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

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?

• Expand the $\operatorname{artanh}$ as $\frac12\log\left(\frac{1+x}{1-x}\right)$ which seems to reduce the problem overall to the evaluation of the two integrals $$\int_0^1\frac{\log(1+x)^2\log(1-x)}x\text{ and }\int_0^1\frac{\log(1+x)\log(1-x)^2}x$$ since the other two integrals can be evaluated exactly in terms of polyloagrithms according to WolframAlpha (and here). – mrtaurho Dec 30 '18 at 17:40
• Mathematica gives something slightly different with NIntegrate[] as well, though gives exact results with Integrate[]. The integrand is divergent at $x=1$, and numerical methods are usually problematic with such integrals. – Ininterrompue Dec 30 '18 at 17:55

As MrTaurho pointed out in the comments, we can rewrite $$\,\displaystyle{\operatorname{arctanh}x=\frac12 \ln\left(\frac{1+x}{1-x}\right)}$$, this gives: $$I=\int_{0}^1 \frac{\operatorname{arctanh}(x)^3}{x}dx=-\frac18\int_0^1 \frac{\ln^3\left(\frac{1-x}{1+x}\right)}{x}dx$$ And we will substitute $$\displaystyle{\frac{1-x}{1+x}=t\Rightarrow dx=-\frac{2}{(1+t)^2}dt}$$ in order to get: $$I=-\frac18\int_0^1 \frac{\ln^3 t}{\frac{1-t}{1+t}}\frac{2}{(1+t)^2}dt=-\frac14 \int_0^1 \frac{\ln^3 t}{1-t^2}dt=-\frac14 \sum_{n=0}^\infty \int_0^1 t^{2n}\ln^3t dt$$ Now we consider the following integral: $$\int_0^1 t^a dt=\frac{1}{a+1}\Rightarrow \int_0^1 t^a \ln^3 tdt=\frac{d^3}{da^3} \left(\frac{1}{a+1}\right)=-\frac{6}{(a+1)^4}$$ $$\Rightarrow I=\frac{6}{4}\sum_{n=0}^\infty \frac{1}{(2n+1)^4}=\frac32\cdot\frac{\pi^4}{96}=\frac{\pi^4}{64}$$

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

$$I(k)=\int_0^1 \frac{\text{arctanh}^kx}{x}dx=\frac{(-1)^k}{2^k}\int_0^1 \frac{\ln^k\left(\frac{1-x}{1+x}\right)}{x}dx\overset{\large\frac{1-x}{1+x}=t}=\frac{2(-1)^k}{2^k}\int_0^1 \frac{\ln^k t}{1-t^2}dt$$ $$=\frac{(-1)^k}{2^{k-1}}\sum_{n=0}^\infty \int_0^1 x^{2n}\ln^k xdx=\frac{(-1)^k}{2^{k-1}}\sum_{n=0}^\infty \frac{(-1)^k k!}{(2n+1)^{k+1}}=\frac{k!}{2^{k-1}} \sum_{n=0}^\infty \frac{1}{(2n+1)^{k+1}}$$ 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: $$I(k)=\frac{k!}{2^{k-1}} \left(1-\frac{1}{2^{k+1}}\right)\zeta(k+1)=\boxed{\frac{k!\left(2^{k+1}-1\right)}{4^k}\zeta(k+1)}$$

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

• Well done! I did not thought about a self-similiar substitution here which reduces the whole problem quite well ^^ (+1) Moreover this explains why the odd powers can be written in terms of $\pi$ since it always and up in values of the Zeta Function. – mrtaurho Dec 30 '18 at 18:23
• @Zacky Best of luck with the notation. You should find, I think, that $\int_0^1\frac{\operatorname{arctanh}^p x}{x}dx=p!\left(2^{1-p}-4^{-p}\right)\zeta(p+1)$. – J.G. Dec 30 '18 at 18:38
• Something like that :D – Nyssa Dec 30 '18 at 18:42
• Your edit makes my own answer generalising your obtained solution redundant and I agree with your solution. Note there is a minor mistake hence you wrote again $\ln^\color{red}{3} t$ in your first line of your generalization. – mrtaurho Dec 30 '18 at 18:45
• Thanks! I was copy-pasting my own latex from above to make it easier :D – Nyssa Dec 30 '18 at 18:49

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 $${3\choose k} (-1)^k\log^{3-k}(1+x)\log^k(1-x)$$ for $$k=0, 1, 2,3,$$). If the power of the log is greater than $$1$$, use differentiation under the integral sign, using $$\frac{\partial^k}{\partial a^k} (1\pm x)^a \Big|_{a=0} = \log^k(1\pm x)$$. 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.

• Hi! What is the meaning of that $3/4$ infinity sign? – Nyssa Dec 30 '18 at 18:56
• @Zacky $\propto$ stands for "proportional to". – mrtaurho Dec 30 '18 at 19:04
• Never heard of it. Thanks! – Nyssa Dec 30 '18 at 19:11
• Yes, and I made it clearer that the constant of proportionality will be the binomial coefficient ${3\choose k}$ for $k=0, 1, 2, 3$. – Zachary Dec 30 '18 at 21:38