UDC 517.5 + 517.6

MATHEMATICS

V. M. KALININ

ANALOGS OF STIRLING’S FORMULA

(Presented by Academician Yu. V. Linnik on 9 X 1967)

Stirling’s formula gives an asymptotic expansion of the gamma function (\Gamma(1+x)) as (x \to \infty) in inverse powers of (x):

[
\Gamma(1+x)=\sqrt{2\pi x}\left(\frac{x}{e}\right)^x
\exp\left{\sum_{j=1}^{\nu-1}\frac{B_{j+1}}{j(j+1)x^j}
+O\left(\frac{1}{x^\nu}\right)\right},
]

[
\Gamma(1+x)=\sqrt{2\pi x}\left(\frac{x}{e}\right)^x
\left{1+\sum_{j=1}^{\nu-1}\frac{b_j}{x^j}
+O\left(\frac{1}{x^\nu}\right)\right},
]

[
\frac{1}{\Gamma(1+x)}=\frac{1}{\sqrt{2\pi x}}\left(\frac{e}{x}\right)^x
\left{1+\sum_{j=1}^{\nu-1}(-)^j\frac{b_j}{x^j}
+O\left(\frac{1}{x^\nu}\right)\right},
]

where the coefficients (b_j) can be found from the recurrence relations

[
b_j=\frac{1}{j}\sum_{k=0}^{j-1}\frac{b_k B_{j-k+1}}{j-k+1},
\qquad b_0=1.
]

Here (B_{j-k+1}) are Bernoulli numbers. (Somewhat different recurrence equations are indicated in (1).) The Bernoulli numbers are easily computed recursively:

[
B_j=(-)^j j\sum_{k=0}^{j-1}\frac{B_k C_{j-1}^k}{j-k+1},
\qquad B_1=-\frac{1}{2}.
]

In applications (for example, in probability theory) it is often convenient to expand the gamma function in powers of a quantity different from (x). The following theorems introduce into Stirling’s formula an arbitrary parameter (\theta), by the choice of which one can obtain the necessary expansions.

Theorem 1. For (x>-1), (\nu=2,3,\ldots), and arbitrary (\theta>-x)

[
\Gamma(1+x)=\sqrt{2\pi}\,(x+\theta)^{x+1/2}
\exp\left{-(x+\theta)+
\sum_{j=1}^{\nu-1}\frac{B_{j+1}(\theta)}{j(j+1)(x+\theta)^j}
+R_\nu\right},
]

where

[
R_\nu=-\frac{1}{\nu}\int_0^1 B_\nu(t)\zeta(\nu,x+t)\,dt
+\frac{1}{\nu}\int_1^\theta \frac{B_\nu(t)}{(x+t)^\nu}\,dt,
]

[
\zeta(\nu,x)=\sum_{k=1}^{\infty}\frac{1}{(x+k)^\nu},
]

(B_{j+1}(\theta)) are Bernoulli polynomials. They may be found, for example, from the following recurrence formulas:

[
(-)^j\frac{B_j(\theta)}{j!}
=
\sum_{k=0}^{j-1}(-)^k\frac{B_k(\theta)}{k!}
\frac{(\theta-1)^{j-k+1}-\theta^{j-k+1}}{(j-k+1)!},
\qquad B_0(\theta)=1.
]

The proof of Theorem 1 is close to the proof of Lemma 1 of paper (2); the difference is that in Taylor’s formula, on which the summation theorem is based, the remainder term is taken in integral form.

Theorem 2. As (x\to\infty), (\nu=1,2,\ldots), uniformly with respect to arbitrary (\theta=O(1)), the expansions

[
\Gamma(1+x)=\sqrt{2\pi}\,(x+\theta)^{x+1/2}
\exp\left{-(x+\theta)+\sum_{j=1}^{\nu-1}
\frac{B_{j+1}(\theta)}{j(j+1)(x+\theta)^j}
+O\left(\frac{1}{(x+\theta)^\nu}\right)\right},
]

[
\Gamma(1+x)=\sqrt{2\pi}\,(x+\theta)^{x+1/2}e^{-(x+\theta)}
\left{1+\sum_{j=1}^{\nu-1}\frac{b_j(\theta)}{(x+\theta)^j}
+O\left(\frac{1}{(x+\theta)^\nu}\right)\right},
]

[
\frac{1}{\Gamma(1+x)}=
\frac{e^{x+\theta}}{\sqrt{2\pi}\,(x+\theta)^{x+1/2}}
\left{1+\sum_{j=1}^{\nu-1}(-1)^j
\frac{b_j(1-\theta)}{(x+\theta)^j}
+O\left(\frac{1}{(x+\theta)^\nu}\right)\right},
]

hold, where (b_j(\theta)) are polynomials of degree (2j), satisfying the recurrence relations

[
b_j(\theta)=\frac{1}{j}\sum_{k=0}^{j-1}
\frac{b_k(\theta)B_{j-k+1}(\theta)}{j-k+1},
\qquad b_0(\theta)=1,
]

[
\Delta b_j(\theta)=\theta b_{j-1}(1+\theta).
]

In particular, from Theorem 2 we find, as (x\to\infty),

[
\Gamma(1+x)=[x]!\,x^{{x}}
\exp\left{\sum_{j=1}^{\nu-1}
\frac{B_{j+1}-B_{j+1}({x})}{j(j+1)x^j}
+O\left(\frac{1}{x^\nu}\right)\right},
]

(here ([x]) and ({x}) are the integer and fractional parts of (x)).

As (n\to\infty) and (x=O(1)), the following equality holds uniformly with respect to (x):

[
\Gamma(x)=
\frac{\sqrt{2\pi}\,n^{\,n+x+1/2}e^{-n}}
{x(x+1)\cdots(x+n)}
\exp\left{\sum_{j=1}^{\nu-1}
\frac{B_{j+1}(-x)}{j(j+1)n^j}
+O\left(\frac{1}{n^\nu}\right)\right}.
]

Theorem 3. As (x\to\infty), uniformly with respect to arbitrary (\theta=O(\sqrt{x})) and arbitrary (a=O(1)), the expansions

[
\Gamma(1+x)=\sqrt{2\pi}\,(x+\theta)^{x+1/2}
\exp\left{-(x+\theta)+\frac{y^2}{2}
+\sum_{j=1}^{\nu-1}
\frac{v_j(y,a)}{(\sqrt{x+\theta})^j}
+O\left(\frac{1}{(\sqrt{x+\theta})^\nu}\right)\right},
]

[
\Gamma(1+x)=\sqrt{2\pi}\,(x+\theta)^{x+1/2}e^{-(x+\theta)-y^2/2}
\left{1+\sum_{j=1}^{\nu-1}
\frac{w_j(y,a)}{(\sqrt{x+\theta})^j}
+O\left(\frac{1}{(\sqrt{x+\theta})^\nu}\right)\right},
]

[
\frac{1}{\Gamma(1+x)}=
\frac{e^{x+\theta-y^2/2}}{\sqrt{2\pi}\,(x+\theta)^{x+1/2}}
\left{1+\sum_{j=1}^{\nu-1}(-i)^j
\frac{w_j(iy,1-a)}{(\sqrt{x+\theta})^j}
+O\left(\frac{1}{(\sqrt{x+\theta})^\nu}\right)\right},
]

hold, where

[
y=\frac{\theta-a}{\sqrt{x+\theta}},
]

[
v_j(y,a)=
\sum_{k=0}^{1+[j/2]}
\frac{B_k(a)C_{j-k+2}^{\,k}y^{j-2k+2}}
{(j-k+1)(j-k+2)},
]

and (w_j(y,a)) are polynomials of degree (3j) in (y) of parity (j), satisfying the recurrence relations

[
w_j(y,a)=\sum_{k=0}^{j-1}
\left(1-\frac{k}{j}\right)w_k(y,a)v_{j-k}(y,a),
\qquad w_0(y,a)=1.
]

(The imaginary unit (i) enters the expression ((-i)^j w_j(iy,1-a)) only in even powers.)

Theorems 2 and 3 are direct consequences of Theorem 1. All the assertions formulated can also be given in a complex version. Thus, Theorem 2 remains valid if (x) is replaced by a complex variable (z) as (|z|\to\infty), uniformly with respect to an arbitrary complex quantity (\theta=O(1)) and with respect to (|\arg z|\le \pi-\varepsilon). Theorem 3 holds when (x) is replaced by (z), uniformly with respect to arbitrary complex quantities (\theta=O(\sqrt[\nu]{z})), (a=O(1)), and with respect to (|\arg z|\le \pi-\varepsilon). This makes it possible, in particular, to describe the asymptotic behavior of (\Gamma(1+z)) for (z=x+iy). For example, one may state the following result:

Theorem 4. For (z=x\pm iy), (y\to\infty), the following equalities hold uniformly with respect to (x=O(1)):

[
\Gamma(1+z)=\sqrt{2\pi}\,y^{x+1/2}\exp\left{-\frac{\pi}{2}y\pm i\left[\frac{\pi}{2}\left(x+\frac12\right)+y(\ln y-1)\right]+
\sum_{j=1}^{\nu-1}\frac{(\mp i)^j B_{j+1}(-x)}{j(j+1)y^j}
+O\left(\frac1{y^\nu}\right)\right},
]

[
\Gamma(1+z)=\sqrt{2\pi}\,y^{x+1/2}\exp\left{-\frac{\pi}{2}y\pm i\left[\frac{\pi}{2}\left(x+\frac12\right)+y(\ln y-1)\right]\right}\times
]

[
\times\left{1+\sum_{j=1}^{\nu-1}(\mp i)^j\frac{b_j(-x)}{y^j}
+O\left(\frac1{y^\nu}\right)\right}.
]

From the complex version of Theorem 3 one can determine the asymptotic behavior of (\Gamma(1+x\pm iy)) as (y\to\infty) and (x=O(\sqrt[\nu]{y})). For example, we immediately obtain

[
\left|\Gamma(1+x\pm iy)\right|
=
\sqrt{2\pi}\,y^{x+1/2}e^{-\frac{\pi}{2}y}
\left{1+O\left(\frac1{\sqrt[\nu]{y}}\right)\right}
]

uniformly with respect to (x=O(\sqrt[\nu]{y})).

We note that, instead of the expansions of (\Gamma(1+x)) and (\Gamma(1+z)), one may everywhere write, with some obvious changes, the expansions of (\Gamma(x)) and (\Gamma(z)).

In the last two theorems the behavior of a frequently occurring ratio is described.

Theorem 5. As (|z|\to\infty), uniformly with respect to an arbitrary complex (\theta=O(1)) and with respect to (|\arg z|\le \pi-\varepsilon), the following expansions hold:

[
\frac{\Gamma(z+\theta)}{\Gamma(z)}
=
z^\theta
\exp\left{
\sum_{j=1}^{\nu-1}(-)^j
\frac{B_{j+1}-B_{j+1}(\theta)}{j(j+1)z^j}
+O\left(\frac1{z^\nu}\right)
\right},
]

[
\frac{\Gamma(z+\theta)}{\Gamma(z)}
=
z^\theta
\left{
1+\sum_{j=1}^{\nu-1}\frac{C_\theta^{(j)}}{z^j}
+O\left(\frac1{z^\nu}\right)
\right},
]

[
\frac{\Gamma(z)}{\Gamma(z+\theta)}
=
z^{-\theta}
\left{
1+\sum_{j=1}^{\nu-1}(-)^j\frac{C_{1-\theta}^{(j)}}{z^j}
+O\left(\frac1{z^\nu}\right)
\right},
]

[
C_\theta^{(j)}
=
\theta(\theta-1)\cdots(\theta-j)
\sum_{k=0}^{j-1}g_{jk}(\theta-j-1)\cdots(\theta-j-k),
]

[
g_{j0}=\frac1{j+1},\qquad
g_{j,j-1}=\frac1{2^j j!},\qquad
g_{jk}=\frac{(j+k)g_{j-1,k}+g_{j-1,k-1}}{j+k+1},
\quad k=1,\ldots,j-2.
]

(The values of the polynomials (C_\theta^{(j)}) for (\theta=j+1,j+2,\ldots) are the Stirling numbers of the first kind (C_\theta^{(j)}), and for (\theta=-1,-2,\ldots) the Stirling numbers of the second kind (\mathfrak{S}_{-\theta}^{(j)}).)

Theorem 6. As $|z|\to\infty$, uniformly with respect to an arbitrary complex $\theta=O(1)$ and with respect to $|\arg z|\leqslant \pi-\varepsilon$, the following expansions hold:
[
\frac{\Gamma(z+\theta\sqrt z)}{\Gamma(z)}
=
z^{\theta\sqrt z}\exp\left{
\frac{\theta^2}{2}
+
\sum_{j=1}^{\nu-1}(-)^j\frac{\alpha_j}{(\sqrt z)^j}
+
O\left(\frac{1}{(\sqrt z)^\nu}\right)
\right},
]
[
\frac{\Gamma(z+\theta\sqrt z)}{\Gamma(z)}
=
z^{\theta\sqrt z}e^{\theta^2/2}
\left{
1+
\sum_{j=1}^{\nu-1}(-)^j\frac{\beta_j}{(\sqrt z)^j}
+
O\left(\frac{1}{(\sqrt z)^\nu}\right)
\right},
]
[
\frac{\Gamma(z)}{\Gamma(z+\theta\sqrt z)}
=
z^{-\theta\sqrt z}e^{-\theta^2/2}
\left{
1+
\sum_{j=1}^{\nu-1}(-)^j\frac{\gamma_j}{(\sqrt z)^j}
+
O\left(\frac{1}{(\sqrt z)^\nu}\right)
\right},
]
where
[
\alpha_j
=
\sum_{k=0}^{[(j+1)/2]}
(-)^k
\frac{C_j^k B_k \theta^{\,j-2k+2}}
{(j-k+1)(j-k+2)},
]
[
\beta_j
=
\sum_{k=0}^{j-1}
\left(1-\frac{k}{j}\right)\beta_k\alpha_{j-k},
\qquad
\beta_0=1,
]
[
\gamma_j
=
-\sum_{k=0}^{j-1}
\left(1-\frac{k}{j}\right)\gamma_k\alpha_{j-k},
\qquad
\gamma_0=1.
]

Another analogue of Stirling’s formula is contained in work (2).

Leningrad Branch
of the V. A. Steklov Mathematical Institute
Academy of Sciences of the USSR

Received
2 X 1967

REFERENCES

1. E. Copson, Asymptotic Expansions, Moscow, 1966.
2. V. M. Kalinin, Theory of Probability and Its Applications, 12, no. 1, 24 (1967).