Calculate this integral containing log inverse tanh: $\int_0^1 x(\arctan x)\ln(\operatorname{arctanh}x)dx$

Can I get some help calculating the following integral

$$I = \int_0^1 x(\arctan x)\ln(\operatorname{arctanh}x)dx$$

By using $$\operatorname{arctanh}x=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$$ we have

$$I=\left(\frac{1}{2}-\frac{\pi}{4}\right)\ln2+\int_{0}^{1}x\left(\arctan x\right)\ln\left(\ln\left(\frac{1+x}{1-x}\right)\right)dx$$

substituting $$x\to\frac{1+t}{1-t}$$ we get

\begin{align} I&=\left(\frac{1}{2}-\frac{\pi}{4}\right)\ln2+2\int_{1}^{\infty}\frac{t-1}{\left(t+1\right)^{3}}\arctan\left(\frac{t-1}{t+1}\right)\ln\left(\ln\left(t\right)\right)dt\\ &=\left(\frac{1}{2}-\frac{\pi}{4}\right)\ln2+2\int_{1}^{\infty}\frac{t-1}{\left(t+1\right)^{3}}\arctan\left(t\right)\ln\left(\ln\left(t\right)\right)dt-\frac{\pi}{2}\int_{1}^{\infty}\frac{t-1}{\left(t+1\right)^{3}}\ln\left(\ln\left(t\right)\right)dt \end{align}

This form bears similarity to Malmsten's integrals detailed by Blagouchine.

The potential key would be finding the anti-derivative of $$\frac{t-1}{\left(t+1\right)^{3}}\ln\left(\ln\left(t\right)\right)$$ then performing integration by parts.

Small update $$\int\frac{t-1}{\left(t+1\right)^{3}}\ln\left(\ln\left(t\right)\right)=\frac{\operatorname{li}(t)-t\ln(\ln(t))}{(t+1)^2}+2\int\frac{\operatorname{li}(t)}{(t+1)^3}dt$$

Edit

I found that

$$2I = \left(2-\frac{\pi}{2}\right)\ln2-\ln\pi+\gamma+\pi\ln\left(\frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\sqrt{2\pi}\right)+\int_{0}^{1}\frac{\arctan t}{\operatorname{arctanh}t}dt$$

where the last integral is the infamous egg. So finding a closed form may prove very difficult.

This comes from the fact that $$\frac{d}{dx}(x^2-1)\arctan(x)\ln(\operatorname{arctanh}(x)) = 2x\arctan(x)\ln(\operatorname{arctanh}(x))+\left(\frac{x^2-1}{x^2+1}\right)\ln(\operatorname{arctanh}(x))-\frac{\arctan(x)}{\operatorname{arctanh}(x)}$$

then evaluating $$\int_0^1 \left(\frac{x^2-1}{x^2+1}\right)\ln(\operatorname{arctanh}(x)) dx$$

• From where did you get this integral? And are you sure that $\operatorname{arctanh}$ is inside the logarithm? Nov 29, 2019 at 22:56
• It came from the calculation of another integral. And yes, the functions are correct. Nov 29, 2019 at 23:08
• Wolfram gives me the value 0.0000381467... x) Nov 30, 2019 at 0:43

What we could do is a Taylor expansion of the integrand around $$x=0$$ and get $$x \tan ^{-1}(x) \log \left(\tanh ^{-1}(x)\right)=\sum_{i=1}^n (a_i+b_i \log(x))x^{2i}$$ where the $$a_i$$'s and $$b_i$$'s are rational numbers.
This would make $$\int_0^1 x \tan ^{-1}(x) \log \left(\tanh ^{-1}(x)\right)\,dx=\sum_{i=1}^n \frac{(2i+1)a_i-b_i}{(2 i+1)^2}$$ As shown in the table below, the convergence is very slow $$\left( \begin{array}{cc} n & result \\ 10 & -0.01638269 \\ 20 & -0.00646257 \\ 30 & -0.00443149 \\ 40 & -0.00299751 \\ 50 & -0.00243324 \\ 60 & -0.00189436 \\ 70 & -0.00164035 \\ 80 & -0.00136305 \\ 90 & -0.00122108 \\ 100 & -0.00105355 \\ 200 & -0.00046489 \\ 300 & -0.00028194 \\ 400 & -0.00019428 \\ 500 & -0.00014327 \end{array} \right)$$
Just for the fun of it, the value given by numerical integration $$(0.0000381466606686)$$seems to be quite close to $$\frac{1}{1000 \sqrt{10} \left(5^{3/4}+11^{2/3}\right)} \approx 0.0000381466605763$$