TAOCP 7.2.1.5 Exercise 48
The integral in equation (23) of the text is the Cauchy integral representation for the Bell numbers obtained from the exponential generating function \(\Pi(z)=e^{e^z-1}\).
Section 7.2.1.5: Generating All Set Partitions
Exercise 48. [HM25] If $n$ is not an integer, the integral in (23) can be taken over a Hankel contour to define a generalized Bell number $\varpi_x$ for all real $x > 0$. Show that, as in (16),
$$ \varpi_x=\frac{1}{e}\sum_{k=0}^{\infty}\frac{k^x}{k!}. $$
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Solution
The integral in equation (23) of the text is the Cauchy integral representation for the Bell numbers obtained from the exponential generating function (\Pi(z)=e^{e^z-1}). For integer (n\ge 0),
[ \varpi_n = \frac{n!}{2\pi i}\oint \frac{e^{e^z-1}}{z^{n+1}},dz, \tag{23} ]
where the contour encircles the origin once counterclockwise.
To define a generalized Bell number (\varpi_x) for real (x>0), we replace the closed contour by a Hankel contour (H) that starts at (-\infty) on the real axis, encircles the origin counterclockwise, and returns to (-\infty). We also replace (n!) by (\Gamma(x+1)) and (n) by (x), interpreting (z^{-x-1}) with the branch cut along the negative real axis and (\arg z\in(-\pi,\pi]) on (H). This gives
[ \varpi_x = \frac{\Gamma(x+1)}{2\pi i}\int_H \frac{e^{e^z-1}}{z^{x+1}},dz. ]
Now expand the exponential generating function as in (15):
[ e^{e^z-1} = e^{-1}\sum_{k=0}^{\infty}\frac{e^{kz}}{k!}. ]
Because (x>0), the series converges absolutely and the integral over (H) converges uniformly for (z) on (H); we may therefore interchange sum and integral:
[ \varpi_x = \frac{\Gamma(x+1)}{2\pi i},e^{-1}\sum_{k=0}^{\infty}\frac{1}{k!} \int_H \frac{e^{kz}}{z^{x+1}},dz. ]
For (k=0) the integrand is (z^{-x-1}); since (x>0) the integral over the Hankel contour is (0). For (k\ge 1) we evaluate
[ I_k = \int_H e^{kz}z^{-x-1},dz. ]
The Hankel contour integral for the reciprocal Gamma function is
[ \frac{1}{\Gamma(z)} = \frac{1}{2\pi i}\int_H t^{-z}e^t,dt ]
for the same contour (H) and the same branch of the logarithm. Substitute (t = kz) (with (k>0)); the contour is invariant under positive scaling, so
[ \frac{1}{\Gamma(z)} = \frac{1}{2\pi i}\int_H (kz)^{-z}e^{kz},k,dz = \frac{k^{1-z}}{2\pi i}\int_H z^{-z}e^{kz},dz. ]
Setting (z = x+1) gives
[ \int_H e^{kz}z^{-x-1},dz = \frac{2\pi i}{\Gamma(x+1)},k^x. ]
Substituting this into the sum yields
[ \varpi_x = \frac{\Gamma(x+1)}{2\pi i},e^{-1}\sum_{k=1}^{\infty}\frac{1}{k!} \cdot\frac{2\pi i}{\Gamma(x+1)},k^x = \frac{1}{e}\sum_{k=0}^{\infty}\frac{k^x}{k!}, ]
where the (k=0) term is (0) for (x>0). This is exactly the generalization of Dobiński’s formula (16).
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