1. Introduction
Let \(z\in \mathbb{C}\) and assume that \(\Re z > 0\).
It is well-known that
\begin{equation} \label{eq:Stirling}
\ln\Gamma(z) = (z-\frac{1}{2})\ln z – z + \frac{1}{2}\ln(2\pi) + J(z),
\end{equation}
(1)
where \(J(z)\) can be written as
\begin{equation} \label{eq:Binet}
J(z) = \frac{1}{\pi}\int_0^\infty\frac{z}{\eta^2+z^2}
\ln\left(\frac{1}{1-e^{-2\pi\eta}}\right)\, d\eta\,.
\end{equation}
(2)
The analytic function \(J(z)\) is known as
Binet’s function and has several equivalent
expressions; see for example, Henrici [
1,
(8.5-7)].
Binet’s function has an asymptotic expansion
\begin{equation} \label{eq:Binet-expansion}
J(z) \sim \frac{\beta_0}{z} – \frac{\beta_1}{z^3}
+ \frac{\beta_2}{z^5} – \cdots,
\end{equation}
(3)
or more precisely, for non-negative integers \(k\),
\begin{equation} \label{eq:Binet-expansion-remainder}
J(z) = \sum_{j=0}^{k-1} (-1)^j\frac{\beta_j}{z^{2j+1}} + r_k(z),
\end{equation}
(4)
where
\begin{equation} \label{eq:beta}
\beta_k = \frac{1}{\pi}\int_0^\infty
\eta^{2k}
\ln\left(\frac{1}{1-e^{-2\pi\eta}}\right)\, d\eta
\end{equation}
(5)
and
\begin{equation} \label{eq:Binet_rk}
r_k(z) = \frac{(-1)^k}{\pi z^{2k-1}}\int_0^\infty
\frac{\eta^{2k}}{z^2+\eta^2}
\ln\left(\frac{1}{1-e^{-2\pi\eta}}\right)\, d\eta\,.
\end{equation}
(6)
It may be shown that
\begin{equation} \label{eq:betakzetaB}
\beta_k = \frac{2(2k)!}{(2\pi)^{2k+2}}\,\zeta(2k+2)
= \frac{(-1)^k}{(2k+1)(2k+2)}\,B_{2k+2}\,,
\end{equation}
(7)
where \(B_{2k+2}\) is a Bernoulli number (\(B_2 = 1/6, B_4 = -1/30\), etc.).
Proofs of these results are given in
Henrici’s book [
1,
§11.1]\(^1\).
Substituting (4) into
(1) gives an asymptotic expansion for \(\ln\Gamma(z)\) that
is usually named after James Stirling,
although some credit is due to Abraham de Moivre.
For the history and early references, see Dutka [2].
It is interesting to note that de Moivre started (about 1721)
by trying to approximate
the central binomial coefficient \(\binom{2n}{n}\), not the factorial (or
Gamma) function- see Dutka [2,pg.227].
It is easy to see from (5) and (6) that
\begin{equation} \label{eq:r_k_theta}
r_k(z) = \theta_k(z)\, (-1)^k\frac{\beta_k}{z^{2k+1}}\,,
\end{equation}
(8)
where
\begin{equation} \label{eq:theta}
\theta_k(z) = \int_0^\infty
\frac{z^2\,\eta^{2k}}{z^2+\eta^2}
\ln\left(\frac{1}{1-e^{-2\pi\eta}}\right)\, d\eta\
\Bigg/
\int_0^\infty
\eta^{2k}
\ln\left(\frac{1}{1-e^{-2\pi\eta}}\right)\, d\eta\,.
\end{equation}
(9)
Suppose now that \(z\) is real and positive.
Since \(z^2/(z^2+\eta^2) \in (0,1)\) and the logarithmic factors
in (9) are positive for all \(\eta\in(0,\infty)\),
we see that
\begin{equation} \label{eq:enveloping}
\theta_k(z) \in (0,1).
\end{equation}
(10)
Thus, the remainder \(r_k(z)\) given by (8) has the same
sign as the next term \((-1)^k\beta_k/z^{2k+1}\)
in the asymptotic series, and is smaller
in absolute value. In the terminology used
by Pólya and Szegö [
3, Ch.4],
the asymptotic series
for \(\ln\Gamma(z)\)
strictly envelops\(^2\)
the function \(\ln\Gamma(z)^3\).
§2 shows that we can deduce a strictly enveloping
asymptotic series for \(\ln(\Gamma(2z+1)/\Gamma(z+1)^2)\) or equivalently,
if \(z=n\) is
a positive integer, for the logarithm of
the central binomial coefficient \(\binom{2n}{n}\).
The series itself is well known,
but we have not found the
enveloping property or the resulting error bound
mentioned in the literature. Henrici
was aware of it, since in his book [1,
§11.2, Problem 6]
he gave the special case \(k=3\) as an exercise, along with a hint
for the solution. Hence, we do not claim any particular originality.
Our purpose is primarily to make some useful asymptotic
series and their associated error bounds readily accessible.
Related results and additional references may be found,
for example, in [4,5,6].
In §2 we consider the central binomial coefficient and its
generalisation to a complex argument. Then, in §3, we
consider some closely related asymptotic series that we can prove to be
strictly enveloping. In §4 we make some remarks on
asymptotic series that are not enveloping.
An Appendix gives numerical values of the coefficients appearing
in three of the asymptotic series.
Finally,
we remark that it is possible to give asymptotic
series related to \(\Gamma(z+\frac12)/\Gamma(z)\)
and \(\binom{2n}{n}\), but in general these
series do not alternate in sign.
See, for example,
[7], [8],
[9, ex.9.60 and pg.602],
[10],
and [11].
2. Asymptotic series for central binomial coefficients
Define
\begin{align*}
\widetilde{\Gamma}(z) :=& \frac{\Gamma(2z+1)}{\Gamma(z+1)^2}\,,\\
\widetilde{J}(z) :=& J(2z) – 2J(z),\\
\widetilde{r}_k(z) :=& r_k(2z) – 2r_k(z),
\end{align*}
and
\begin{equation} \label{eq:betatilde}
\widetilde{\beta}_k := (2-2^{-2k-1})\beta_k =
(-1)^{k}\frac{(1-4^{-k-1})}{(k+1)(2k+1)}B_{2k+2}\,.
\end{equation}
(11)
As noted above,
the central binomial coefficient \(\binom{2n}{n}\) is simply
\(\widetilde{\Gamma}(n)\).
Using elementary properties of the Gamma function, we have
\begin{equation} \label{eq:Gammatilde}
\widetilde{\Gamma}(z) = \frac{2}{z} \frac{\Gamma(2z)}{\Gamma(z)^2}\,.
\end{equation}
(12)
Thus, from (1) and the same equation with \(z \mapsto 2z\),
we have
\begin{equation} \label{eq:lnGammatildeJtilde}
\ln\widetilde{\Gamma}(z) = \ln\left(\frac{4^z}{\sqrt{\pi z}}\right) + \widetilde{J}(z).
\end{equation}
(13)
Also, from (4)
and the definition of \(\widetilde{J}(z)\), we have an asymptotic series
for \(\widetilde{J}(z)\), namely:
\begin{equation} \label{eq:Jtilde-series}
\widetilde{J}(z) = -\sum_{j=0}^{k-1} (-1)^j\frac{\widetilde{\beta}_j}{z^{2j+1}} +
\widetilde{r}_k(z).
\end{equation}
(14)
Since \(\binom{2n}{n} = \widetilde{\Gamma}(n)\),
equations (13)-(14)
give an asymptotic series for \(\ln\binom{2n}{n}\).
Lemma 1 shows that the
remainder \(\widetilde{r}_k(z)\) can be expressed as an integral analogous to
the integral (6) for \(r_k(z)\).
Lemma 1.
For \(z\in\mathbb{C}, \Re z > 0\), and \(k\) a non-negative integer,
\begin{equation} \label{eq:lemma1a}
\widetilde{\beta}_k = – \frac{1}{\pi}\int_0^\infty
\eta^{2k}
\ln\tanh(\pi\eta)d\eta\,,
\end{equation}
(15)
\begin{equation} \label{eq:lemma1b}
\widetilde{r}_k(z) = \frac{(-1)^k}{\pi z^{2k-1}}\int_0^\infty
\frac{\eta^{2k}}{z^2+\eta^2}
\ln\tanh(\pi\eta)d\eta\,,
\end{equation}
(16)
and
\begin{equation} \label{eq:Jr}
\widetilde{J}(z) = \widetilde{r}_0(z).
\end{equation}
(17)
Proof.
Making the change of variables \(z \mapsto 2z\) and
\(\eta \mapsto 2\eta\) in (6), we obtain
\[
r_k(2z) = \frac{(-1)^k}{\pi z^{2k-1}}\int_0^\infty
\frac{\eta^{2k}}{z^2+\eta^2}
\ln\left(\frac{1}{1-e^{-4\pi\eta}}\right)\,d\eta\,.
\]
Now
\begin{equation*}
\ln\left(\frac{1}{1-e^{-4\pi\eta}}\right) –
2\ln\left(\frac{1}{1-e^{-2\pi\eta}}\right)
= \ln\left(\frac{1-e^{-2\pi\eta}}{1+e^{-2\pi\eta}}\right)
= \ln\tanh(\pi\eta),
\end{equation*}
so (16)-(17) follow from the
definitions of \(\widetilde{r}_k(z)\) and \(\widetilde{J}(z)\).
The proof of (15) is similar.
Corollary 1 gives a result analogous
to Equations (8)-(9).
Corollary 1.
For \(z\in\mathbb{C}, \Re z > 0\), and \(k\) a non-negative integer,
\begin{equation} \label{eq:rtildek}
\widetilde{r}_k(z) = \widetilde{\theta}_k(z)(-1)^{k+1}\frac{\widetilde{\beta}_k}{z^{2k+1}}
\,,
\end{equation}
(18)
where
\begin{equation} \label{eq:thetatilde}
\widetilde{\theta}_k(z) = \int_0^\infty \frac{z^2\,\eta^{2k}}{z^2+\eta^2}
\ln\tanh(\pi\eta)d\eta
\Bigg/
\int_0^\infty \eta^{2k}
\ln\tanh(\pi\eta)d\eta\,.
\end{equation}
(19)
Proof.
This is straightforward from Equations (15)-(16)
of Lemma 1.
Corollary 2 gives a result analogous to
the bound (10).
Corollary 2.
If \(z\) is real and positive, then \(\widetilde{\theta}_k(z) \in (0,1)\).
Proof.
We write (19) as
\begin{equation} \label{eq:thetatilde2}
\widetilde{\theta}_k(z) = \frac{\displaystyle
\int_0^\infty \frac{z^2\,\eta^{2k}}{z^2+\eta^2}
\ln\coth(\pi\eta)d\eta}
{\displaystyle
\int_0^\infty \eta^{2k}
\ln\coth(\pi\eta)d\eta}\,.
\end{equation}
(20)
Observe that
\(\coth(y) = \cosh(y)/\sinh(y) > 1\) for \(y\in(0,\infty)\),
so \(\ln\coth(y) > 0\) for \(y = {\pi\eta} > 0\).
Since \(z^2/(z^2+\eta^2) \in (0,1)\) for real positive \(z\) and \(\eta\),
the result follows.
Corollary 3.
If \(z\) is real and positive, then the asymptotic
series (14) for \(\widetilde{J}(z)\) is strictly enveloping.
Proof.
This is immediate from Corollary 2.
Remark 1.
We may compare
Corollary 2 with (the proof of)
Lemma 2.7 of [12].
The latter, after allowing for different notation,
gives the bound
\[
\frac{-1}{4^{k+1}-1} < \widetilde{\theta}_k(z) < \frac{4^{k+1}}{4^{k+1}-1}.
\]
This is clearly weaker than the bound of Corollary 2,
and not sufficient to prove Corollary 3.
3. Some related asymptotic series
Lemma 2.
If \(z\in\mathbb{C}, \Re z > 0\), then
\[
\widetilde{J}(z) = \ln\left(\frac{\Gamma(z+\frac12)}{z^{1/2}\Gamma(z)}\right)
.
\]
Proof.
This follows from Equations (12)-(13)
and the duplication formula
\(\Gamma(z)\Gamma(z+\frac{1}{2}) = 2^{1-2z}\pi^{1/2}\Gamma(2z)\).
From Lemma 2 and (14) we immediately
obtain an asymptotic expansion
\begin{equation} \label{eq:logGrat}
\ln\left(\frac{\Gamma(z+\frac12)}{\Gamma(z)}\right)
\sim \frac{\ln z}{2} + \sum_{j \ge 0}(-1)^{j+1}
\frac{\widetilde{\beta}_j}{z^{2j+1}}
\end{equation}
(21)
which is strictly enveloping if \(z\) is real and positive.
Define
\begin{equation} \label{eq:betahat}
\widehat{\beta}_j = \widetilde{\beta}_j – \beta_j = (1-2^{-2j-1})\beta_j.
\end{equation}
(22)
Using the asymptotic expansion for \(\ln\Gamma(z)\) given by Equations
(1) and (3), we see
from (21) that \(\ln\Gamma(z+\frac12)\) has an asymptotic
expansion
\begin{equation} \label{eq:lnGammahalf}
\ln\Gamma(z+\frac{1}{2}) \sim z\ln z – z +
\frac{1}{2}\ln(2\pi)
+ \sum_{j \ge 0}(-1)^{j+1}\,\frac{\widehat{\beta}_j}{z^{2j+1}}
\,.
\end{equation}
(23)
In fact, the expansion (23) was already obtained
by Gauss [
13, Eqn.[59] of
Art.29] in 1812.
However, Gauss did not explicitly bound the truncation error.
Writing (23) as
\begin{equation} \label{eq:lnGammahalf2}
\ln\Gamma(z+\frac{1}{2}) = z\ln z – z +
\frac{1}{2}\ln(2\pi)
+ \sum_{j=0}^{k-1}(-1)^{j+1}\,\frac{\widehat{\beta}_j}{z^{2j+1}}
+ \widehat{r}_k(z)
\,,
\end{equation}
(24)
we have an unsurprising result for the truncation error \(\widehat{r}_k(z)\):
the error is of the same sign as the first neglected term
\((-1)^{k+1}\widehat{\beta}_k/z^{2k+1}\),
and is bounded in magnitude by this term. This is shown in
Lemma 3 and Corollaries 4-5
below.
Lemma 3.
For \(z\in\mathbb{C}, \Re z > 0\), and \(k\) a non-negative integer,
\begin{equation} \label{eq:lemma3a}
\widehat{\beta}_k = \frac{1}{\pi}\int_0^\infty
\eta^{2k}
\ln\left(1+e^{-2\pi\eta}\right)d\eta
\end{equation}
(25)
and
\begin{equation} \label{eq:lemma3b}
\widehat{r}_k(z) = \frac{(-1)^{k+1}}{\pi z^{2k-1}}\int_0^\infty
\frac{\eta^{2k}}{z^2+\eta^2}
\ln\left(1+e^{-2\pi\eta}\right)d\eta\,.
\end{equation}
(26)
Proof.
This is similar to the proof of Lemma 1.
Corollary 4.
For \(z\in\mathbb{C}, \Re z > 0\), and \(k\) a non-negative integer,
\begin{equation} \label{eq:rhatk}
\widehat{r}_k(z) = \widehat{\theta}_k(z)(-1)^{k+1}\frac{\widehat{\beta}_k}{z^{2k+1}}\,,
\end{equation}
(27)
where
\begin{equation} \label{eq:thetahat}
\widehat{\theta}_k(z) = \int_0^\infty \frac{z^2\,\eta^{2k}}{z^2+\eta^2}
\ln\left(1+e^{-2\pi\eta}\right)d\eta
\Bigg/
\int_0^\infty \eta^{2k}
\ln\left(1+e^{-2\pi\eta}\right)d\eta\,.
\end{equation}
(28)
Proof.
This is a straightforward consequence of Lemma 3.
Corollary 5.
If \(z\) is real and positive, then the asymptotic expansion
for \(\ln\Gamma(z+\frac{1}{2})\) given in (24) is
strictly enveloping.
Proof.
From (28) we have \(\widehat{\theta}_k(z) \in (0,1)\).
Remark 2.
If we make the change of variables \(z \mapsto n+\frac12\)
in (23), and assume that \(n\) is a positive integer,
we obtain an asymptotic series for \(n!\) in negative powers of \((n+\frac{1}{2})\):
\begin{equation} \label{eq:deMoivre}
\ln n! \sim (n+\frac{1}{2})\ln(n+\frac{1}{2}) – (n+\frac{1}{2}) + \frac{1}{2}\ln(2\pi)
+ \sum_{j\ge 0}(-1)^{j+1}\,\frac{\widehat{\beta}_j}{(n+\frac{1}{2})^{2j+1}}
\,.
\end{equation}
(29)
In fact, (29) was stated (without proof) by
de Moivre [
14,
15] as early as 1730,
see Dutka [
1,(5), pg.233].
4. Non-enveloping asymptotic series
Lest the reader has gained the impression that all “naturally occurring”
asymptotic series are enveloping (for real positive arguments),
we give two classes of examples to show that this
is not the case. In fact, enveloping series are the exception, not the
rule. Our first class of examples is given by the following Lemma.
Lemma 4.
Let \(x\in(0,+\infty)\) and \(f(x) := J(x) + \exp(-bx)\)
for some constant \(b \in (0, 2\pi)\).
Then \(f(x)\) has an asymptotic series
\begin{equation} \label{eq:f-asym}
f(x) \sim \sum_{j=0}^{\infty} (-1)^j\frac{\beta_j}{x^{2j+1}}\,.
\end{equation}
(30)
However, the series (30) does not envelop \(f\).
Proof.
For all \(k \ge 0\),
\(\exp(-bx) = O(x^{-2k-1})\) as \(x \to +\infty\).
Thus, it follows from (4) that
\(f(x)\) has the claimed asymptotic series (in fact the same series
as the Binet function \(J\).)
This proves the first claim.
To prove the final claim,
suppose, by way of contradiction, that
the series (30) envelops \(f\).
For each integer \(k > 0\),
define \(x_k := k/\pi\).
From (7), the \(\beta_k\) grow like
\((2k)!/(2\pi)^{2k}\),
and from Stirling’s approximation we see that
\begin{equation} \label{eq:constraint}
\beta_k/x_k^{2k+1} = O(\exp(-2\pi x_k))
\;\text{ as }\; k \to \infty.
\end{equation}
(31)
Since the same series envelops both \(f\) and \(J\),
(31) implies that
\[
|f(x_k)-J(x_k)| = O(\exp(-2\pi x_k))
\;\text{ as }\; k \to \infty.
\]
Since \(\exp(-2\pi x) = o(\exp(-bx))\), it follows that,
for sufficiently large \(k\),
\[
|f(x_k)-J(x_k)| < \exp(-bx_k).
\]
This contradicts the definition of \(f\),
so the assumption that the series (30) envelops \(f\)
must be false.
Remark 3.
Lemma 4 can be generalised. For example, the
conclusion holds if \(f(x) = J(x) + g(x)\),
where \(g(x) = O(x^{-k})\) for all positive integers \(k\),
but \(g(x) \ne O(\exp(-2\pi x))\).
Also, we can replace the function \(J(x)\) by a different function
that has an enveloping asymptotic series whose terms grow at
the same rate as those of \(J(x)\).
Our second class of examples involves asymptotic expansions where all
(or all but a finite number) of the terms are of the same sign
(assuming a positive real argument \(x\)).
Such series can not be strictly enveloping [
3,
Ch.4].
As examples, we mention the Bessel function \(I_0(x)\)
(see Olver and Maximon [
16,§10.40.1]),
the product of two Bessel functions \(I_0(x)K_0(x)\)
(see [
16,
§10.40.6] and [
17, Lemma 3.1]),
and the Riemann-Siegel theta function (see [
18, §6.5]).
In all these examples the terms have constant sign, so the remainder changes
monotonically as the number of terms increases with the argument \(x\) fixed.
Eventually the remainder changes sign and starts increasing in absolute
value. Often the point where the remainder changes sign
is close to where the terms are smallest in absolute value,
but this is not always true-
see for example [
19, §§4-5].
5. Concluding remarks
We have considered three different but related asymptotic series
that can all be proved to be strictly enveloping. Our proofs depend on the
fact that the three relevant functions
\(-\ln(1-e^{-2\pi\eta})\), \(\ln\coth(\pi\eta)\),
and \(\ln(1+e^{-2\pi\eta})\) are positive for all \(\eta\in(0,\infty)\).
We remark that these three functions are linearly dependent,
since
\[
\coth(\pi\eta) = \frac{1+e^{-2\pi\eta}}{1-e^{-2\pi\eta}}\,.
\]
It follows that the sequences \((\beta_k)_{k\ge 0}\),
\((\widetilde{\beta}_k)_{k\ge 0}\) and \((\widehat{\beta}_k)_{k\ge 0}\)
are linearly dependent. In fact,
\(\widetilde{\beta}_k = \beta_k + \widehat{\beta}_k\) for all \(k\ge 0\).
A table of numerical values is given in the Appendix.
Acknowledgments
We thank an anonymous referee
for simplifying the proof of Lemma 4 and
for noting that the domain of validity of (2)
is the right half-plane \(\Re z > 0\)
(although the Binet function \(J(x)\)
may be continued into the left half-plane by
analytic continuation, see [
1, Thm.8.5a]).
Conflicts of Interest
“The author declares no conflict of interest”.
