be useful.

I will prove OP's claim at the end. First some general remarks. Let $$F(\alpha) = \int_0^1 \frac{\log(1+x^\alpha)}{1+x}dx \qquad G(\alpha) = \int_0^1 \frac{\arctan(x^\alpha)}{1+x^2}dx$$ for $\alpha$ real quadratic irrational, they have the same nature arithmetically. $F(2+\sqrt{3})$ is the simplest example, you should read answers there first.

Let $\mathcal{C}$ be the ray class group of a certain modulus $\mathfrak{m}$ of $K=\mathbb{Q}(\alpha)$, this $\mathfrak{m}$ depends on $\alpha$ as well as whether your focus is $F$ or $G$. $F(\alpha), G(\alpha)$ are closely related to values of Hecke L functions of $\mathfrak{m}$ at $1$.

Such integrals are first investigated by Kronecker mentioned in comments, also by Muzaffar and Williams linked comments. Their arguments rely on genera theory, so only applies to the case $\mathcal{C}$ is $2$-torsion.

Actually, regardless of shape of ray class group $\mathcal{C}$, we have closed form like following: $$\tag{1}\sum_{i=1}^N a_i \log b_i \log c_i \quad a_i,b_i,c_i \in \bar{\mathbb{Q}}$$

Precise statement is: let $\alpha\neq 1,-1, \alpha\in \mathcal{O}_K$, $N_{K/\mathbb{Q}}(\alpha) = 1$. Assuming a conjecture in algebraic number theory [a] , then

  • when $\text{Tr}_{K/\mathbb{Q}}(\alpha) \equiv 0 \pmod{2}$, then $F(\alpha)$ can be written in form of $(1)$.
  • when $\text{Tr}_{K/\mathbb{Q}}(\alpha) \equiv 2 \pmod{4}$, then $G(\alpha)$ can be written in form of $(1)$.

It predicts the closed form of $F(2+\sqrt{3}),F(4+\sqrt{15}),F(6+\sqrt{15})$ and $G(3+2\sqrt{2}),G(5+2\sqrt{6})$, their explicit closed forms have been worked out by Muzaffar and Williams. The above result for other $\alpha$ such as $$F(18 + \sqrt {323})\quad F((\frac{1+\sqrt{5}}{2})^{6})\quad F(3588 \sqrt{46}+24335)\quad G(23+4\sqrt{33})$$ seem not to succumb under genera theory, nonetheless, they have result like $(1)$. For the first one in above list, explicitly, $$F(18 + \sqrt {323} ) = \frac{{{\pi ^2}}}{{12}}(9 - \sqrt {323} ) + \log \frac{{8 - \sqrt {19} - \sqrt {79 - 16\sqrt {19} } }}{2}\log \frac{{2 - \sqrt {19} + \sqrt {27 - 4\sqrt {19} } }}{2} \\ - \log \frac{{2 + \sqrt {19} + \sqrt {27 + 4\sqrt {19} } }}{2}\log \frac{{8 + \sqrt {19} - \sqrt {79 + 16\sqrt {19} } }}{2} + \frac{{{{\log }^2}2}}{2} + \frac{1}{2}\log 2\log (18 + \sqrt {323} ) $$ the corresponding $\mathcal{C}$ in the case is $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}$, the narrow Hilbert class group of $\mathbb{Q}(\sqrt{323})$.

Let $\alpha>0$, $$G(\alpha) = \int_0^1 \frac{\arctan t^\alpha}{1+t^2}dt = \sum_{m\geq 0, n\geq 0} \frac{(-1)^{m+n}}{(2m+1)(\alpha(2m+1)+(2n+1))} $$ integration by parts say $G(\alpha)+G(\alpha^{-1}) = \frac{\pi^2}{16}$. If $\alpha$ is a totally positive unit in a real quadratic field $K$, $\bar{\alpha}$ its conjugate, then $\alpha\bar{\alpha} = 1$ and $$\frac{G(\alpha)-G(\bar{\alpha})}{\bar{\alpha} - \alpha} = \sum_{m\geq 0, n\geq 0} \frac{(-1)^{m+n}}{N((\alpha(2m+1)+(2n+1))} = \lim_{s\to 1} L(s)$$ where [b] $$L(s) = \sum_{m\geq 0, n\geq 0} \frac{(-1)^{m+n}}{N((\alpha(2m+1)+(2n+1))^s} = \sum_{m\geq 0, n\geq 0} \frac{f(m+n\alpha)}{N(m+n\alpha)^s}$$ here $$f(m+n\alpha) = \begin{cases} 0 &\text{ if } mn \text{ is even} \\ (-1)^{(m-1)/2 + (n-1)/2} &\text{ if } mn \text{ is odd} \\ \end{cases}$$ Let $E$ be the subgroup of units in $\mathcal{O}_K$ generated by $\alpha$, then it can be shown, for $x\in \mathbb{Z}[\alpha]$, $$f(xE) = f(x) \iff Tr_{K/\mathbb{Q}}(x)\equiv 2 \pmod{4} $$ if this holds, $L(s)$ can be rewritten as $$L(s) = \sum_{x\in \mathcal{O}_K^+ / E} \frac{f(x)}{N(x)^s} $$ here $\mathcal{O}_K^+$ is set of non-zero totally positive element of $\mathcal{O}_K$.

Let's derive OP's result for the simple case $G(2+\sqrt{3})$.

Now let $K=\mathbb{Q}(\sqrt{2}), \alpha = 3+2\sqrt{2}$. $\mathfrak{m}$ be the modulus of $K$ consisting of all real places, plus ideal $(2)$. Then the ray class group $\mathcal{C}$ has order 2, associated class field is $K(\sqrt{-1})$. Let $\chi_1, \chi_2$ be characters of $\mathcal{C}$, with $\chi_1$ being trivial, $$L(s,\chi_i) = \sum_{\mathfrak{a},(\mathfrak{a},2) = 1} \frac{\chi_i(\mathfrak{a})}{N(\mathfrak{a})^s}$$ The non-trivial ray class is represented by $I = (5+3\sqrt{2})$.

Let $$g(x) = \begin{cases} (-1)^{m+n} \quad &\text{ if } x = (n+3m)+(2n+1)\sqrt{2}, n,m\in \mathbb{Z} \\ 0 \quad &\text{ otherwise }\end{cases}$$ since $g(x) = 0$ for $2\mid x$, $$L(s) = 4^{-s}\sum_{x\in \mathcal{O}_K^+ / E} \frac{g(x)}{N(x)^s} = 4^{-s}\sum_{\substack{x\in \mathcal{O}_K^+ / E \\ (x,2)=1}} \frac{g(x)}{N(x)^s} + 4^{-s}\sum_{\substack{x\in \mathcal{O}_K^+ / E \\ (x,2)=1}} \frac{g(\sqrt{2}x)}{N(\sqrt{2}x)^s}$$ the second term equals $$8^{-s}\sum_{\substack{x\in \mathcal{O}_K^+ / E \\ (x,2)=1}} \frac{1}{N(x)^s} = 8^{-s}L(s,\chi_1)$$

Note that, when writing $x=a+b\sqrt{2}, (x,2)=1$, one has $b \text{ odd} \iff (x)= I \in \mathcal{C}$, so first term equals $-4^{-s}(L(s,\chi_1)/2-L(s,\chi_2)/2)$.

Let $L_D(s) = \sum_{n\geq 1} \left(\frac{D}{n}\right)n^{-s}$ the Dirichlet L function for quadratic field of discriminant $D$. Then, by looking at primes in $K$ that split in $K(\sqrt{-1})$, one sees $$L(s,\chi_1) = (1-2^{-s})\zeta_K(s) = (1-2^{-s})\zeta(s)L_8(s) \qquad L(s,\chi_2) = L_{-4}(s)L_{-8}(s) $$

Hence $$L(s) = 4^{-s} \left(\left(1-2^{-s}\right) \left(2^{-s}-\frac{1}{2}\right) L_{8}(s) \zeta (s)+\frac{1}{2} L_{-4}(s) L_{-8}(s)\right)$$ Letting $s\to 1$, $$\frac{G(\alpha)-G(\bar{\alpha})}{\bar{\alpha} - \alpha} = 4^{-1} (-\frac{\log 2}{4} L_{8}(1)+\frac{1}{2} L_{-4}(1) L_{-8}(1)) = \frac{\pi ^2}{64 \sqrt{2}}-\frac{\log (2) \log \left(\sqrt{2}+1\right)}{16 \sqrt{2}}$$ here $L_D(1)$ can be evaluated either by class number formula, or using explicit formulae for general Dirichlet-L functions. together with $G(\alpha)+G(\bar{\alpha}) = \frac{\pi^2}{16}$ solves the value of $G(\alpha)$. QED.

[a]: the Stark's conjecture regarding value of Artin L-functions at $1$.

[b]: the series when $s=1$ is not absolutely convergent, still, this analytic issue can be justified.

  • $\begingroup$ I am getting lost from the second sentence, "Let $\mathcal{C}$ be the ray class group... " what material should I read to understand those prerequisite? $\endgroup$
    – MathFail
    Jul 27, 2022 at 20:51
  • $\begingroup$ Mathfail: Class Field Theory. See the book Class Field Theory of Nancy Childress. $\endgroup$
    – FDP
    Jul 28, 2022 at 0:57

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