Evaluating $\int_0^1 \frac{x\cdot \tanh^{-1}(x)}{3+x^2} dx$

Question

I was trying to evaluate the integral

$$ I = \int_0^1 \frac{x \tanh^{-1}x}{3+x^2} dx$$

To solve this, I considered the family of integrals

$$ I(a) = \int_0^1 \frac{ x \tanh^{-1}ax}{3+x^2} dx$$

Note that $I(0) = 0$.

Using Leibnitz rule, we get:

$$ I'(a) = \int_0^1 \frac{(x^2+3)-(3)}{(3+x^2)(1-a^2x^2)}dx$$

$$I'(a) = \int_0^1 \frac{dx}{1-a^2x^2} + 3\cdot \int_0^1 \frac{dx}{(x^2+3)(a^2x^2-1)}$$

Multiplying and dividing by $a^2$ for the second integral, we get:

$$I'(a) = \frac{1}{2a} \cdot \ln(\frac{1+a}{1-a}) + 3\cdot a \cdot \int_0^1 \frac{adx}{(a^2x^2+3a^2)(a^2x^2-1)}$$

For the second integral, I substitute $ax = v$

$$I_0(a) = 3a\cdot \int_0^a \frac{dv}{(v^2+3a^2)(v^2-1)}$$

$$ = \frac{3a}{3a^2+1} \cdot (\int_0^a \frac{dv}{v^2-1} - \int_0^a \frac{dv}{v^2+3a^2})$$

$$I_0(a) = \frac{3a}{3a^2+1} \cdot ( \frac{1}{2}\cdot \ln(\frac{1-a}{1+a}) - \frac{1}{\sqrt3 a}\cdot \frac{\pi}{6})$$

Adding the integrals, we get:

$$I'(a) = \frac{1}{2} \cdot (\frac{1}{a} - \frac{3a}{3a^2+1}) \ln(\frac{1+a}{1-a}) - \frac{\sqrt3}{3a^2+1} \cdot \frac{\pi}{6} $$

This means that

$$ I(b) = \int_0^b I'(a) da$$

But I'm not sure how to integrate the first part of $I'(a)$ - that is, I am struggling to evaluate:

$$I_1(b) = \int_0^b \frac{1}{2} \cdot (\frac{1}{a}-\frac{3a}{3a^2+1}) \ln(\frac{1+a}{1-a}) da$$

$$I_1(b) = \int_0^b \frac{1}{2} \cdot \frac{1}{a(3a^2+1)} \ln(\frac{1+a}{1-a}) da$$

We proceed by substituting $\frac{1-a}{1+a} = t$ in the integrand and obtain:

$$I_1(b) = -\frac{1}{4}\cdot \int_{\frac{1-b}{1+b}}^1 \frac{\operatorname{ln}t\cdot (1+t)}{(t^2-t+1)\cdot (1-t)}dt$$

Can anyone help me in evaluating this specific integral? I would like to know the strategies used to evaluate integrals of such kinds.

Any help is appreciated. Thanks for reading!

Answer 1

Alternatively, reduce the integral to the familiar $\int_0^1\frac{\ln(1-x)}x dx= \frac{\pi^2}6$ as follows

\begin{align}\int_0^1\frac{x \tanh ^{-1}x}{x^2+3}\overset{x\to\frac{1-x}{1+x}}{dx} =& \ \frac14 \int_0^1\ln x\ d \left( \ln\frac{(1+x)^2}{x^2+x+1}\right)\\ \overset{ibp}= & \ \frac14 \int_0^1\frac1x \ln\frac{(1-x^2)^2}{(1-x)(1-x^3)}\ dx\\ =& \ \frac1{12} \int_0^1\frac{\ln(1-x)}x dx=\frac{\pi^2}{72} \end{align} where $\int^1_0 \frac{\ln(1-x)}{x}dx = 2\int^1_0 \frac{\ln(1-x^2)}{x}dx = 3\int^1_0 \frac{\ln(1-x^3)}{x}dx $ is used in the simplification.

Answer 2

$$I(a)=\int_0^1\frac{x \tanh ^{-1}(a x)}{x^2+3}\,dx$$ $$I'(a)=\int_0^1\frac{x^2}{\left(x^2+3\right) \left(1-a^2 x^2\right)}\,dx$$ $$I'(a)=\frac{1}{3 a^2+1}\int_0^1 \Bigg(\frac{1}{1-a^2 x^2}-\frac{3}{x^2+3} \Bigg)\,dx$$ $$I'(a)=\frac{1}{2(3 a^2+1)}\Bigg(\frac 1a \log \left(\frac{1+a}{1-a}\right)-\frac{\pi }{\sqrt{3}} \Bigg)$$ $$I(a)=-\frac{\pi ^2}{18}+\frac 12 \int_0^1 \frac 1 {a(3a^2+1)} \log \left(\frac{1+a}{1-a}\right)\,da$$ $$J=\int_0^1 \frac 1 {a(3a^2+1)} \log \left(\frac{1+a}{1-a}\right)\,da$$ $$J=\int_0^1\left(\frac{1}{a}-\frac{3 a}{3 a^2+1}\right)\log \left(\frac{1+a}{1-a}\right)\,da$$ $$J=\frac{\pi ^2}{4}-\frac{\pi ^2}{9}=\frac{5\pi ^2}{36}$$ Recombining everything $$I(1)=-\frac{\pi ^2}{18}+\frac{5\pi ^2}{72}=\frac{\pi ^2}{72}$$

Answer 3

I'll write you the final result (the process is quite long if you're interested I can write you the basic steps in the comments)

For $a\in(0,1)$ $$\int_{0}^{1}\frac{x\cdot\operatorname{tanh^{-1}}(ax)}{x^2+3}\mathrm{d}x=\frac{\text{Li}_2\left(\frac{a-1}{a+1}\right)+\operatorname{cot^{-1}}\left(\sqrt{3}a\right)^{2}-K(a)}{2}+\operatorname{tanh^{-1}}\left(a\right)\ln\left(\frac{2a}{\sqrt{3a^{2}+1}}\right)$$

Where:

• $\text{Li}_2(x)$ is the polylogarithm function
• $\displaystyle K(a):=\Re\left[\text{Li}_{2}\left(\frac{(a-1)(\sqrt{3}a-i)}{(a+1)(\sqrt{3}a+i)}\right)\right]$

You can easily compute $K(a)$ through the polylogarithm series: $$\text{Li}_2(z)=\sum_{k=1}^{\infty}\frac{z^k}{k^2}\qquad |z|\leq1$$ For $a\in[0,1]$ $$K(a)=\sum_{k=1}^{\infty}\frac{\cos\left(2k\operatorname{tan^{-1}}\left(\sqrt{3}a\right)\right)}{k^{2}}\left(\frac{1-a}{1+a}\right)^{k}$$

I'll leave you the Desmos link where I wrote the final result.

Answer 4

$$I_1(b) = -\frac{1}{4}\, \int_{\frac{1-b}{1+b}}^1 \frac{(1+t)\,\log(t)}{(t^2-t+1)\,(1-t)}dt$$

Let $$t^2-t+1=(t-\alpha)(t-\beta)\qquad \text{where} \qquad(\alpha,\beta)=\frac{1\pm i \sqrt{3}}{2} $$ Use partial fraction decomposition $$I_1(b)=\int_{\frac{1-b}{1+b}}^1 \left(\frac{\log (t)}{2 (t-1)}-\frac{\log (t)}{4 (t-\alpha )}-\frac{\log (t)}{4 (t-\beta )}\right)\,dt$$ to face three integrals $$J(\gamma)=\int_c^1 \frac{\log (t)}{t-\gamma }\,dt$$ $$J(\gamma)=-\log (c) \log \left(1-\frac{c}{\gamma }\right)-\text{Li}_2\left(\frac{c}{\gamma }\right)+\text{Li}_2\left(\frac{1}{\gamma }\right)$$

As a total, with $c=\frac{1-b}{1+b}$ $$I_1(b) =-\frac{\pi ^2}{72}+\frac{1}{4} \log (c) \log \left(c^2-c+1\right)+\frac{1}{2}\text{Li}_2(1-c)+$$ $$\frac 14\left(\text{Li}_2\left(\frac{1+i \sqrt{3}}{2} c\right)+\text{Li}_2\left(\frac{1-i \sqrt{3}}{2} c\right)\right)$$

Expanded as a series around $b=1$ $$I_1(b) =\frac{5\pi ^2}{72}+\frac 18\sum_{n=1}^\infty (-1)^{n+1}\, P_n \, (b-1)^n$$ where $P_n$ write $$P_n=\gamma_n\, L+\delta_n \qquad \text{with} \qquad L=\log \left(\frac{2}{1-b}\right)$$

The first of them are $$\left( \begin{array}{cc} n & P_n \\ 1 & L+1 \\ 2 & \frac{5}{4}L+\frac{3}{8} \\ 3 & \frac{4 }{3}L-\frac{1}{72} \\ 4 & \frac{41 }{32}L-\frac{103}{384} \\ 5 & \frac{91 }{80}L-\frac{1963}{4800} \\ 6 & \frac{91 }{96}L-\frac{1759}{3840} \\ \end{array} \right)$$