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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

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  • $\begingroup$ You are citing the formula for $B_n$ incorrectly, but the formula on the Wolfram site is also incorrect. The corrected version is $$ B_n = \frac{{2n!}}{{\pi {\rm e}}}\int_0^\pi {\exp ({\rm e}^{\cos t} \cos (\sin t))\sin ({\rm e}^{\cos t} \sin (\sin t))\sin (nt)\,{\rm d}t} . $$ You can check it numerically. $\endgroup$
    – Gary
    Commented May 26, 2023 at 2:34
  • $\begingroup$ Updated thanks! $\endgroup$
    – BBadman
    Commented Jun 15, 2023 at 5:22

1 Answer 1

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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.

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    $\begingroup$ Sorry for the late reply and thank you for the thoughtful response! This is definitely a much more fruitful approach. After some more digging, I found out bell polynomials are also called Touchard polynomials, which have the integral representation as you’ve proven. Unsure about the equality of the two last ones, but ya it seems they’re the same. $\endgroup$
    – BBadman
    Commented Jun 6, 2023 at 15:52

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