10

I cannot prove that:

$$ \int_0^1 \! \frac{x}{\operatorname{artanh} x} \, {\rm d}x = \frac{7 \, \zeta(3)}{\pi^2} $$

where $\operatorname{artanh}$ is the inverse hyperbolic tangent function, and $\zeta$ is the Rieman zeta function. This integral is equivalent to the following integrals:

$$ \int_1^\infty \! \frac{{\rm d}x}{x^3 \operatorname{arcosh}x} ,\quad \int_0^1 \! \frac{x}{\operatorname{arsech}x} \, {\rm d}x $$

Also, increasing the order of the integrand yields:

$$ \begin{align} \int_0^1 \! \frac{x^2}{\operatorname{artanh}^2 x} \, {\rm d}x &= \frac{124 \, \zeta(5)}{\pi^4} - \frac{14 \, \zeta(3)}{3\pi^2} \\[3pt] \int_0^1 \! \frac{x^3}{\operatorname{artanh}^3 x} \, {\rm d}x &= \frac{1905 \, \zeta(7)}{\pi^6} - \frac{124 \, \zeta(5)}{\pi^4} \\[3pt] \int_0^1 \! \frac{x^4}{\operatorname{artanh}^4 x} \, {\rm d}x &= \frac{28616 \, \zeta(9)}{\pi^8} - \frac{2540 \, \zeta(7)}{\pi^6} + \frac{124 \, \zeta(5)}{5\pi^4} \end{align} $$

Sorry for my bad English.

Lai
  • 20,421
Kei Tojo
  • 404
  • 2
    Your English is essentially perfect here – FShrike Aug 16 '23 at 13:45
  • 4
    Maybe there could be a way to transform the integral so that it's $\pi$ times the one in the following link? Evaluating integral $\int_0^1 \frac{x-x^2}{\sin \pi x} dx = \frac{7 \zeta(3)}{\pi^3}$ By the way, currently there's a discrepancy between your title and the formula at the beginning of your post, $\pi$ versus $\pi^2$. – Bruno B Aug 16 '23 at 13:52
  • 1
    Might be related to integrals like $\int_0^\infty t^{s-1}/(e^t+1)dt$, which involve $\zeta(s)$ and $\Gamma(s)$. – eyeballfrog Aug 16 '23 at 14:04
  • 2
    Minor nitpick on an otherwise interesting question: it would be nice to know whether you obtained this identity (or alleged identity) from an engine, reference, conjecture, etc. For example, If you conjectured it based on a numerical search and verified it to 100 places with Wolfram. – Jake Mirra Aug 16 '23 at 14:05
  • Hello @JakeMirra , I found these integrals in my high school math club journal. There was no proof given, so I tried to give one, but it was completely toothless. The journals only describe the results up to $$ \int_0^1 ! \frac{x^9}{\operatorname{artanh}^9 x} , {\rm d}x $$ and I verified that they all agree to 50 decimal places with using Wolfram. Sorry. – Kei Tojo Aug 16 '23 at 14:22
  • 1
    It’s worth noting that my method is an option. There are easier approaches, see the comments under my answer. Of course anyone who wants can and should write up more elementary answers. I’m just a fan of contour-bashing – FShrike Aug 17 '23 at 20:05
  • 1
    Towards a general formula: $$\int_{0}^{1}\frac{x^n}{\operatorname{artanh}^n(x)},\mathrm{d}x=\sum_{r=0}^{n}\binom{n+r}{r+1}\frac{(-1)^{(n-r)/2}}{\pi^{n+r}} \left(2^{n+r+1}-1\right)\zeta\left(n+r+1\right)\operatorname{Res}_{x=0}\left[x^{r+1}\coth^n(x)\operatorname{csch}^2(x)\right]$$ – KStarGamer Aug 23 '23 at 00:20

1 Answers1

10

$\newcommand{\d}{\,\mathrm{d}}\newcommand{\artanh}{\operatorname{artanh}}$We have: $$I=\int_0^1\frac{x}{\artanh x}\d x=\int_0^\infty\frac{\sinh t}{t\cdot\cosh^3t}\d t=\frac{1}{2}\int_\Bbb R\frac{\sinh t}{t\cdot\cosh^3t}\d t$$

It suffices to show: $$\int_{\Bbb R}\frac{\sinh t}{t\cdot\cosh^3t}\d t=\frac{14\zeta(3)}{\pi^2}$$

Which is amenable to contour integration. By multiplying with a regularising term $e^{i\cdot\delta z}$ and integrating around a box contour, the box part vanishes and we just collect the residues. By taking $\delta\to0^+$ we recover the original integral.

What are the poles? These will be whenever $\cosh z=0$ so they occur at $z=\frac{\pi i}{2}(2n+1)$ for $n\ge0$. The poles are all of third order; that means we have our work cut out for us. This method will also generalise to the second, third, fourth, etc. order integrands, it will just get extremely tedious.

The Taylor expansion of $\cosh$ at $\zeta_n:=\frac{\pi i}{2}(2n+1)$ will be: $i\cdot(-1)^n((z-\zeta_n)+(1/3!)(z-\zeta_n)^3+\cdots)$ so the Taylor expansion of $\cosh^3$ at $\zeta_n$ is $i\cdot(-1)^{n+1}((z-\zeta_n)^3+(1/2)(z-\zeta_n)^5+\cdots)$. Taking reciprocals: $$\frac{1}{\cosh^3z}=\frac{i(-1)^n}{(z-\zeta_n)^3}+\frac{i(-1)^{n+1}}{2(z-\zeta_n)}+o(1)$$In a neighbourhood of $\zeta_n$. The Taylor expansion of $\sinh$ at $\zeta_n$ will be $i(-1)^n(1+(1/2)(z-\zeta_n)^2+(1/4!)(z-\zeta_n)^4+\cdots)$. The Taylor expansion of $\frac{1}{z}$ at $\zeta_n$ is $\zeta_n^{-1}-\zeta_n^{-2}(z-\zeta_n)+\zeta_n^{-3}(z-\zeta_n)^2+\cdots$.

By multiplying series it follows the residue of the integrand at $\zeta_n$ is: $$i(-1)^n\cdot i(-1)^n(1/2)\cdot\zeta_n^{-1}+i(-1)^n\cdot i(-1)^n\cdot\zeta_n^{-3}+(i(-1)^{n+1}/2)\cdot i(-1)^n\cdot\zeta_n^{-1}\\=-\zeta_n^{-3}=\frac{8}{i\cdot\pi^3(2n+1)^3}$$So that: $$I=2\pi i\sum_{n=0}^\infty\frac{8}{i\cdot\pi^3(2n+1)^3}=\frac{16}{\pi^2}\sum_{n=0}^\infty\frac{1}{(2n+1)^3}$$The sum over the reciprocal of all odd cubes is $\zeta(3)-2^{-3}\zeta(3)=\frac{7}{8}\zeta(3)$. So we should have: $$\int_\Bbb R\frac{\sinh t}{t\cdot\cosh^3t}\d t=\frac{16\cdot7}{8\pi^2}\zeta(3)=\frac{14}{\pi^2}\zeta(3)$$As expected.

Rigorously, when the modifier $e^{i\cdot\delta z}$ is present, this has Taylor series $e^{i\cdot\delta\zeta_n}(1+i\delta(z-\zeta_n)+\cdots)$ and contributes to the residues a factor which $\to1$ as $\delta\to0^+$. As long as one is happy exchanging the limits with the series, we can ignore this in the residue calculation.

About the box contour. For $N\in\Bbb N$ large consider the box contour with vertices $\pm2\pi N,\pm2\pi N+2\pi N\cdot i$. On the lowest horizontal strip we get an approximation to the desired integral. On the upper horizontal strip the imaginary parts are fixed to $2\pi N\cdot i$; for $z=x+2\pi N\cdot i$, $\cosh(z)=\cosh(x)\cosh(2\pi N\cdot i)+\sinh(x)\sinh(2\pi N\cdot i)=\cosh(x)$ and $\sinh(z)=\sinh(x)\cosh(2\pi N\cdot i)+\cosh(x)\sinh(2\pi N\cdot i)=\sinh(x)$. Therefore, with the modified integrand $f(z)=\frac{\sinh z}{z\cdot\cosh^3z}e^{i\cdot\delta z}$, $z\cdot f(z)$ is bounded in modulus by $e^{-2\pi N\cdot\delta}$ on the upper horizontal. Thus the integral on the upper horizontal vanishes.

On the vertical segments $\cosh(z)$ is comparable in magnitude to $\cosh(2\pi N)$ and similarly for the $\sinh$ term; it follows the integral on the verticals vanishes.

FShrike
  • 40,125
  • 1
    After integration by parts twice we have \begin{equation} \int_{\mathbb{R}}\dfrac{\sinh(t)}{t\cosh^3(t)},\mathrm{d}t = \int_{\mathbb{R}}\dfrac{\sinh^2(t)}{2\cosh^2(t)}\cdot \dfrac{1}{t^2},\mathrm{d}t = \int_{\mathbb{R}}\dfrac{1-\cosh(t)}{2\cosh(t)}\cdot \dfrac{t\cosh(t)-2\sinh(t)}{t^3},\mathrm{d}t. \end{equation} Now all the poles $\frac{\pi i}{2}(2n+1)$ are simple. – JanG Aug 17 '23 at 14:29
  • 3
    @JanG Your second integral is likely the simplest as $\int_0^\infty\frac{\tanh^2t}{t^2},dt$ has a real-analytic evaluation using Feynman's trick repeatedly. Source from Quanto (unsurprisingly) – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Aug 17 '23 at 15:30
  • @TheSimpliFire Thank you for a good idea. However, why not skip all integrations by parts. Via the substitution $e^{2t}=s$ we get \begin{equation} \int_{\mathbb{R}}\dfrac{\sinh(t)}{t\cosh^3(t)},\mathrm{d}t = 4\int_{0}^{\infty}\dfrac{s-1}{(1+s)^3}\dfrac{ds}{\ln s} \end{equation} which is treated by Quanto. – JanG Aug 17 '23 at 19:50