Conjecture: $\lim\limits_{x\to\infty}\operatorname{Re}\text W_x(x)\mathop=\limits^?-\ln(2\pi)$

Question

The inspiration for the question is

Closed form of $$\frac{d}{dk}\text W_k(z)$$ Derivative of W-Lambert function with respect to its branch cuts experiment.

I also like making functions central to my work. So I though about “centralizing” the Generalized W-Lambert function so that the branch cut is the same as the argument:

$$\text W_k(z)=\text W_x(x)$$

I then considered what the value of $$\text W_{\pm \infty}(\pm \infty)$$ would be. Assume that the signs can be chosen in any order. I found a conjectured closed form. For simplicity, let’s consider the convergent real part of the expression. Let’s also further constrain our problem by choosing signs to be

$$\text{Re}(\text W_{ \infty}( \infty))$$

$$\lim_{x\to\pm \infty}\text {Re}(\text W_x(x))\mathop=^{\large ?} -\ln(2\pi)$$

The imaginary part turn out to be asymptotic to $$\pm 2\pi i \infty=\pm\infty i$$ so it is just an infinite complex number.

I found the possible result after using a discrete limit. For example, the value at $x=10^{10}$ is the following. I used the the real an imaginary part as well as different signs for completion.

$$\text W_x(x)= \text W_x(-x)= \text W_{x}(\pm ix) =-1.8378770663843454835603092354803385245074740939... + 6.2831853070225068442428720324366974719089270342... × 10^{10} i, \text W_{-x}(-x)= \text W_{-x}(x)=\text W_{-x}(\pm ix)=-1.8378770663343454835578092354801887222706941730... - 6.2831853067083475788838927085903664574425628653... × 10^{10} i $$

Note the $2\pi$ approximation in the scientific notation if the imaginary part. As said before, the real part is approximately:

$$-\ln(2\pi)= -1.8378770663843454835603092354803385245074740939... $$

Here are the Inverse Symbolic Calculator results.

How can you formally evaluate $$\text{Re}\lim_{x\to \infty} \text W_x(x)$$ using a discrete limit? Please correct me and give me feedback!

Just for fun, the Wright Omega function can also have the interesting identity that:

$$ω(z)\mathop=^\text{def} \text W_{\left \lceil\frac{\text{Im}(z)}{2\pi}-\frac12\right\rceil}\left(e^z\right)\implies -\ln(2\pi)=\lim_{x\to\infty} ω(\ln(x)+2i\pi x)$$

Where $$\left\lceil\frac{\text{Im}(z)}{2\pi}-\frac12\right\rceil =\frac{i\left(\ln\left(e^z\right)-z\right)}{2\pi}=\text{unwindK(z)}=\text K(z)$$

is called the Unwinding Number

Answer 1

$\underline{\text{A Reformulation}:-}$

I think we can use one of the asymptotic series expansion given in wolfram alpha:

$\textstyle\displaystyle{\Re W_x(x)=\Re\left(2x\pi i+\ln(x)-\ln(2x\pi i+\ln(x))-\sum_{n=1}^{\infty}(2x\pi i+\ln(x))^{-n}\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\ln^m(2x\pi i+\ln(x))\right)}$

For, $\ln(2x\pi i+\ln(x))$ we can write the expression inside the logarithm in the polar form:

$\ln(re^{i\theta})$ where $r=\sqrt{4x^2\pi^2+\ln^2(x)}$ and since $x>0$ we have $\theta=\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)$

So, $\ln(2x\pi i+\ln(x))=\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)$

And, $\textstyle\displaystyle{(2x\pi i+\ln(x))^{-n}=\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n}$

So, $\textstyle\displaystyle{\Re W_x(x)=\ln(x)-\frac{1}{2}\ln(4x^2\pi^2-\ln^2(x))-\Re\left(\sum_{n=1}^{\infty}\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\right)}$

$\textstyle\displaystyle{\lim_{x\rightarrow\infty}\Re W_x(x)=\ln\left(\lim_{x\rightarrow\infty}\frac{x}{\sqrt{4x^2\pi^2-\ln^2(x)}}\right)-\lim_{x\rightarrow\infty}\left(\Re\left(\sum_{n=1}^{\infty}\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\right)\right)}$

Let's do the limit inside the natural log first-

Let, $\textstyle\displaystyle{L=\lim_{x\rightarrow\infty}\frac{x}{\sqrt{4x^2\pi^2+\ln^2(x)}}}$

$\textstyle\displaystyle{L^2=\lim_{x\rightarrow\infty}\frac{x^2}{4x^2\pi^2+\ln^2(x)}}$

$\textstyle\displaystyle{\overset{\text{L.H.}\frac{\infty}{\infty}}{=}\lim_{x\rightarrow\infty}\frac{2x}{8x\pi^2+\frac{2}{x}\ln(x)}}$

$\textstyle\displaystyle{=\lim_{x\rightarrow\infty}\frac{x^2}{4x^2\pi^2+\ln(x)}}$

$\textstyle\displaystyle{\overset{\text{L.H.}\frac{\infty}{\infty}}{=}\lim_{x\rightarrow\infty}\frac{2x}{8x\pi^2+\frac{1}{x}}}$

$\textstyle\displaystyle{=\lim_{x\rightarrow\infty}\frac{2x}{8x\pi^2}}$

$\textstyle\displaystyle{=\frac{1}{2\pi}}$

$\implies\ln(L)=-\ln(2\pi)$

Which leaves us with the conclusion that your conjecture will be true if:

$\textstyle\displaystyle{\lim_{x\rightarrow\infty}\left(\sum_{n=1}^{\infty}\Re\bigg[\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\bigg]\right)=0}$

---

$\underline{\text{The Proof}:-}$

From this post we have that-

Assume $\sum_m f(m,1)>-\infty$. If for each $m$ we have $f(m,n)\le f(m,n')$ whenever $n\le n'$, then $$\lim_n\sum_{m=1}^\infty f(m,n) = \sum_{m=1}^\infty\lim_n f(m,n).\tag1$$

Define, $\textstyle\displaystyle{f(n,x):=\Re\bigg[\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\bigg]}$

To apply the theorem we see that the function is not defined at $1$ because of the $\ln(x)$ in the denominator inside the inverse tangent, but we can define it in terms of a limit-

$f(n,1):=\lim_{x\rightarrow 1}f(n,x)$

$\textstyle\displaystyle{=\Re\lim_{x\rightarrow 1}\bigg[\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\bigg]}$

$\textstyle\displaystyle{=\Re\left[\lim_{x\rightarrow 1}\bigg[\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n\bigg]\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\lim_{x\rightarrow 1}\bigg[\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\bigg]\right]}$

$\textstyle\displaystyle{=\Re\left[0×\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\left(\ln(2\pi)+i\frac{\pi}{2}\right)^m\right]}$

$\implies f(n,1)=0$

$\implies\sum_{n}f(n,1)=0>-\infty$

For the next condition I am not sure is the following reasoning correct or not, I would request the readers to correct me if I am wrong:

Since we are taking the discrete limit we have

$f(n,x)\leq f(n,x+1)$

$f(n,x+1)-f(n,x)\geq 0$

Since $x\rightarrow\infty$, we have-

$\textstyle\displaystyle{\lim_{x\rightarrow\infty}(f(n,x+1)-f(n,x))\geq 0}$

Notice that $f(n,x+1)\sim f(n,x)$ as $x\rightarrow\infty$, so

$\textstyle\displaystyle{\lim_{x\rightarrow\infty}(f(n,x+1)-f(n,x))\geq 0}$

$0\geq 0$

Which is trivially true.

So we can apply the theorem meaning that we can interchange the sum and the limit giving us-

$\textstyle\displaystyle{\lim_{x\rightarrow\infty}\left(\sum_{n=1}^{\infty}\Re\bigg[\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\bigg]\right)}$

$\textstyle\displaystyle{=\sum_{n=1}^{\infty}\sum_{m=1}^{n}\frac{(-1)^n}{m!}S_n^{(1+n-m)}\lim_{x\rightarrow\infty}\left(\Re\bigg[\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\bigg]\right)}$

Notice that $n\geq m$, so for $m=n$ we see that the limit is $0$ from Wolfram Alpha. And for $n>m$ we can write $n=m+k$ so we have-

$\textstyle\displaystyle{\lim_{x\rightarrow\infty}\left(\Re\bigg[\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^n\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\bigg]\right)}$

$\textstyle\displaystyle{=\Re\bigg[\left(\lim_{x\rightarrow\infty}\left(\frac{\ln(x)-2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^k\right)\left(\lim_{x\rightarrow\infty}\left(\frac{\ln(x)+2x\pi i}{4x^2\pi^2+\ln^2(x)}\right)^m\left(\frac{1}{2}\ln(4x^2\pi^2+\ln^2(x))+i\tan^{-1}\left(\frac{2x\pi}{\ln(x)}\right)\right)^m\right)\bigg]}$

Which is $0$ by this and this.

Which proves our statement.

$\underline{\text{Conclusion}:-}$

$\boxed{\textstyle\displaystyle{\lim_{x\rightarrow\infty}\Re\operatorname{W}_x(x)=-\ln(2\pi)}}$

$\underline{\text{Comment}:-}$

This was a really interesting equality. I never expected the branch cuts of the Lambert W Function to be this interesting(have a connection with $\pi$), that's why I never really looked into the branches of the productlog. And remember-

Every Time You're Not Running $\pi$ Is Getting closer.

Answer 2

This is not an answer since just based on numerical calculations.

Using $x=10^k$, define $$R_k=10^{k+1} \left(1+\frac{\Re\left(W_{10^k}\left(10^k\right)\right)}{\log (2 \pi )}\right)$$ $$I_k=10^{k+1}\left(10^{k+1} \left(1-\frac{\Im\left(W_{10^k}\left(10^k\right)\right)}{ 2\pi 10^k}\right)-\frac{5}{2}\right)$$ Some results for $0 \leq k \leq 9$ (I have not been able to compute for $k=10$)

$$\left( \begin{array}{ccc} k & R_k & I_k \\ 0 & 1.65388508388988 & 5.36676559479282 \\ 1 & 1.38001939449002 & 4.70977390795459 \\ 2 & 1.36217751498421 & 4.66071573895105 \\ 3 & 1.36045571658805 & 4.65592771371741 \\ 4 & 1.36028414014521 & 4.65545005440653 \\ 5 & 1.36026698851516 & 4.65540229987385 \\ 6 & 1.36026527341227 & 4.65539752453454 \\ 7 & 1.36026510190259 & 4.65539704700174 \\ 8 & 1.36026508475162 & 4.65539699924848 \\ 9 & 1.36026508303653 & 4.65539699447315 \end{array} \right)$$

These numbers have not been identified by inverse symbolic calculators.