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

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    $\begingroup$ From where did you get this integral? And are you sure that $\operatorname{arctanh}$ is inside the logarithm? $\endgroup$ – Zacky Nov 29 '19 at 22:56
  • $\begingroup$ It came from the calculation of another integral. And yes, the functions are correct. $\endgroup$ – tyobrien Nov 29 '19 at 23:08
  • $\begingroup$ Wolfram gives me the value 0.0000381467... x) $\endgroup$ – Kermatoni Nov 30 '19 at 0:43
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I am very skeptical about a possible closed form for this integral.

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 $$\left( \begin{array}{cc} n & result \\ 10 & -0.01638269 \\ 20 & -0.00646257 \\ 30 & -0.00443149 \\ 40 & -0.00299751 \\ 50 & -0.00243324 \\ 60 & -0.00189436 \\ 70 & -0.00164035 \\ 80 & -0.00136305 \\ 90 & -0.00122108 \\ 100 & -0.00105355 \\ 200 & -0.00046489 \\ 300 & -0.00028194 \\ 400 & -0.00019428 \\ 500 & -0.00014327 \end{array} \right)$$

As you can see, even with so many terms, we even do not get the correct sign.

So, forget all of that and just use numerical integration.

Just for the fun of it, the value given by numerical integration $(0.0000381466606686)$seems to be quite close to $$\frac{1}{1000 \sqrt{10} \left(5^{3/4}+11^{2/3}\right)} \approx 0.0000381466605763$$

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