The Laplace Limit Constant $\lambda$ is well know constant which is the $y$ value of the global extrema of:
$$x\,\text{sech}(x):$$
Therefore:
$$x=\max(x\,\text{sech}(x))=-\min(x\,\text{sech}(x))=1.19967…\implies y= 1.19967… \text{sech}(1.19967… )=\lambda= 0.6627434193491815809747420971092…$$
Similarly, the Laplace Limit is also the radius of convergence of the Lagrange Inversion Theorem solution for Kepler’s Equation with the following expansion. Note there are other interpretations of the constant:
$$E-\epsilon\sin(E)=M\implies E = M \;+\; \sum_{n=1}^{+\infty} \frac{\varepsilon^n}{2^{n-1}\,n!} \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k\,\binom{n}{k}\,(n-2k)^{n-1}\,\sin((n-2k)\,M)$$
converging if
$$\epsilon<\lambda$$
It can be shown that:
$$\lambda=x:\frac{xe^\sqrt{x^2+1}}{\sqrt{x^2+1}+1}=1$$
“Closed Form $1$”:
Now define $\sqrt{x^2+1}=u$:
$$e^{2u}=\frac{u+1}{u-1}\implies u=\frac12\ln\left(\frac{u+1}{u-1}\right)=\coth^{-1}(u)\implies \coth(\sqrt{x^2+1})=\sqrt{x^2+1}$$
so the Laplace Limit is a transformed fixed point of the hyperbolic cotangent. Note that:
$$\coth(u)=u\iff\cot(iu)=-iu\mathop\implies^{iu=v}\cot(v)+v=0\iff-\sqrt{\frac2{\pi v}}\left(\frac{\cos(v)}v+\sin(v)\right)=\text Y_\frac32(v)=0$$
Where appears the BesselY function. Now use the Bessel Y Zeroes function $\text y_{n,k}\implies \text Y_n(\text y_{n,k})=0$ where $k\in\Bbb N$ gives the $k$th largest root even though Wolfram Alpha sometimes evaluates the function with real $k$:
$$\text Y_\frac32(v)=0 \implies \boxed{x=\lambda,\pm i\sqrt{\text y_{\frac32,k\in\Bbb N}^2+1}}$$
Which are solutions. Even though we have infinite solutions for $\frac{xe^\sqrt{x^2+1}}{\sqrt{x^2+1}+1}=1$, they are all complex and not the real solution for the equation, which would be the Laplace Limit.
Attempt 2:
Take the Regularized Beta function:
$$\boxed{\frac{xe^{\sqrt{x^2+1}}}{\sqrt{x^2+1}+1}=1\iff\frac{\pi i}2\left(\text I_{\cosh^2(\sqrt{x^2+1})}\left(\frac32,-\frac12\right)-1\right)=0}$$
Here is the experimental result using the Inverse Beta Regularized Series $\text I^{-1}_z(a,b)$ from Wolfram functions which extends the domain of the function to many values.
$$\frac{\pi i}2\left(\text I_{\cosh^2(\sqrt{x^2+1})}\left(\frac32,-\frac12\right)-1\right)=0 \implies x=\pm\sqrt{\cosh^{-1}\left(\pm\sqrt{\text I^{-1}_1\left(\frac32,-\frac12\right)}\right)^2-1}$$
$$\text I^{-1}_z(a,b)\mathop=^{a>0}(az\text B(a,b))^\frac1a+\frac{b-1}{a+1} (az\text B(a,b))^\frac 2a+ \frac{(b-1)(a^2+3ab-a+5b-4}{2(a+1)^2(a+2)}(az\text B(a,b))^\frac 3a +…$$
However, using $4$ of the terms from the series gives the wrong number for various attempts, so we cannot use the $\text I^{-1}_z(a,b)$ function nor its expansion to get a solution. It is still possible to use this technique, but it has not worked yet.
Conclusion:
However, neither of these solutions gives a closed form without stretching the definition of a function or finding a different value for the Laplace Limit equation $\frac{xe^\sqrt{x^2+1}}{\sqrt{x^2+1}+1}=1$. Is there any closed form for the Laplace Limit using library functions, those implemented in softwares like SciPy or Wolfram Functions among others, besides “Laplace Limit=$\lambda$? Please correct me and give me feedback!
Explicit series solution attempt:
Using @Claude Leibovici’s reference we get a possible sum representation for $\text{coth}(x_*)=x_*=\sqrt{1-\lambda^2}\implies \lambda=\sqrt{x_*^2+1}$: for some radius of convergence. See page $3$ of the link:
$$e^x\frac{x-t}{x-s}=a\implies x= x=t+(s-t)\sum_{n=1}^\infty \frac{\text L_{n-1}^1((t-s)n)a^n}{ne^{nt}} = x=t+(s-t)\sum_{n=1}^\infty \,_1\text F_1(1-n;2; (t-s)n ) e^{-nt} a^n =t-\sum_{n=1}^\infty\sum_{k=1}^\infty \binom nk\frac{(ae^{-t})^n(n(s-t))^k}{n^2(k-1)!}$$
Therefore we should find that $x\to 2x,s=-2,t=2,a=1$:
$$\boxed{\text{coth}(x)=x\iff e^{2x}\frac{2x-2}{2x+2}=1\mathop\implies x= 1+2\sum_{n=1}^\infty \frac{\text L_{n-1}^1(4n)}{ne^{2n}}=1+2\sum_{n=1}^\infty \,_1\text F_1(1-n;2; 4n ) e^{-2n}=2-\sum_{n=1}^\infty\sum_{k=1}^\infty \binom nk\frac{4^k n^{k-2}}{(k-1)! e^{2n}} }$$
where the series converges, but Wolfram Alpha mistakenly says that it diverges. Note the [Confluent Hypergeomtric function](The Laplace Limit Constant $\lambda$ is well know constant which is the $y$ value of the global extrema of: $$x\,\text{sech}(x):$$ Therefore: $$x=\max(x\,\text{sech}(x))=-\min(x\,\text{sech}(x))=1.19967…\implies y= 1.19967… \text{sech}(1.19967… )=\lambda= 0.6627434193491815809747420971092…$$ Similarly, the Laplace Limit is also the radius of convergence of the Lagrange Inversion Theorem solution for Kepler’s Equation with the following expansion. Note there are other interpretations of the constant: $$E-\epsilon\sin(E)=M\implies E = M \;+\; \sum_{n=1}^{+\infty} \frac{\varepsilon^n}{2^{n-1}\,n!} \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k\,\binom{n}{k}\,(n-2k)^{n-1}\,\sin((n-2k)\,M)$$ converging if $$\epsilon<\lambda$$ It can be shown that: $$\lambda=x:\frac{xe^\sqrt{x^2+1}}{\sqrt{x^2+1}+1}=1$$ “Closed Form $1$”: Now define $\sqrt{x^2+1}=u$: $$e^{2u}=\frac{u+1}{u-1}\implies u=\frac12\ln\left(\frac{u+1}{u-1}\right)=\coth^{-1}(u)\implies \coth(\sqrt{x^2+1})=\sqrt{x^2+1}$$ so the Laplace Limit is a transformed fixed point of the hyperbolic cotangent. Note that: $$\coth(u)=u\iff\cot(iu)=-iu\mathop\implies^{iu=v}\cot(v)+v=0\iff-\sqrt{\frac2{\pi v}}\left(\frac{\cos(v)}v+\sin(v)\right)=\text Y_\frac32(v)=0$$ Where appears the BesselY function. Now use the Bessel Y Zeroes function $\text y_{n,k}\implies \text Y_n(\text y_{n,k})=0$ where $k\in\Bbb N$ gives the $k$th largest root even though Wolfram Alpha sometimes evaluates the function with real $k$: $$\text Y_\frac32(v)=0 \implies \boxed{x=\lambda,\pm i\sqrt{\text y_{\frac32,k\in\Bbb N}^2+1}}$$ Which are solutions. Even though we have infinite solutions for $\frac{xe^\sqrt{x^2+1}}{\sqrt{x^2+1}+1}=1$, they are all complex and not the real solution for the equation, which would be the Laplace Limit. Attempt 2: Take the Regularized Beta function: $$\boxed{\frac{xe^{\sqrt{x^2+1}}}{\sqrt{x^2+1}+1}=1\iff\frac{\pi i}2\left(\text I_{\cosh^2(\sqrt{x^2+1})}\left(\frac32,-\frac12\right)-1\right)=0}$$ Here is the experimental result using the Inverse Beta Regularized Series $\text I^{-1}_z(a,b)$ from Wolfram functions which extends the domain of the function to many values. $$\frac{\pi i}2\left(\text I_{\cosh^2(\sqrt{x^2+1})}\left(\frac32,-\frac12\right)-1\right)=0 \implies x=\pm\sqrt{\cosh^{-1}\left(\pm\sqrt{\text I^{-1}_1\left(\frac32,-\frac12\right)}\right)^2-1}$$ $$\text I^{-1}_z(a,b)\mathop=^{a>0}(az\text B(a,b))^\frac1a+\frac{b-1}{a+1} (az\text B(a,b))^\frac 2a+ \frac{(b-1)(a^2+3ab-a+5b-4}{2(a+1)^2(a+2)}(az\text B(a,b))^\frac 3a +…$$ However, using $4$ of the terms from the series gives the wrong number for various attempts, so we cannot use the $\text I^{-1}_z(a,b)$ function nor its expansion to get a solution. It is still possible to use this technique, but it has not worked yet. Conclusion: However, neither of these solutions gives a closed form without stretching the definition of a function or finding a different value for the Laplace Limit equation $\frac{xe^\sqrt{x^2+1}}{\sqrt{x^2+1}+1}=1$. Is there any closed form for the Laplace Limit using library functions, those implemented in softwares like SciPy or Wolfram Functions among others, besides “Laplace Limit=$\lambda$? Please correct me and give me feedback! Explicit series solution attempt: Using @Claude Leibovici’s reference we get a possible sum representation for $\text{coth}(x_*)=x_*=\sqrt{1-\lambda^2}\implies \lambda=\sqrt{x_*^2+1}$: for some radius of convergence. See page $3$ of the link: >$$e^x\frac{x-t}{x-s}=a\implies x= x=t+(s-t)\sum_{n=1}^\infty \frac{\text L_{n-1}^1((t-s)n)a^n}{ne^{nt}} = x=t+(s-t)\sum_{n=1}^\infty \,_1\text F_1(1-n;2; (t-s)n ) e^{-nt} a^n =t-\sum_{n=1}^\infty\sum_{k=1}^\infty \binom nk\frac{(ae^{-t})^n(n(s-t))^k}{n^2(k-1)!}$$ Therefore we should find that $x\to 2x,s=-2,t=2,a=1$: $$\boxed{\text{coth}(x)=x\iff e^{2x}\frac{2x-2}{2x+2}=1\mathop\implies x= 1+2\sum_{n=1}^\infty \frac{\text L_{n-1}^1(4n)}{ne^{2n}}=1+2\sum_{n=1}^\infty \,_1\text F_1(1-n;2; 4n ) e^{-2n}=2-\sum_{n=1}^\infty\sum_{k=1}^\infty \binom nk\frac{4^k n^{k-2}}{(k-1)! e^{2n}} }$$ where the series converges, but Wolfram Alpha mistakenly says that it diverges. Here is the real solution, the Laplace Limit: with rewritten
Here is the real solution, the Laplace Limit:
$$\boxed{\lambda=\left(1+2\sum_{n=1}^\infty \,_1\text F_1(1-n;2; 4n ) e^{-2n}\right)\text{sech}\left(1+2\sum_{n=1}^\infty \,_1\text F_1(1-n;2; 4n ) e^{-2n}\right) =\sqrt{\left(1+2\sum_{n=1}^\infty \frac{\text L_{n-1}^1(4n)}{ne^{2n}} \right)^2-1}}$$
The squared sums could be rewritten. Maybe we can simplify?
The function $f(x)=x\, \text{sech}(x)$ is, as you wrote, maximum when $$(x-1)\,e^x=(x+1)\,e^{-x} \implies e^{-2x}=\frac{x-1}{x+1}$$ So, more than likely, the only solution is given in terms of the generalized Lambert function (have a look at equation $(4)$ in the linked paper). So, from a formal point of view, there is a closed form for this constant.
In fact, it looks to me better to write that
$$\lambda=\sqrt{x^2_*-1}\quad \text{where} \quad x_*\tanh(x_*)-1=0$$ Using series expansion around $x=a$ and then series reversion, we have $$x_*=a + t +\frac{1-a b-b^2+a b^3 }{-a-b+a b^2 }t^2+$$ $$\frac{\left(a^2+6\right)-8 a b+\left(a^2-9\right) b^2+14 a b^3+\left(3-5 a^2\right) b^4-6 a b^5+3 a^2 b^6 }{3(-a-b+a b^2)^3 }t^3+O(t^4)$$ where $b=\tanh(a)$ and $t=\frac{1-ab}{a+b-ab^2}$.
Using this very truncated expansion and $a=2W(1)$, we have
$$x_*=1.199680\cdots$$ while the "exact" value is $x_*= 1.199678\cdots$
This approximation would give $\lambda\sim 0.662747$ while the exact value is $0.662743$
Update
Around the solution, the function which is the closest to linearity is $$h(x)=1- x\tanh(x)$$ So, a recursive definition of its solution $x_*$ could be $$x_{n+1}=\frac{2 x_n^2+\cosh (2 x_n)+1}{2 x_n+\sinh (2 x_n)}\qquad \text{with}\qquad x_0=2\Omega$$
$$\left( \begin{array}{cc} n & x_n \\ 0 & 1.13428658082 \\ 1 & 1.19974041144 \\ 2 & 1.19967864026 \end{array} \right)$$
@AaronHendrickson's contribution: As pointed out by @Claude, we want the value $x_\ast$ solving $$ 1=e^{2x_\ast}\frac{x_\ast-1}{x_\ast+1}. $$ For convenience we can make the substitution $z=2x_\ast$ so that we instead seek a solution $x_\ast=z/2$ with $z$ solving $$ 1=e^{z}\frac{z-2}{z+2}. $$ According to the linked arXiv article we have by definition $$ x_\ast=\frac{1}{2}W\left({2\atop -2};1\right), $$ with $W({\mathbf t\atop\mathbf s};z)$ being the generalized Lambert function. Employing the series expansion in Theorem 4 of the linked article (using $a=1$, $t=2$, $s=-2$, and $T=4$) as well as the relation $-L^\prime_n(w)=L_{n-1}^{(1)}(w)$ gives $$ x_\ast=1+2\sum_{n=1}^\infty\frac{L_{n-1}^{(1)}(4n)}{n}e^{-2n}. $$ This series can then be implemented in Mathematica as follows: