# Evaluating $\int_0^{1/\sqrt{3}}\frac{\arctan(x)\ln(1-3x^2)}{1+x^2}\,\mathrm{d}x$

Recently while working on an interesting problem, I'm stuck on evaluating the following daunting but interesting integral:

$$\int_0^{1/\sqrt{3}}\dfrac{\arctan(x)\ln(1-3x^2)}{1+x^2}\,\mathrm{d}x$$

I have been working on it since quite some time, I've tried some trivial substitutions and integration by parts but it didn't get me anywhere. Any help would be highly appreciated.

• Isn’t this improper because of the log term? Jul 4, 2021 at 3:22
• Also, doesn’t this fail to converge since the log term goes to $-\infty$? Jul 4, 2021 at 3:27
• @Randall logarithmic singularitities are integrable Jul 4, 2021 at 3:35
• @NinadMunshi ah you are right. Jul 4, 2021 at 3:43
• Let $x=\tan u$ and that clears the denominator. Then note that $$1-3\tan^2 u = \frac{2\cos(2u)-1}{\cos^2 u}.$$ This should be able to split up the log. Jul 4, 2021 at 4:25

Define the function $$\mathcal{I}:\mathbb{R}_{>0}\rightarrow\mathbb{R}$$ via the (convergent) improper integral

$$\mathcal{I}{\left(a\right)}:=-8\int_{0}^{a}\mathrm{d}x\,\frac{\arctan{\left(x\right)}\ln{\left(1-a^{-2}x^{2}\right)}}{1+x^{2}}.\tag{1}$$

This answer will consider the problem of the general evaluation of $$\mathcal{I}$$ and obtain a closed-form expression for it in terms of elementary functions and polylogarithms.

Note that this includes the OP's integral as the following particular value:

$$-\frac18\mathcal{I}{\left(\frac{1}{\sqrt{3}}\right)}=\int_{0}^{\frac{1}{\sqrt{3}}}\mathrm{d}x\,\frac{\arctan{\left(x\right)}\ln{\left(1-3x^{2}\right)}}{1+x^{2}}\approx-0.124354.$$

Recall that the standard Clausen functions are defined by the following Fourier series:

$$\operatorname{Cl}_{2m}{\left(\theta\right)}:=\sum_{n=1}^{\infty}\frac{\sin{\left(n\theta\right)}}{n^{2m}};~~~\small{m\in\mathbb{N}\land\theta\in\mathbb{R}},\tag{2a}$$

$$\operatorname{Cl}_{2m+1}{\left(\theta\right)}:=\sum_{n=1}^{\infty}\frac{\cos{\left(n\theta\right)}}{n^{2m+1}};~~~\small{m\in\mathbb{N}\land\theta\in\mathbb{R}}.\tag{2b}$$

The Clausen function of order $$2$$ may also be given by the integral representation

$$\operatorname{Cl}_{2}{\left(\theta\right)}=-\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(\left|2\sin{\left(\frac{\varphi}{2}\right)}\right|\right)};~~~\small{\theta\in\mathbb{R}}.$$

The following pair of special values and pair of derivatives are obtained directly from the Fourier series definitions $$(2)$$:

$$\operatorname{Cl}_{2m}{\left(\theta\right)}=0;~~~\small{m\in\mathbb{N}},$$

$$\operatorname{Cl}_{2m+1}{\left(0\right)}=\sum_{n=1}^{\infty}\frac{1}{n^{2m+1}}=\zeta{\left(2m+1\right)};~~~\small{m\in\mathbb{N}},$$

$$\frac{d}{d\theta}\operatorname{Cl}_{2m+1}{\left(\theta\right)}=-\operatorname{Cl}_{2m}{\left(\theta\right)};~~~\small{m\in\mathbb{N}\land\theta\in\mathbb{R}},$$

$$\frac{d}{d\theta}\operatorname{Cl}_{2m+2}{\left(\theta\right)}=\operatorname{Cl}_{2m+1}{\left(\theta\right)};~~~\small{m\in\mathbb{N}\land\theta\in\mathbb{R}}.$$

Then, the following integration formulas for higher order Clausen functions follow quite simply from the fundamental theorem of calculus:

$$\int_{0}^{\theta}\mathrm{d}\varphi\,\operatorname{Cl}_{2m}{\left(\varphi\right)}=\zeta{\left(2m+1\right)}-\operatorname{Cl}_{2m+1}{\left(\theta\right)};~~~\small{m\in\mathbb{N}\land\theta\in\mathbb{R}},$$

$$\int_{0}^{\theta}\mathrm{d}\varphi\,\operatorname{Cl}_{2m+1}{\left(\varphi\right)}=\operatorname{Cl}_{2m+2}{\left(\theta\right)};~~~\small{m\in\mathbb{N}\land\theta\in\mathbb{R}}.$$

Suppose $$a\in\mathbb{R}_{>0}$$, and set $$\alpha:=\arctan{\left(a\right)}$$. Then, $$0<\alpha<\frac{\pi}{2}\land a=\tan{\left(\alpha\right)}$$, and we find

\begin{align} \mathcal{I}{\left(a\right)} &=-8\int_{0}^{a}\mathrm{d}x\,\frac{\arctan{\left(x\right)}\ln{\left(1-a^{-2}x^{2}\right)}}{1+x^{2}}\\ &=-8\int_{0}^{\tan{\left(\alpha\right)}}\mathrm{d}x\,\frac{\arctan{\left(x\right)}\ln{\left(1-\cot^{2}{\left(\alpha\right)}x^{2}\right)}}{1+x^{2}}\\ &=-8\int_{0}^{\alpha}\mathrm{d}\varphi\,\varphi\ln{\left(1-\cot^{2}{\left(\alpha\right)}\tan^{2}{\left(\varphi\right)}\right)};~~~\small{\left[\arctan{\left(x\right)}=\varphi\right]}\\ &=-8\int_{0}^{\alpha}\mathrm{d}\varphi\,\varphi\ln{\left(\frac{\sin^{2}{\left(\alpha\right)}\cos^{2}{\left(\varphi\right)}-\cos^{2}{\left(\alpha\right)}\sin^{2}{\left(\varphi\right)}}{\sin^{2}{\left(\alpha\right)}\cos^{2}{\left(\varphi\right)}}\right)}\\ &=-8\int_{0}^{\alpha}\mathrm{d}\varphi\,\varphi\ln{\left(\csc^{2}{\left(\alpha\right)}\sec^{2}{\left(\varphi\right)}\sin{\left(\alpha-\varphi\right)}\sin{\left(\alpha+\varphi\right)}\right)}\\ &=8\int_{0}^{\alpha}\mathrm{d}\varphi\,\varphi\left[\ln{\left(\sin^{2}{\left(\alpha\right)}\right)}+\ln{\left(\cos^{2}{\left(\varphi\right)}\right)}-\ln{\left(\sin{\left(\alpha-\varphi\right)}\right)}-\ln{\left(\sin{\left(\alpha+\varphi\right)}\right)}\right]\\ &=8\int_{0}^{\alpha}\mathrm{d}\varphi\,\varphi\ln{\left(\sin^{2}{\left(\alpha\right)}\right)}\\ &~~~~~+8\int_{0}^{\alpha}\mathrm{d}\varphi\,\varphi\left[2\ln{\left(\cos{\left(\varphi\right)}\right)}-\ln{\left(\sin{\left(\alpha-\varphi\right)}\right)}-\ln{\left(\sin{\left(\alpha+\varphi\right)}\right)}\right]\\ &=4\alpha^{2}\ln{\left(\sin^{2}{\left(\alpha\right)}\right)}\\ &~~~~~+8\int_{0}^{\alpha}\mathrm{d}\varphi\int_{0}^{\varphi}\mathrm{d}\vartheta\,\left[2\ln{\left(\cos{\left(\varphi\right)}\right)}-\ln{\left(\sin{\left(\alpha-\varphi\right)}\right)}-\ln{\left(\sin{\left(\alpha+\varphi\right)}\right)}\right]\\ &=4\alpha^{2}\ln{\left(\sin^{2}{\left(\alpha\right)}\right)}\\ &~~~~~+8\int_{0}^{\alpha}\mathrm{d}\vartheta\int_{\vartheta}^{\alpha}\mathrm{d}\varphi\,\left[2\ln{\left(\cos{\left(\varphi\right)}\right)}-\ln{\left(\sin{\left(\alpha-\varphi\right)}\right)}-\ln{\left(\sin{\left(\alpha+\varphi\right)}\right)}\right].\\ \end{align}

Then,

\begin{align} \mathcal{I}{\left(a\right)} &=4\alpha^{2}\ln{\left(\sin^{2}{\left(\alpha\right)}\right)}\\ &~~~~~+8\int_{0}^{\alpha}\mathrm{d}\vartheta\int_{\vartheta}^{\alpha}\mathrm{d}\varphi\,\left[2\ln{\left(\cos{\left(\varphi\right)}\right)}-\ln{\left(\sin{\left(\alpha-\varphi\right)}\right)}-\ln{\left(\sin{\left(\alpha+\varphi\right)}\right)}\right]\\ &=4\alpha^{2}\ln{\left(\sin^{2}{\left(\alpha\right)}\right)}\\ &~~~~~+8\int_{0}^{\alpha}\mathrm{d}\vartheta\int_{\vartheta}^{\alpha}\mathrm{d}\varphi\,\left[-\ln{\left(2\sin{\left(\alpha-\varphi\right)}\right)}-\ln{\left(2\sin{\left(\alpha+\varphi\right)}\right)}+2\ln{\left(2\cos{\left(\varphi\right)}\right)}\right]\\ &=4\alpha^{2}\ln{\left(\sin^{2}{\left(\alpha\right)}\right)}\\ &~~~~~+8\int_{0}^{\alpha}\mathrm{d}\vartheta\,\bigg{[}-\int_{\vartheta}^{\alpha}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\alpha-\varphi\right)}\right)}-\int_{\vartheta}^{\alpha}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\alpha+\varphi\right)}\right)}\\ &~~~~~+2\int_{\vartheta}^{\alpha}\mathrm{d}\varphi\,\ln{\left(2\cos{\left(\varphi\right)}\right)}\bigg{]}\\ &=4\alpha^{2}\ln{\left(\sin^{2}{\left(\alpha\right)}\right)}\\ &~~~~~+8\int_{0}^{\alpha}\mathrm{d}\vartheta\,\bigg{[}-\int_{0}^{\alpha-\vartheta}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\varphi\right)}\right)}-\int_{\alpha+\vartheta}^{2\alpha}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\varphi\right)}\right)}\\ &~~~~~+2\int_{\frac{\pi}{2}-\alpha}^{\frac{\pi}{2}-\vartheta}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\varphi\right)}\right)}\bigg{]}\\ &=4\alpha^{2}\ln{\left(\sin^{2}{\left(\alpha\right)}\right)}+4\int_{0}^{\alpha}\mathrm{d}\vartheta\,\bigg{[}-\int_{0}^{2\alpha-2\vartheta}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\varphi}{2}\right)}\right)}\\ &~~~~~-\int_{2\alpha+2\vartheta}^{4\alpha}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\varphi}{2}\right)}\right)}+2\int_{\pi-2\alpha}^{\pi-2\vartheta}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\varphi}{2}\right)}\right)}\bigg{]};~~~\small{\left[\varphi\mapsto\frac{\varphi}{2}\right]}\\ &=4\alpha^{2}\ln{\left(\sin^{2}{\left(\alpha\right)}\right)}+4\int_{0}^{\alpha}\mathrm{d}\vartheta\,\bigg{[}\operatorname{Cl}_{2}{\left(2\alpha-2\vartheta\right)}\\ &~~~~~+\operatorname{Cl}_{2}{\left(4\alpha\right)}-\operatorname{Cl}_{2}{\left(2\alpha+2\vartheta\right)}+2\operatorname{Cl}_{2}{\left(\pi-2\alpha\right)}-2\operatorname{Cl}_{2}{\left(\pi-2\vartheta\right)}\bigg{]},\\ \end{align}

and then,

\begin{align} \mathcal{I}{\left(a\right)} &=4\alpha^{2}\ln{\left(\sin^{2}{\left(\alpha\right)}\right)}+4\int_{0}^{\alpha}\mathrm{d}\vartheta\,\bigg{[}\operatorname{Cl}_{2}{\left(2\alpha-2\vartheta\right)}\\ &~~~~~+2\operatorname{Cl}_{2}{\left(2\alpha\right)}-\operatorname{Cl}_{2}{\left(2\alpha+2\vartheta\right)}-2\operatorname{Cl}_{2}{\left(\pi-2\vartheta\right)}\bigg{]}\\ &=4\alpha^{2}\ln{\left(\sin^{2}{\left(\alpha\right)}\right)}+8\alpha\operatorname{Cl}_{2}{\left(2\alpha\right)}\\ &~~~~~+4\int_{0}^{\alpha}\mathrm{d}\vartheta\,\left[\operatorname{Cl}_{2}{\left(2\alpha-2\vartheta\right)}-\operatorname{Cl}_{2}{\left(2\alpha+2\vartheta\right)}-2\operatorname{Cl}_{2}{\left(\pi-2\vartheta\right)}\right]\\ &=4\alpha^{2}\ln{\left(\sin^{2}{\left(\alpha\right)}\right)}+8\alpha\operatorname{Cl}_{2}{\left(2\alpha\right)}\\ &~~~~~+2\int_{0}^{2\alpha}\mathrm{d}\vartheta\,\left[\operatorname{Cl}_{2}{\left(2\alpha-\vartheta\right)}-\operatorname{Cl}_{2}{\left(2\alpha+\vartheta\right)}-2\operatorname{Cl}_{2}{\left(\pi-\vartheta\right)}\right];~~~\small{\left[\vartheta\mapsto\frac{\vartheta}{2}\right]}\\ &=4\alpha^{2}\ln{\left(\sin^{2}{\left(\alpha\right)}\right)}+8\alpha\operatorname{Cl}_{2}{\left(2\alpha\right)}\\ &~~~~~+2\left[\int_{0}^{2\alpha}\mathrm{d}\vartheta\,\operatorname{Cl}_{2}{\left(2\alpha-\vartheta\right)}-\int_{0}^{2\alpha}\mathrm{d}\vartheta\,\operatorname{Cl}_{2}{\left(2\alpha+\vartheta\right)}-2\int_{0}^{2\alpha}\mathrm{d}\vartheta\,\operatorname{Cl}_{2}{\left(\pi-\vartheta\right)}\right]\\ &=4\alpha^{2}\ln{\left(\sin^{2}{\left(\alpha\right)}\right)}+8\alpha\operatorname{Cl}_{2}{\left(2\alpha\right)}\\ &~~~~~+2\left[\int_{0}^{2\alpha}\mathrm{d}\vartheta\,\operatorname{Cl}_{2}{\left(\vartheta\right)}-\int_{2\alpha}^{4\alpha}\mathrm{d}\vartheta\,\operatorname{Cl}_{2}{\left(\vartheta\right)}-2\int_{\pi-2\alpha}^{\pi}\mathrm{d}\vartheta\,\operatorname{Cl}_{2}{\left(\vartheta\right)}\right]\\ &=4\alpha^{2}\ln{\left(\sin^{2}{\left(\alpha\right)}\right)}+8\alpha\operatorname{Cl}_{2}{\left(2\alpha\right)}\\ &~~~~~+2\left[\operatorname{Cl}_{3}{\left(0\right)}-\operatorname{Cl}_{3}{\left(2\alpha\right)}+\operatorname{Cl}_{3}{\left(4\alpha\right)}-\operatorname{Cl}_{3}{\left(2\alpha\right)}+2\operatorname{Cl}_{3}{\left(\pi\right)}-2\operatorname{Cl}_{3}{\left(\pi-2\alpha\right)}\right]\\ &=4\alpha^{2}\ln{\left(\sin^{2}{\left(\alpha\right)}\right)}+8\alpha\operatorname{Cl}_{2}{\left(2\alpha\right)}\\ &~~~~~+2\operatorname{Cl}_{3}{\left(0\right)}+4\operatorname{Cl}_{3}{\left(\pi\right)}+2\operatorname{Cl}_{3}{\left(4\alpha\right)}-4\operatorname{Cl}_{3}{\left(2\alpha\right)}-4\operatorname{Cl}_{3}{\left(\pi-2\alpha\right)}\\ &=4\alpha^{2}\ln{\left(\sin^{2}{\left(\alpha\right)}\right)}+8\alpha\operatorname{Cl}_{2}{\left(2\alpha\right)}-\zeta{\left(3\right)}+\operatorname{Cl}_{3}{\left(4\alpha\right)}.\blacksquare\\ \end{align}

For the particular case in which $$a=\frac{1}{\sqrt{3}}$$, we then have $$\alpha=\arctan{\left(\frac{1}{\sqrt{3}}\right)}=\frac{\pi}{6}$$, and hence

\begin{align} \mathcal{I}{\left(\frac{1}{\sqrt{3}}\right)} &=4\left(\frac{\pi}{6}\right)^{2}\ln{\left(\sin^{2}{\left(\frac{\pi}{6}\right)}\right)}+8\cdot\frac{\pi}{6}\operatorname{Cl}_{2}{\left(2\cdot\frac{\pi}{6}\right)}-\zeta{\left(3\right)}+\operatorname{Cl}_{3}{\left(4\cdot\frac{\pi}{6}\right)}\\ &=-\frac{2\pi^{2}}{9}\ln{\left(2\right)}+\frac{4\pi}{3}\operatorname{Cl}_{2}{\left(\frac{\pi}{3}\right)}-\zeta{\left(3\right)}+\operatorname{Cl}_{3}{\left(\frac{2\pi}{3}\right)}\\ &=\frac{4\pi}{3}G-\frac{13}{9}\zeta{\left(3\right)}-\frac{2\pi^{2}}{9}\ln{\left(2\right)},\\ \end{align}

where here $$G$$ denotes Gieseking's constant.

This is not a full answer.

Consider using Feynman trick $$I(a)=\int_0^{\frac 1 {\sqrt 3}}\frac{ \tan ^{-1}(a x)\log \left(1-3 x^2\right)}{1+x^2}\,dx$$ $$I'(a)=\int_0^{\frac 1 {\sqrt 3}}\frac{x \log \left(1-3 x^2\right)}{\left(x^2+1\right) \left(a^2 x^2+1\right)}\,dx=\frac{\text{Li}_2\left(\frac{1}{4}\right)-\text{Li}_2\left(\frac{a^2}{a^2+3}\right)}{ 2 \left(a^2-1\right)}$$ We cannot split the next integral but a CAS finds the antiderivative (very long !).

The computation of $$\int_0^1 \frac{\text{Li}_2\left(\frac{1}{4}\right)-\text{Li}_2\left(\frac{a^2}{a^2+3}\right)}{2 \left(a^2-1\right)}\,da$$ took a very long time because we need to take the limits (not just apply the formula which gives indeterminate at both bounds). I have an awful result containing a bunch of polylogarithms with complex argements.

The last integrand is quite nice and, by the end, I used a series expansion of it built around $$a=0$$. This gives $$\frac{\text{Li}_2\left(\frac{1}{4}\right)-\text{Li}_2\left(\frac{a^2}{a^2+3}\right)}{2 \left(a^2-1\right)}=k+\sum_{n=1}^p \left({b_n}+k\right)\, a^{2n}+O(a^{2p+2}) \qquad \text{with} \qquad k=-\frac{1}{2}\,\text{Li}_2\left(\frac{1}{4}\right)$$ where the $$b_n$$'s are $$\left\{\frac{1}{6},\frac{1}{8},\frac{265}{1944},\frac{115}{864},\frac{78173}{583200}, \frac{234029}{1749600},\frac{11474681}{85730400},\frac{137669537}{1028764800},\frac {33455693611}{249989846400},\cdots\right\}$$

Now, as a function of $$p$$, the value of the integral $$\left( \begin{array}{ccc} 1 & \frac{1}{18}-\frac{2 }{3}\text{Li}_2\left(\frac{1}{4}\right)& -0.1228795372 \\ 2 & \frac{29}{360}-\frac{23 }{30}\text{Li}_2\left(\frac{1}{4}\right) & -0.1246448011 \\ 3 & \frac{3403}{34020}-\frac{88 }{105}\text{Li}_2\left(\frac{1}{4}\right) & -0.1242890078 \\ 4 & \frac{31249}{272160}-\frac{563 }{630}\text{Li}_2\left(\frac{1}{4}\right) & -0.1243695042 \\ 5 & \frac{178228}{1403325}-\frac{3254}{3465} \text{Li}_2\left(\frac{1}{4}\right) & -0.1243499428 \\ 6 & \frac{240448777}{1751349600}-\frac{88069 }{90090} \text{Li}_2\left(\frac{1}{4}\right) & -0.1243549342 \\ 7 & \frac{3361000121}{22986463500}-\frac{45536 }{45045}\text{Li}_2\left(\frac{1}{4}\right) & -0.1243536145 \\ 8 & \frac{1926817784779}{12504636144000}-\frac{1593269 }{1531530}\text{Li}_2\left(\frac{1}{4}\right) & -0.1243539730 \\ 9 & \frac{2325692917041587}{14433476269212000}-\frac{15518938 }{14549535} \text{Li}_2\left(\frac{1}{4}\right) & -0.1243538735 \\ 10 & \frac{203084487005661923}{1212412006613808000}-\frac{31730711 }{29099070} \text{Li}_2\left(\frac{1}{4}\right)& -0.1243539016 \\ 11 & \frac{1661411146332456037}{9585632427290419500}-\frac{372177944 }{334639305}\text{Li}_2\left(\frac{1}{4}\right) & -0.1243538935 \\ 12 & \frac{822107289750189315107}{4601103565099401360000}-\frac{3788707301 }{3346393050}\text{Li}_2\left(\frac{1}{4}\right) & -0.1243538959 \end{array} \right)$$