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