Simplify/re-express a potential expression for the inverse of $\frac{\sinh(x)}x$

Question

$\DeclareMathOperator\M M \DeclareMathOperator\csch {csch}$

An attempt to invert the sinhc function $\frac{\sinh(x)}x$uses Mellin inversion. Define $f(x)$ as the inverse of $y=x\csch(x),0<y\le1$ and $0$ otherwise, graphed here. Therefore, $f(x)$’s Mellin transform is:

$$\M_t(f(t))=\int_0^1t^{s-1}f(t)dt=\int_0^\infty t(t\csch(t))^{s-1} d(t\csch(t))=-\frac1s\int_0^\infty (t\csch(t))^sdt$$

after substituting $f(t)\to t$ as well as integration by parts on $t$ and $(t\csch(t))^{s-1} d(t\csch(t))$ with $\left.\frac ts(t\csch(t))^s\right|_0^\infty=0$. The final integral appears in:

Compute integral of general form $ \int_0^\infty \left(\frac{x}{\sinh x}\right)^n d x $

We apply binomial series, converging on $e^x>1\iff 0<x$, to get:

$$\M_t(f(t))=-\frac{2^s}s\int_0^\infty\left(\frac t{e^t-e^{-t}}\right)^sdt=-\frac{2^s}s\sum_{n=0}^\infty\binom{-s}n (-1)^n\int_0^\infty t^s e^{-(2n+s)t}dt=-2^s\Gamma(s)\sum_{n=0}^\infty\binom{-s}n \frac{(-1)^n}{(2n+s)^{s+1}}$$

Expanding/simplifying the binomial and applying the inverse Mellin transform finally gives:

$$\boxed{\operatorname{sinhc}^{-1}(x)=\pm\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}(2x)^sg(s)ds;g(s)=2^{-s}\M_t(f(t))(s)=\sum_{n=0}^\infty\frac{\Gamma(n+s)}{(2n+s)^{s+1}n!},x\ge1}$$

shown here:

The result is accurate to $2$ decimal places if one truncates the integral/sum at $\pm130$. The boxed result is an integral of a sum which looks a bit cumbersome to use. However, it could possibly be a sum of a sum if Ramanujan master theorem, which also uses the Mellin transform, is applied. Another idea is a closed form for $\M_t(f(t))$. Both would make the boxed result cleaner.

If the boxed result it correct, what is a cleaner expression for the inverse of $\frac{\sinh(x)}x$ possibly using the Ramanujan master theorem or getting a closed form for $\M_t(f(t))$?

Answer 1

This isn't really an answer, but perhaps provides some insight with respect to a closed-form representation for $F(s)=\mathcal{M}_t[f(t)](s)$ where $f(t)$ is the inverse function of $x\, \text{csch}(x)$.

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The Mellin transform result can be expressed as

$$F(s)=\mathcal{M}_x[f(t)](s)=2^s\, \Gamma(s)\, G(s)\tag{1}$$

where

$$G(s)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=0}^N \binom{-s}{n}\, \frac{(-1)^n}{(2 n+s)^{s+1}}\right)\tag{2}.$$

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$G(s)$ seems to have a closed-form representation at non-zero integer values of $s$ as illustrated in the following table. I'm wondering if Mathematica understands $G(s)$ in terms other known functions at least when $G(s)$ is evaluated at non-zero integer values of $s$.

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$$\begin{array}{ccc} s & G(s)=\sum\limits_{n=0}^{\infty} \binom{-s}{n}\, \frac{(-1)^n}{(2 n+s)^{s+1}} & F(s)=2^s\, \Gamma (s)\, G(s) \\ -5 & 0 & - \\ -4 & 0 & - \\ -3 & 0 & - \\ -2 & 0 & - \\ -1 & 0 & - \\ 0 & - & - \\ 1 & \frac{\pi ^2}{8} & \frac{\pi ^2}{4} \\ 2 & \frac{\pi ^2}{48} & \frac{\pi ^2}{12} \\ 3 & -\frac{1}{768} \pi ^2 \left(\pi ^2-12\right) & -\frac{1}{48} \pi ^2 \left(\pi ^2-12\right) \\ 4 & -\frac{\pi ^2 \left(\pi ^2-15\right)}{17280} & -\frac{1}{180} \pi ^2 \left(\pi ^2-15\right) \\ 5 & \frac{\pi ^2 \left(120-100 \pi ^2+9 \pi ^4\right)}{368640} & \frac{1}{480} \pi ^2 \left(120-100 \pi ^2+9 \pi ^4\right) \\ 6 & \frac{\pi ^2 \left(315-105 \pi ^2+8 \pi ^4\right)}{29030400} & \frac{\pi ^2 \left(315-105 \pi ^2+8 \pi ^4\right)}{3780} \\ 7 & \frac{6720 \pi ^2-19600 \pi ^4+14504 \pi ^6-1275 \pi ^8}{2477260800} & \frac{6720 \pi ^2-19600 \pi ^4+14504 \pi ^6-1275 \pi ^8}{26880} \\ 8 & \frac{1575 \pi ^2-1470 \pi ^4+490 \pi ^6-36 \pi ^8}{24385536000} & \frac{1575 \pi ^2-1470 \pi ^4+490 \pi ^6-36 \pi ^8}{18900} \\ 9 & \frac{\pi ^2 \left(40320-282240 \pi ^2+663264 \pi ^4-439144 \pi ^6+37975 \pi ^8\right)}{3329438515200} & \frac{\pi ^2 \left(40320-282240 \pi ^2+663264 \pi ^4-439144 \pi ^6+37975 \pi ^8\right)}{161280} \\ 10 & \frac{\pi ^2 \left(3465-6930 \pi ^2+6006 \pi ^4-1804 \pi ^6+128 \pi ^8\right)}{15450675609600} & \frac{\pi ^2 \left(3465-6930 \pi ^2+6006 \pi ^4-1804 \pi ^6+128 \pi ^8\right)}{41580} \\ \end{array}$$

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Its worth noting a couple of points about the results in the table above.

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First, when evaluated at positive integers $G(s)$ seems to be of the form

$$G(s)=\frac{1}{b_s} \sum\limits_{k=1}^{\left\lceil \frac{s}{2}\right\rceil} a_{s,k}\, \pi^{2 k}\tag{3}$$

where there's an extra term introduced in the numerator at every successive odd positive integer which is perhaps a clue.

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Second, when $s$ is a negative integer its sufficient to evaluate the sum in formula (2) above using the upper evaluation limit $N=|s|$ since for the term of $G(s)$ one has

$$\binom{-s}{n}\, \frac{(-1)^n}{(2 n+s)^{s+1}}=0\,,\quad s\in \mathbb{Z}\land s<0\land n\in \mathbb{Z}\land n>|s|\tag{4}$$

which is analogous to how some globally convergent series for the Riemann zeta function $\zeta(s)$, the Dirichlet eta function $\eta(s)$, and formulas (5c) and (5d) below for $H_a(s)$ defined in formula (5a) below (which is more closely related to this question) converge at finite evaluation limits at negative integer values of $s$. In formula (5a) below $\zeta(s,a)$ is the Hurwitz zeta function.

$$H_a(s)=4^{-s-1} \left(\zeta\left(s+1,\frac{a}{4}\right)-\zeta\left(s+1,\frac{a+2}{4}\right)\right)\tag{5a}$$

$$H_a(s)=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=0}^K \frac{(-1)^n}{(2 n+a)^{s+1}}\right),\quad\Re(s)>0\tag{5b}$$

$$H_a(s)=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{n=0}^K \frac{1}{2^{n+1}} \sum\limits_{k=0}^n \binom{n}{k}\frac{(-1)^k}{(2 k+a)^{s+1}}\right)\tag{5c}$$

$$H_a(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^{K+1}} \sum\limits_{n=0}^K \frac{(-1)^n}{(2 n+a)^{s+1}} \sum\limits_{k=0}^{K-n} \binom{K+1}{K-k-n}\right)\tag{5d}$$

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Ignoring the binomial term $\binom{-s}{n}$ in formula (2) above for $G(s)$ one has

$$\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{(2 n+s)^{s+1}}=4^{-s-1} \left(\zeta \left(s+1,\frac{s}{4}\right)-\zeta \left(s+1,\frac{s+2}{4}\right)\right)\tag{6}$$

which is equivalent to formulas (5a) and (5b) above for $H_a(s)$ evaluated at $a=s$, but I'm not sure how to account for the binomial term $\binom{-s}{n}$ in formula (2) above. I believe formulas (5c) and (5d) for $H_a(s)$ above are globally convergent for the case $a=s$ as well as when $a$ is a constant.

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Figures (1) and (2) below illustrate the real and imaginary parts of formula (2) for $G(s)$ above evaluated at $N=100$ and $N=1000$ in blue and orange respectively. Note $G(s)$ seems to have a pole at $s=0$, converge to zero as $s\to\infty$, and typically has both imaginary and real components when evaluated for $s<0$, but seems to evaluate to zero at negative integer values of $s$ consistent with the table above. I'm not sure if formula (2) for $G(s)$ above is globally convergent, but it if is evaluations seem to indicate it converges more slowly for $\Re(s)<0$ than for $\Re(s)>0$.

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Figure (1): Illustration of real part of formula (2) for $G(s)$ evaluated at $N=100$ and $N=1000$ in blue and orange respectively

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Figure (2): Illustration of imaginary part of formula (2) for $G(s)$ evaluated at $N=100$ and $N=1000$ in blue and orange respectively

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I noticed splitting the evaluation of $G(s)$ into two steps including

$$H(z)=\sum\limits_{n=0}^{\infty} \binom{-z}{n} \frac{(-1)^n}{(2 n+z)^{s+1}}\tag{7}$$

and

$$G(s)=\underset{z\to s}{\text{lim}}\, H(z)\tag{8}$$

leads to a functional representation of $H(z)$ at positive integer values of $s$ (and hence $G(s)$ in a limit sense) as illustrated in the following table.

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$$\begin{array}{cccc} s & G(s)=\sum\limits_{n=0}^{\infty} \binom{-s}{n} \frac{(-1)^n}{(2 n+s)^{s+1}} & G(s)=\underset{z\to s}{\text{lim}}\, H(z) & H(z)=\sum\limits_{n=0}^{\infty} \binom{-z}{n} \frac{(-1)^n}{(2 n+z)^{s+1}} \\ 1 & \frac{\pi ^2}{8} & \frac{\pi ^2}{8} & \sqrt{\pi } 2^{-z-2} \cot \left(\frac{\pi z}{2}\right) \Gamma \left(\frac{1}{2}-\frac{z}{2}\right) \Gamma \left(\frac{z}{2}\right) \\ 2 & \frac{\pi ^2}{48} & \frac{\pi ^2}{48} & \frac{2^{-z-4} \Gamma \left(\frac{1}{2}-\frac{z}{2}\right) \Gamma \left(\frac{z}{2}\right) \left(\pi ^2 \cot ^2\left(\frac{\pi z}{2}\right)+\psi ^{(1)}\left(\frac{z}{2}\right)-\psi ^{(1)}\left(1-\frac{z}{2}\right)\right)}{\sqrt{\pi }} \\ 3 & -\frac{1}{768} \pi ^2 \left(\pi ^2-12\right) & -\frac{1}{768} \pi ^2 \left(\pi ^2-12\right) & \frac{1}{3} \sqrt{\pi } 2^{-z-5} \Gamma \left(\frac{1}{2}-\frac{z}{2}\right) \Gamma \left(\frac{z}{2}\right) \left(3 \cot \left(\frac{\pi z}{2}\right) \left(\psi ^{(1)}\left(\frac{z}{2}\right)-\psi ^{(1)}\left(1-\frac{z}{2}\right)\right)+\pi ^2 \left(\cot ^3\left(\frac{\pi z}{2}\right)+\sin (\pi z) \csc ^4\left(\frac{\pi z}{2}\right)\right)\right) \\ 4 & -\frac{\pi ^2 \left(\pi ^2-15\right)}{17280} & -\frac{\pi ^2 \left(\pi ^2-15\right)}{17280} & \frac{2^{-z-8} \Gamma \left(\frac{1}{2}-\frac{z}{2}\right) \Gamma \left(\frac{z}{2}\right) \left(\psi ^{(0)}\left(\frac{z}{2}\right)^4+\psi ^{(0)}\left(1-\frac{z}{2}\right)^4+3 \left(\psi ^{(1)}\left(1-\frac{z}{2}\right)-\psi ^{(1)}\left(\frac{z}{2}\right)\right)^2-4 \psi ^{(0)}\left(1-\frac{z}{2}\right)^3 \psi ^{(0)}\left(\frac{z}{2}\right)+6 \psi ^{(0)}\left(\frac{z}{2}\right)^2 \left(\psi ^{(1)}\left(\frac{z}{2}\right)-\psi ^{(1)}\left(1-\frac{z}{2}\right)\right)+6 \psi ^{(0)}\left(1-\frac{z}{2}\right)^2 \left(\psi ^{(0)}\left(\frac{z}{2}\right)^2+\psi ^{(1)}\left(\frac{z}{2}\right)-\psi ^{(1)}\left(1-\frac{z}{2}\right)\right)-4 \psi ^{(0)}\left(1-\frac{z}{2}\right) \left(\psi ^{(0)}\left(\frac{z}{2}\right)^3+3 \left(\psi ^{(1)}\left(\frac{z}{2}\right)-\psi ^{(1)}\left(1-\frac{z}{2}\right)\right) \psi ^{(0)}\left(\frac{z}{2}\right)+\psi ^{(2)}\left(\frac{z}{2}\right)-\psi ^{(2)}\left(1-\frac{z}{2}\right)\right)+\psi ^{(3)}\left(\frac{z}{2}\right)-\psi ^{(3)}\left(1-\frac{z}{2}\right)-4 \pi ^3 \sin (\pi z) \csc ^4\left(\frac{\pi z}{2}\right) \psi ^{(0)}\left(\frac{z}{2}\right)\right)}{3 \sqrt{\pi }} \\ \end{array}$$

Answer 2

EDIT 1/2:

An attempt made here is by using simple calculus.

A simple parabola function and its inverse function in order to outline the steps is included.

$$y= 1+x^2,~ y'= 2x,~\text{first order derivative reciprocal and }$$

by variables swap $ y'\to 1/y',~ x\to y $

$$ \frac{1}{y'}=2y~; \frac {dx}{dy}=2y~, $$

Relabeling for inverse function,

$$ \frac {dx1}{dy1}=2y1~;$$

Integrate

$$ x1=y1^2 + c,$$

with Boundary conditions $ (x1=1, y1=0 \to c=1~) $

$$x1=y1^2+1~, y1=\sqrt{x1-1}~$$

The inverse function is gray, parabola symmetrical to the x-axis, inverse functions always reflect about straight line $x=y.$

We adopt exactly the same procedure for obtaining the inverse function of the given (red curve) function.

$$ y=\frac{\sinh x }{x} $$

Differentiate to find ode

$$\frac{dy}{dx} =\left(\frac{x \cosh x-\sinh x}{x^2}\right); $$

Interchanging variables

$$\frac{dx}{dy} =\left(\frac{y \cosh y-\sinh y}{y^2}\right); $$

Relabel for inverse function clear identification:

$$\frac{dx1}{dy1} =\left(\frac{y1 \cosh y1-\sinh y1}{y1^2}\right); $$

The RHS is seen to be (by inspection) directly the full derivative of $\left(\dfrac{\sinh y1}{y1}\right) $ with respect to $y1$ by the Quotient Rule of differentiation

$$ \frac{dx1}{dy1}=d\left(\frac{\sinh y1}{y1} \right)/dy1 ; $$

Integrating

$$ x1=\frac{\sinh y1}{y1} + C_1 $$

Boundary condition $x1=1,y1=0\to C_1=0,$ using L'Hospital's Rule

$$ \boxed{x1=\left(\frac{\sinh y1}{y1}\right) } $$

which is the equation of the sketched (green ) inverted function curve in closed form, reflected about the straight line $x=y$.