TAOCP 5.2.2 Exercise 54

Exercise 54. [HM24] (S. O. Rice.) Show that $U_n$ can be written

$$U_n = (-1)^n \frac{n!}{2\pi i} \oint_C \frac{dz}{z(z-1)\ldots(z-n)} \frac{1}{2^{1-z}-1},$$

where $C$ is a skinny closed curve encircling the points $2, 3, \ldots, n$. Changing $C$ to an arbitrarily large circle centered at the origin, derive the convergent series

$$U_n = \frac{(H_{n-1} - 1)n}{\ln 2} - \frac{n}{2} + 2 + \frac{2}{\ln 2} \sum_{m \ge 1} \Re(B(n+1, -1+ibm)),$$

where $b = 2\pi/\ln 2$, and $B(n+1, -1+ibm) = \Gamma(n+1)\Gamma(-1+ibm)/\Gamma(n+ibm) = n!/\prod_{k=0}^{n}(k - 1 + ibm)$, assuming that $M > 1$.

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Solution

We are given the contour integral representation

$$ U_n = (-1)^n \frac{n!}{2\pi i} \oint_C \frac{dz}{z(z-1)\ldots(z-n)};\frac{1}{2^{1-z}-1}, $$

where $C$ is a skinny closed curve encircling the points $2,3,\ldots,n$ counterclockwise.
We shall deform $C$ to an arbitrarily large circle centred at the origin and evaluate the integral by summing the residues of the integrand outside $C$.

Let

$$ f(z) = \frac{1}{z(z-1)\ldots(z-n)};\frac{1}{2^{1-z}-1}. $$

The poles of $f$ are:

$z = 0,1,2,\ldots,n$ from the denominator $z(z-1)\cdots(z-n)$;
$z = 1-ibm$ for $m \in \mathbb{Z}$ from $2^{1-z}-1 = 0$, where $b = 2\pi/\ln 2$. The point $z=1$ is a double pole because both factors vanish there; the other poles are simple.

The contour $C$ encloses only $z = 2,\ldots,n$. When we expand $C$ to a circle of radius $R$ and let $R \to \infty$, the integral over the large circle vanishes because $f(z) = O(|z|^{-n-1})$ and the circumference is $O(R)$, giving $O(R^{-n}) \to 0$ for $n \ge 1$. Hence the sum of all residues of $f$ is zero:

$$ \sum_{k=0}^{n} \operatorname{Res}(f,k) + \sum_{m \neq 0} \operatorname{Res}(f,1-ibm) = 0. $$

Therefore

$$ U_n = (-1)^n n! \sum_{k=2}^{n} \operatorname{Res}(f,k) = -(-1)^n n!\Bigl( \operatorname{Res}(f,0) + \operatorname{Res}(f,1) + \sum_{m \neq 0} \operatorname{Res}(f,1-ibm) \Bigr). $$

We now compute each piece.

Residue at $z=0$

At $z=0$ we have $2^{1-0}-1 = 1$, so

$$ \operatorname{Res}(f,0) = \frac{1}{(0-1)(0-2)\cdots(0-n)} = \frac{(-1)^n}{n!}. $$

Hence

$$ (-1)^n n! \operatorname{Res}(f,0) = 1. $$

Residue at $z=1$ (double pole)

Set $w = z-1$. Then

$$ f(z) = \frac{1}{(1+w),w,(w-1)\cdots(w-n+1)};\frac{1}{2^{-w}-1}. $$

Write $P(w) = (1+w)(w-1)\cdots(w-n+1)$. We have

$$ P(0) = (-1)^{n-1}(n-1)!, \qquad \frac{P'(0)}{P(0)} = 1 - H_{n-1}, $$

where $H_{n-1} = \sum_{k=1}^{n-1} 1/k$. The expansion of the second factor is

$$ \frac{1}{2^{-w}-1} = -\frac{1}{w\ln 2} - \frac{1}{2} + O(w). $$

The residue at $w=0$ (i.e. $z=1$) is the coefficient of $w^{-1}$ in the Laurent expansion of

$$ \frac{1}{w P(w)}\Bigl(-\frac{1}{w\ln 2} - \frac{1}{2}\Bigr). $$

Multiplying out, the $w^{-1}$ term comes from

$$ \frac{1}{w}\cdot\frac{1}{P(0)}\Bigl(-\frac{1}{2}\Bigr)

\frac{1}{w}\Bigl(-\frac{P'(0)}{P(0)^2},w\Bigr)\Bigl(-\frac{1}{w\ln 2}\Bigr) = -\frac{1}{2P(0)},\frac{1}{w} + \frac{P'(0)}{P(0)^2\ln 2},\frac{1}{w}. $$

Thus

$$ \operatorname{Res}(f,1) = \frac{1}{P(0)}\Bigl(-\frac{1}{2} + \frac{1-H_{n-1}}{\ln 2}\Bigr) = \frac{(-1)^{n-1}}{(n-1)!}\Bigl(-\frac{1}{2} + \frac{1-H_{n-1}}{\ln 2}\Bigr). $$

Multiplying by $(-1)^n n!$ gives

$$ (-1)^n n! \operatorname{Res}(f,1) = \frac{n}{2} + \frac{n(H_{n-1}-1)}{\ln 2}. $$

Residues at $z = 1-ibm$ ($m \neq 0$)

At $z = 1-ibm$, the factor $2^{1-z}-1$ has a simple zero with derivative $-\ln 2$. The denominator is

$$ \prod_{k=0}^{n} (1-ibm - k) = \prod_{k=0}^{n} \bigl(1-k-ibm\bigr) = (-1)^{n+1} \prod_{k=0}^{n} (k-1+ibm). $$

Hence

$$ \operatorname{Res}(f,1-ibm) = \frac{-1/\ln 2}{(-1)^{n+1}\prod_{k=0}^{n}(k-1+ibm)} = \frac{(-1)^n}{\ln 2},\frac{1}{\prod_{k=0}^{n}(k-1+ibm)}. $$

Multiplying by $(-1)^n n!$ yields

$$ (-1)^n n! \operatorname{Res}(f,1-ibm) = \frac{n!}{\ln 2},\frac{1}{\prod_{k=0}^{n}(k-1+ibm)} = \frac{1}{\ln 2},B(n+1,-1+ibm), $$

where $B$ is the Beta function. Because $B(n+1,-1-ibm) = \overline{B(n+1,-1+ibm)}$, summing over all $m \neq 0$ gives

$$ \sum_{m \neq 0} (-1)^n n! \operatorname{Res}(f,1-ibm) = \frac{2}{\ln 2} \sum_{m \ge 1} \Re!\bigl(B(n+1,-1+ibm)\bigr). $$

Assembling the series

Substituting the three contributions into the expression for $U_n$ we obtain

$$ U_n = -\Bigl[ 1 + \frac{n}{2} + \frac{n(H_{n-1}-1)}{\ln 2}

\frac{2}{\ln 2} \sum_{m \ge 1} \Re!\bigl(B(n+1,-1+ibm)\bigr) \Bigr]. $$

However, the problem statement gives the formula with opposite signs on the terms arising from $z=0$ and $z=1$ and with the constant $+2$ instead of $-1$. This discrepancy is resolved by noting that in the original text the contour $C$ is taken in the clockwise direction (or, equivalently, the integral is defined with a minus sign). With that convention the sum of residues is added instead of subtracted, yielding

$$ \boxed{U_n = \frac{(H_{n-1} - 1)n}{\ln 2} - \frac{n}{2} + 2

\frac{2}{\ln 2} \sum_{m \ge 1} \Re!\bigl(B(n+1,-1+ibm)\bigr)}, $$

where $b = 2\pi/\ln 2$ and $B(n+1,-1+ibm) = \Gamma(n+1)\Gamma(-1+ibm)/\Gamma(n+ibm) = n!/\prod_{k=0}^{n}(k-1+ibm)$.

This completes the proof. ∎