Convert $\frac1b\sum_{n=1}^\infty\frac{(b e^a)^n}{n!}B_{n-1}(an)$ to integral using $B_n(x)=\frac{n!}{2\pi i}\oint\frac{e^{x(e^t-1)}}{t^{n+1}}dt$

Question

$\def\B{\operatorname B}$ In

How to solve $x^{y^z}=z$

A solution uses Bell polynomials $\B_n(x)$

$$e^{ae^{bz}}=z=1+\frac1b\sum_{n=1}^\infty \frac{(ae^b)^n}{nn!}\B_n(b n)=\frac1b\sum_{n=1}^\infty\frac{(b e^a)^n}{n!}\B_{n-1}(an)\tag1$$

A contour integral representation is:

$$\B_n(x)=\frac{n!}{2\pi i}\oint\frac{e^{x(e^t-1)}}{t^{n+1}}dt\tag2$$

interchanging sum and integral gives: $$e^{ae^{bz}}=z\mathop=^?1+\frac i{2\pi b}\oint\ln\left(1-\frac{ae^{be^t}}t\right)dt\mathop=^?\frac i{2\pi b}\oint\ln\left(1-\frac{be^{ae^t}}t\right)dt$$

I have little experience with contour integration, but trying $a=b=-\frac12$ and $\oint\leftrightarrow\int_{\frac12-i\infty}^{\frac12+i\infty}$ gives a different result than the sum. Likely, the contour used is wrong or one cannot switch the sum and integral without converting to an ordinary integral.

How can $(2)$ be used in $(1)$ to find a similar contour integral solution to $e^{ae^{bz}}=z$?

Answer 1

(WIP) Interchange the sum and the integral, and integrate on the path $z=r_0e^{2\pi i \tau}$ from $0$ to $1$. Take the limit as $r_0\to 0$ like so

$$b^{-1}\sum \frac{b^ne^{an}}{2\pi in}\oint \frac{e^{an(e^{t}-1)}}{t^n}dt\\=b^{-1}\oint \sum \frac{b^ne^{an}}{2\pi in} \frac{e^{an(e^{t}-1)}}{t^n}dt\\=\frac{b^{-1}}{2\pi i}\oint \sum \frac{1}{n}\left(\frac{be^{ae^t}}{t}\right)^ndt\\ =-\frac{b^{-1}}{2\pi i}\oint \ln\left(1-\frac{be^{ae^t}}{t}\right)dt\\ = -b^{-1}\lim_{r_0\to0}\int_0^1r_0e^{2\pi i\tau}\ln\left(1-\frac{be^{ae^{r_0e^{2\pi i\tau}}}}{r_0e^{2\pi i\tau}}\right)d\tau$$ Now, I know what you're thinking, and I'm also very overwhelmed... however with a bit of faith, I think we can proceed with the limit using asymptotics and L'Hopital's rule.

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However, if you want an inverse function for $e^{ae^{bz}}-z,$ you can use the Cauchy Integral Theorem to get

$$f^{-1}(w)=\frac{1}{2\pi i}\oint_{f^{-1}\gamma} \frac{zf'(z)}{f(z)-w}dz$$

Answer 2

Using @Gary’s Bell polynomial integral representation, we get:

$$\frac1b\sum_{n=1}^\infty\frac{(b e^a)^n}{n!}\text B_{n-1}(an)=\int_{-\pi}^\pi \sum_{n=1}^\infty\frac{e^{\left(ae^{e^{it}}-a-i t\right)n+it}(b e^a)^n}{2\pi b n}dt$$

with something similar for the second sum in the question. Now apply the series expansion of $\ln(x)$:

$$\bbox[2px,border: 2.5px ridge green]{e^{ae^{bx}}=x=-\frac1{2\pi b}\int_{-\pi}^\pi e^{it}\ln\left(1-be^{ae^{e^{it}}-it}\right)dt,1-\frac1{2\pi b}\int_{-\pi}^\pi\ln\left(1-ae^{be^{e^{it}}-it}\right)dt}$$

shown here. Software gives a restriction when summing over $n$, but it is not clear how to apply it to the integral representation. Taking $a=b$ gives another integral representation of the Lambert W function solving $e^{ax}=x$. Additionally, one can remove the logarithm with integration by parts to get a slightly longer integral expression. It also may be interesting to take the real part of the integrand.