Jtdylktuy

I'm trying to evaluate the following integral:

$$int_0^1 frac{operatorname{arctanh}^3(x)}{x}dx$$

I was playing around trying to numerically approximate the answer with known constants and found that the integral is almost exactly $frac{pi^4}{64}$. The integral seems to break down after about $11$ decimal places.

I have a suspicion that this stems from the integral:
$$int_0^1 frac{operatorname{arctanh}(x)}{x}dx$$
since this is exactly equal to $frac{pi^2}{8}$.

Also for the integral:
$$int_0^1 frac{operatorname{arctanh}^5(x)}{x}dx$$
This is suspicously close to $frac{pi^6}{128}$ but not exactly.
For some reason the above integrals diverges slightly from the some from of $frac{pi^n}{2^m}$ for some $n$ and $m$.

My question is: Why does this happen? And what are the true values of those integrals?

As MrTaurho pointed out in the comments, we can rewrite $,displaystyle{operatorname{arctanh}x=frac12 lnleft(frac{1+x}{1-x}right)}$, this gives:
$$I=int_{0}^1 frac{operatorname{arctanh}(x)^3}{x}dx=-frac18int_0^1 frac{ln^3left(frac{1-x}{1+x}right)}{x}dx$$
And we will substitute $displaystyle{frac{1-x}{1+x}=tRightarrow dx=-frac{2}{(1+t)^2}dt}$ in order to get:
$$I=-frac18int_0^1 frac{ln^3 t}{frac{1-t}{1+t}}frac{2}{(1+t)^2}dt=-frac14 int_0^1 frac{ln^3 t}{1-t^2}dt=-frac14 sum_{n=0}^infty int_0^1 t^{2n}ln^3t dt$$
Now we consider the following integral: $$int_0^1 t^a dt=frac{1}{a+1}Rightarrow int_0^1 t^a ln^3 tdt=frac{d^3}{da^3} left(frac{1}{a+1}right)=-frac{6}{(a+1)^4}$$
$$Rightarrow I=frac{6}{4}sum_{n=0}^infty frac{1}{(2n+1)^4}=frac32cdotfrac{pi^4}{96}=frac{pi^4}{64}$$

Of course this can be generalized for any power, so let's do that. Consider:

$$I(k)=int_0^1 frac{text{arctanh}^kx}{x}dx=frac{(-1)^k}{2^k}int_0^1 frac{ln^kleft(frac{1-x}{1+x}right)}{x}dxoverset{largefrac{1-x}{1+x}=t}=frac{2(-1)^k}{2^k}int_0^1 frac{ln^k t}{1-t^2}dt$$
$$=frac{(-1)^k}{2^{k-1}}sum_{n=0}^infty int_0^1 x^{2n}ln^k xdx=frac{(-1)^k}{2^{k-1}}sum_{n=0}^infty frac{(-1)^k k!}{(2n+1)^{k+1}}=frac{k!}{2^{k-1}} sum_{n=0}^infty frac{1}{(2n+1)^{k+1}}$$
Where above we used the following result: $$int_0^1 t^a dt=frac{1}{a+1}Rightarrow int_0^1 t^a ln^k tdt=frac{d^k}{da^k} left(frac{1}{a+1}right)=frac{(-1)^k k!}{(a+1)^{k+1}}$$
And finally it reduces to: $$I(k)=frac{k!}{2^{k-1}} left(1-frac{1}{2^{k+1}}right)zeta(k+1)=boxed{frac{k!left(2^{k+1}-1right)}{4^k}zeta(k+1)}$$

One can verify the result by comparing to the one announced by Maple: $$I(5)=int_0^1 frac{text{arctanh}^5 x}{x}dx=frac{5!(2^6-1)}{4^5}zeta(6)=frac{945}{128}cdotfrac{pi^6}{945}=frac{pi^6}{128}$$

An alternative route to the general result: For $z in mathbb{C}$ with $operatorname{Re}(z) > 0$ we can let $x = tanh(t/2)$ and use the geometric series to find
begin{align}
int limits_0^1 frac{operatorname{artanh}^z (x)}{x} , mathrm{d} x &= frac{1}{2^z} int limits_0^infty frac{t^z}{sinh(t)} , mathrm{d} t = 2^{1-z} sum limits_{k=0}^infty int limits_0^infty t^z mathrm{e}^{-(2k+1) t} , mathrm{d} t \
&= 2^{1-z} int limits_0^infty u^z mathrm{e}^{-u} , mathrm{d} u sum limits_{k=0}^infty frac{1}{(2k+1)^{1+z}} = 2^{1-z} Gamma(1+z) lambda(1+z) \
&= frac{2^{1+z}-1}{4^z} Gamma(1+z) zeta(1+z) , ,
end{align}
where $lambda$ is the Dirichlet lambda function.

Here's another method. After expanding $text{arctanh}^3(x)=frac{1}{8}Big(log(1+x)-log(1-x)Big)^3$, you will get terms of the form $propto log^a(1+x)log^b(1-x)$(more precisely ${3choose k} (-1)^klog^{3-k}(1+x)log^k(1-x)$ for $k=0, 1, 2,3,$). If the power of the log is greater than $1$, use differentiation under the integral sign, using $frac{partial^k}{partial a^k} (1pm x)^a Big|_{a=0} = log^k(1pm x)$. With powers of log equal to $1$, expand the log into its Taylor series. You can also combine these methods if both types of powers appear in the integral.

According to Maple,
$$ int_0^1 frac{text{arctanh}(x)^3}{x}; dx = frac{pi^4}{64} $$
and
$$ int_0^1 frac{text{arctanh}(x)^5}{x}; dx = frac{pi^6}{128} $$

In fact, it seems $int_0^1 dfrac{text{arctanh}(x)^k}{x}; dx$ is a rational number times $zeta(k+1)$ for each positive integer $k$, with denominator a power of $2$.