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$\newcommand{\Ti}{\operatorname{Ti}}$Considering the integral

$$\int_{0}^{\frac{1}{\sqrt{3}}}\frac{\arctan(x)}{1-x^2}\,dx$$

Here is my work, I want a fact check form my fellow MSE users since wolfram doesn’t give a closed form for the answer.

$$I=\int_{0}^{\frac{1}{\sqrt{3}}}\frac{\arctan(x)}{1-x^2}\,dx$$

IBP yields

$$I=\frac{\pi}{12}\log(2+\sqrt{3})-\frac{1}{2}\int_{0}^{\frac{1}{\sqrt{3}}}\frac{\log(\frac{1+x}{1-x})}{x^2+1}\,dx$$

Now let $$x\longrightarrow{\frac{1-x}{1+x}}$$

$$I=\frac{\pi}{12}\log(2+\sqrt{3})+\frac{1}{2}\int_{2-\sqrt{3}}^{1}\frac{\log(x)}{x^2+1}\,dx$$

$$I=\frac{\pi}{12}\log(2+\sqrt{3})+\frac{1}{2}I_{1}$$

Now to evaluate

$$I_{1}=\int_{2-\sqrt{3}}^{1}\frac{\log(x)}{x^2+1}\,dx$$

$$I_{1}=\int_{2-\sqrt{3}}^{1}\log(x)\sum_{n=0}^{\infty}(-1)^nx^{2n}\,dx$$

swapping operators on an interval of convergence gives

$$I_{1}=\sum_{n=0}^{\infty}(-1)^n\int_{2-\sqrt{3}}^{1}x^{2n}\log(x)\,dx$$

IBP yields

$$I_{1}=\sum_{n=0}^{\infty}(-1)^n[-\frac{(2-\sqrt{3})^{2n+1}}{2n+1}\log(2-\sqrt{3})-\frac{1}{(2n+1)^2}+\frac{(2-\sqrt{3})^{2n+1}}{(2n+1)^2}]$$

All of these are standard sums and with simplification the sum evaluates to

$$I_{1}=\frac{\pi}{12}\log(2+\sqrt{3})-G+\Ti_{2}(2-\sqrt{3})$$

Pluggin this value back into I we get the final result:

$$\int_{0}^{\frac{1}{\sqrt{3}}}\frac{\arctan(x)}{1-x^2}\,dx=\frac{\pi}{8}\log(2+\sqrt{3})-\frac{1}{2}G+\frac{1}{2}\Ti_{2}(2-\sqrt{3})$$

Where $G$ denotes Catalan’s constant, and $\Ti_{2}(x)$ denotes the inverse tangent integral

Is this right? Also, do any of you have a different solution method? I would be delighted to see!

FShrike
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Person
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3 Answers3

7

Continue with \begin{align} &\int_{0}^{\frac{1}{\sqrt{3}}}\frac{\arctan x}{1-x^2}\,dx\\ =& \ \frac{\pi}{12}\ln(2+\sqrt{3})+\frac{1}{2}\int_{2-\sqrt{3}}^{1}\frac{\ln x}{x^2+1}\,\overset{x=\tan t}{dx}\\ =& \ \frac{\pi}{12}\ln(2+\sqrt{3})+\frac{1}{2} \int_{0}^{\pi/4}\ln(\tan t)dt - \frac12 \int^{\pi/12}_{0}\ln(\tan t)dt\\ =& \ \frac{\pi}{12}\ln(2+\sqrt{3})+\frac{1}{2}(-G)- \frac{1}{2}\cdot (-\frac23 G)\\ =& \ \frac{\pi}{12}\ln(2+\sqrt{3})-\frac{1}6 G\\ \end{align} where $\int^{\pi/12}_{0}\ln(\tan t)dt=-\frac23G$

Quanto
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Integration by parts yields $$ \int_0^{\frac{1}{\sqrt{3}}} \frac{\arctan x}{1-x^2} d x= \frac{\pi}{12} \ln (2+\sqrt{3})+\frac{1}{2} \int_0^{\frac{1}{\sqrt{3}}} \frac{1}{1+x^2} \ln \left(\frac{1-x}{1+x}\right) d x $$ $$\int_0^{\frac{1}{\sqrt{3}}} \frac{1}{1+x^2} \ln \left(\frac{1-x}{1+x}\right) dx \stackrel{x=\tan \theta}{=} \int_0^{\frac{\pi}{6}} \ln \left(\frac{1-\tan \theta}{1+\tan \theta}\right) d \theta\stackrel{\theta\mapsto\frac{ \pi}{4}-\theta}{=} -\int_{\frac{\pi}{4}}^{\frac{\pi}{12}} \ln (\tan \theta) d \theta \\ = \int_0^{\frac{\pi}{4}} \ln (\tan \theta) d \theta-\int_0^{\frac{\pi}{12}} \ln (\tan \theta )d\theta=-G+\frac{2G}{3} =-\frac{G}{3} $$ Hence we can conclude that $$ \boxed{\int_0^{\frac{1}{\sqrt{3}}} \frac{\arctan x}{1-x^2} d x=\frac{\pi}{12} \ln (2+\sqrt{3})-\frac{1}{6} G\,} $$

Lai
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For the antiderivative, you could start with $$\tan^{-1}(x)=\frac i 2 \log \left(\frac{i+x}{i-x}\right)$$ and $$\frac 1 {1-x^2}=\frac{1}{2 (x+1)}-\frac{1}{2 (x-1)}$$ Decomposing the logarithm, you face four integrals looking like $$I(a,b)=\int \frac{\log (i+a x)}{x+b}\,dx$$ $$I(a,b)=\text{Li}_2\left(\frac{a x+i}{i-a b}\right)+\log (a x+i) \, \log\left(\frac{a (b+x)}{a b-i}\right)$$ The problem is that you will face four nasty polylogarithms with complex arguments (not pleasant).

For the given bounds, it would write $$\frac{\pi}{24} \, \log \left(\frac{27\left(7+4 \sqrt{3}\right)}{8} \right)-C-\frac i 4 \Big[\cdots \Big]$$ where $$\Big[\cdots \Big]=-\text{Li}_2\left(\frac{1}{2}+\frac{1}{2 \sqrt{3}}-i \left(\frac{1}{2}-\frac{1}{2 \sqrt{3}}\right)\right)+$$ $$\text{Li}_2\left(\frac{1}{2}+\frac{1}{ 2 \sqrt{3}}+i \left(\frac{1}{2}-\frac{1}{2 \sqrt{3}}\right)\right)-$$ $$\text{Li}_2\left(\frac{1}{2}-\frac{1}{ 2 \sqrt{3}}-i \left(\frac{1}{2}+\frac{1}{2 \sqrt{3}}\right)\right)+$$ $$\text{Li}_2\left(\frac{1}{2}-\frac{1}{ 2 \sqrt{3}}+i \left(\frac{1}{2}+\frac{1}{2 \sqrt{3}}\right)\right)$$ which, numerically, is @Quanto's nice and simple result.

Claude Leibovici
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