Integral representation of Bell Polynomial?

Question

From Wolfram Alpha: https://functions.wolfram.com/IntegerFunctions/BellB/07/01/0001/,

we have an integral representation for Bell numbers as:

$B_n = \frac{2n!}{\pi e} \displaystyle{\int_{0}^{\pi}} e^{e^{\cos(t)} \cos(\sin(t))} \sin(e^{\cos(t)}\sin(\sin(t))) \sin(nt) dt$

I've attempted to extend this to Bell polynomials $B_n(x)$, and I have a result, although, ideally there would be something more elegant (wishful thinking maybe). I'll outline the steps and result gotten at the bottom of this post, but is there possibly a less ugly/shorter closed form integral representation of the bell polynomials? Alternatively, if anyone could help me simplify my result, as I lack the computational resources, that would also be great.

My results:

Notation: Here $B_{n}(x)$ denotes the $n$th Bell polynomial, $n \in \mathbb{N}$, of $x$, $x \in \mathbb{R}^*$, $S(n,k)$ denotes the Stirling number of the second kind.

From https://mathworld.wolfram.com/BellPolynomial.html (14),

$B_{n}(x) = \displaystyle{\sum_{k=0}^{n}} S(n,k) x^k$

From https://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html (10),

$S(n,k) = \frac{1}{k!} \displaystyle{\sum_{j=0}^{k} (-1)^j {k \choose j} (k-j)^n}$

Lastly, the first formula here: https://en.wikipedia.org/wiki/Reciprocal_gamma_function#Integral_representations_at_the_positive_integers states

$\frac{1}{n!} = \frac{1}{2 \pi} \displaystyle{\int_{-\pi}^{\pi}} e^{-n it} e^{e^{it}} dt$, for all $n \in \mathbb{N}$; however, this only seems to be partially correct. It seems the correct formula is

$\frac{1}{n!} = \frac{1}{2 \pi} \displaystyle{\int_{-\pi}^{\pi}} \Re[e^{-n it} e^{e^{it}}] dt$ for $n \in \mathbb{N} \cup$ {$0$} or $n \in \mathbb{Z}^*$ . If anyone could kindly help verify. Nonetheless, that is the formula I will be using.

Derivation:

$B_{n}(x) = \displaystyle{\sum_{k=0}^{n}} x^k S(n,k)$

$= \displaystyle{\sum_{k=1}^{n}} x^k S(n,k)$, since $S(n,0) = 0$ for $n > 0$

$ = \displaystyle{\sum_{k=1}^{n}} \displaystyle{\sum_{j=0}^{k} (-1)^j {k \choose j} (k-j)^n} x^k \frac{1}{k!}$

$ = \displaystyle{\sum_{k=1}^{n}} \displaystyle{\sum_{j=0}^{k} (-1)^j {k \choose j} (k-j)^n} x^k \frac{1}{2 \pi} \displaystyle{\int_{-\pi}^{\pi}} \Re[e^{-k it} e^{e^{it}}]dt $

$ = \frac{1}{2 \pi} \Re \left[ \displaystyle{\int_{-\pi}^{\pi}} e^{e^{it}} \displaystyle{\sum_{k=1}^{n}} x^k e^{-k it} \displaystyle{\sum_{j=0}^{k} (-1)^j {k \choose j} (k-j)^n} dt \right] $

$ = \frac{1}{2 \pi} \Re \left[ \displaystyle{\int_{-\pi}^{\pi}} e^{e^{it}} \displaystyle{\sum_{k=1}^{n}} x^k e^{-k it} k! S(n,k) dt \right] $

$= \frac{-1}{2 \pi} \Re \left[ \displaystyle{\int_{-\pi}^{\pi}} e^{e^{it}} \frac{\left(\frac{i}{2}\right)^{n+1} \left(n! \left(i \coth ^{-1}\left(e^{-i t} \left(2 x+e^{i t}\right)\right)\right)^{-n-1}+\pi ^{-n-1} \left((-1)^n \psi ^{(n)}\left(\frac{i \coth ^{-1}\left(e^{-i t} \left(2 x+e^{i t}\right)\right)}{\pi }\right)-\psi ^{(n)}\left(-\frac{i \coth ^{-1}\left(e^{-i t} \left(2 x+e^{i t}\right)\right)}{\pi }\right)\right)\right)}{1+e^{-i t} x} dt \right]$

$= \frac{-1}{2 \pi} \displaystyle{\int_{-\pi}^{\pi}} \Re \left[ e^{e^{it}} \frac{\left(\frac{i}{2}\right)^{n+1} \left(n! \left(i \coth ^{-1}\left(e^{-i t} \left(2 x+e^{i t}\right)\right)\right)^{-n-1}+\pi ^{-n-1} \left((-1)^n \psi ^{(n)}\left(\frac{i \coth ^{-1}\left(e^{-i t} \left(2 x+e^{i t}\right)\right)}{\pi }\right)-\psi ^{(n)}\left(-\frac{i \coth ^{-1}\left(e^{-i t} \left(2 x+e^{i t}\right)\right)}{\pi }\right)\right)\right)}{1+e^{-i t} x}\right] dt $

I have also tried computing the real part $\Re[e^{-k it} e^{e^{it}}] = e^{\cos(t)} \cos(k t - \sin(t))$ first, and then expanding; however, this is not of great help in trying to simplify the final expression, if possible, from my own experience.

Thanks!

Best,

Ben

Answer 1

Partial answer. Since $$ \exp (({\rm e}^t - 1)x) = \sum\limits_{n = 0}^\infty {\frac{{B_n (x)}}{{n!}}t^n } , $$ by Cauchy's formula, \begin{align*} B_n (x) & = \frac{{n!}}{{2\pi {\rm i}}}\oint_{|w| = 1} {\exp (({\rm e}^w - 1)x)\frac{{{\rm d}w}}{{w^{n + 1} }}} \\ & = \frac{{n!}}{{2\pi }}\int_{ - \pi }^\pi {\exp (({\rm e}^{{\rm e}^{\mathrm{i}t} } - 1)x - {\rm i}tn){\rm d}t} \\ & = \frac{{n!}}{{2\pi {\rm e}^x }}\int_{ - \pi }^\pi {\exp (({\rm e}^{\cos t + {\rm i}\sin t} )x - {\rm i}tn){\rm d}t} \\ & = \frac{{n!}}{{2\pi {\rm e}^x }}\int_{ - \pi }^\pi {\exp (x{\rm e}^{\cos t} \cos (\sin t))\exp ({\rm i}x{\rm e}^{\cos t} \sin (\sin t) - {\rm i}nt){\rm d}t} . \end{align*} Taking real parts on both sides yields \begin{align*} B_n (x) & = \frac{{n!}}{{2\pi {\rm e}^x }}\int_{ - \pi }^\pi {\exp (x{\rm e}^{\cos t} \cos (\sin t))\cos (x{\rm e}^{\cos t} \sin (\sin t) - nt){\rm d}t} \\& = \frac{{n!}}{{2\pi {\rm e}^x }}\int_{ - \pi }^\pi {\exp (x{\rm e}^{\cos t} \cos (\sin t))\cos (x{\rm e}^{\cos t} \sin (\sin t))\cos (nt){\rm d}t} \\& \quad \;+ \frac{{n!}}{{2\pi {\rm e}^x }}\int_{ - \pi }^\pi {\exp (x{\rm e}^{\cos t} \cos (\sin t))\sin (x{\rm e}^{\cos t} \sin (\sin t))\sin (nt){\rm d}t} \\& = \frac{{n!}}{{\pi {\rm e}^x }}\int_0^\pi {\exp (x{\rm e}^{\cos t} \cos (\sin t))\cos (x{\rm e}^{\cos t} \sin (\sin t))\cos (nt){\rm d}t} \\& \quad\;+ \frac{{n!}}{{\pi {\rm e}^x }}\int_0^\pi {\exp (x{\rm e}^{\cos t} \cos (\sin t))\sin (x{\rm e}^{\cos t} \sin (\sin t))\sin (nt){\rm d}t} . \end{align*} It seems that the last two integrals are equal. If that is true, then $$ B_n (x) = \frac{{2 \cdot n!}}{{\pi {\rm e}^x }}\int_0^\pi {\exp (x{\rm e}^{\cos t} \cos (\sin t))\cos (x{\rm e}^{\cos t} \sin (\sin t))\cos (nt){\rm d}t} $$ and $$ B_n (x) = \frac{{2 \cdot n!}}{{\pi {\rm e}^x }}\int_0^\pi {\exp (x{\rm e}^{\cos t} \cos (\sin t))\sin (x{\rm e}^{\cos t} \sin (\sin t))\sin (nt){\rm d}t} , $$ respectively. The case $x=1$ of the second formula would yield the corrected version of the one on the Wolfram Functions site.