6
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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). $\endgroup$ – mrtaurho Dec 30 '18 at 17:40
  • $\begingroup$ 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. $\endgroup$ – Ininterrompue Dec 30 '18 at 17:55
12
$\begingroup$

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: $$I(5)=\int_0^1 \frac{\text{arctanh}^5 x}{x}dx=\frac{5!(2^6-1)}{4^5}\zeta(6)=\frac{945}{128}\cdot\frac{\pi^6}{945}=\frac{\pi^6}{128}$$

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – mrtaurho Dec 30 '18 at 18:23
  • 3
    $\begingroup$ @Zacky Best of luck with the notation. You should find, I think, that $\int_0^1\frac{\operatorname{arctanh}^p x}{x}dx=p!\left(2^{1-p}-4^{-p}\right)\zeta(p+1)$. $\endgroup$ – J.G. Dec 30 '18 at 18:38
  • $\begingroup$ Something like that :D $\endgroup$ – Nyssa Dec 30 '18 at 18:42
  • 1
    $\begingroup$ 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. $\endgroup$ – mrtaurho Dec 30 '18 at 18:45
  • $\begingroup$ Thanks! I was copy-pasting my own latex from above to make it easier :D $\endgroup$ – Nyssa Dec 30 '18 at 18:49
3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.