§ 2. The gamma function in the complex domain

1. Extending the gamma function to c

Let us return to Weierstrass’ formula which gives the expression

\frac{1}{\Gamma(x)} = x\, e^{\gamma x} \prod_{n=1}^{\infty} \left(1 + \frac{x}{n}\right) e^{-x/n}

for $1/\Gamma(x)$ for all real $x$ , and consider the infinite product with general term $\left(1 + \frac{z}{n}\right) e^{-z/n}$ for arbitrary complex $z$ . One can write $e^{-z/n} = 1 - \frac{z}{n} + h(z)$ , with $|h(z)| \leq \frac{|z|^2}{2n^2} e^{|z/n|}$ (III, p. 106, formula (8)), whence

\left(1 + \frac{z}{n}\right) e^{-z/n} = 1 + v_n(z)

with $|v_n(z)| \leq \frac{|z|^2}{n^2} \left(1 + \frac{e^{|z|}}{2}(1 + |z|)\right)$ ; thus the infinite product under consideration is absolutely and uniformly convergent on every compact subset of $\mathbf{C}$ ; further, its value is zero only at the points $z = -n$ (\emph{Gen. Top.}, IX, p. 214, corollary). In view of the formula (1) of VII, p. 315, for every complex $z$ one puts

\frac{1}{\Gamma(z)} = z\, e^{\gamma z} \prod_{n=1}^{\infty} \left(1 + \frac{z}{n}\right) e^{-z/n}.

The function $\Gamma(z)$ is thus defined for every point $z \in \mathbf{C}$ apart from the points $-n$ ( $n \in \mathbf{N}$ ); it is continuous on this set, and $(z + n)\Gamma(z) \sim \frac{(-1)^n}{n!}$ on a neighbourhood of $-n$ . Formula (2) shows that one has $\Gamma(\overline{z}) = \overline{\Gamma(z)}$ for every $z$ different from a negative integer.

The argument that allows one to pass from Gauss’ formula (VII, p. 307, formula (8)) to Weierstrass’ formula, on reversing the steps, applies also to complex $z$ and shows that, for $z \neq -n$ ( $n \in \mathbf{N}$ ), one has

\Gamma(z) = \lim_{n \to \infty} \frac{n^z n!}{z(z+1)\ldots(z+n)}

on agreeing to put $n^z = e^{z \log n}$ . Since

\frac{n^{z+1} n!}{(z+1)(z+2)\ldots(z+n+1)} = z \frac{n}{n+1+z} \frac{n^z n!}{z(z+1)\ldots(z+n)}

one again has, on passing to the limit, the fundamental functional equation

\Gamma(z+1) = z\, \Gamma(z)

for every $z \neq -n$ ( $n \in \mathbf{N}$ ).

Let $p$ be an arbitrary integer $> 0$ , and $K_p$ the open disc $|z| < p$ ; for every $z \in K_p$ , and every integer $n > p$ , $1 + \frac{z}{n}$ is not a negative real number, so $\log \left( 1 + \frac{z}{n} \right)$ is defined, and it follows from the above that the series with general term $\log \left( 1 + \frac{z}{n} \right) - \frac{z}{n}$ ( $n > p$ ) is normally convergent on $K_p$ ; the same holds for the series obtained by differentiating the general term a finite number of times, since one has

\left| \frac{1}{n} - \frac{1}{z+n} \right| \leq \frac{p}{n(n-p)} \quad \text{and} \quad \left| \frac{1}{(z+n)^k} \right| \leq \frac{1}{(n-p)^k} \qquad (k > 1)

for $z \in K_p$ and $n > p$ . One then sees (cf. II, p. 59, Remark 3) that $\Gamma(z)$ is indefinitely differentiable at all the points $z \in \mathbf{C}$ apart from the points $-n$ , and at these points one has

\frac{\Gamma'(z)}{\Gamma(z)} = -\gamma - \frac{1}{z} + \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{z+n} \right)

(5)

D^{k-1} \left( \frac{\Gamma'(z)}{\Gamma(z)} \right) = \sum_{n=0}^{\infty} \frac{(-1)^k (k-1)!}{(z+n)^k} \qquad \text{for} \quad k \geq 2,

(6)

the right-hand sides of (5) and (6) being normally convergent on every compact subset of $\mathbf{C}$ that does not contain any integer $\leq 0$ . Further, one can write

\log \Gamma(z) \equiv -\gamma z - \log z + \sum_{n=1}^{\infty} \left( \frac{z}{n} - \log \left( 1 + \frac{z}{n} \right) \right) \qquad (\text{mod. } 2\pi i)

(7)

agreeing that when a logarithm in this formula is that of a real negative number it has one or the other limit values (differing by $2\pi i$ ) of $\log z$ at this point; the series on the right-hand side of (7) is then normally convergent on every compact subset of $\mathbf{C}$ not containing any integer $\leq 0$ .

2. The complements’ relation and the legendre-gauss multiplication formula

One derives immediately from formula (2) of VII, p. 315, that, for every $z \in \mathbf{C}$ ,

\frac{1}{\Gamma(z)\Gamma(-z)} = -z^2 \prod_{n=1}^{\infty} \left( 1 - \frac{z^2}{n^2} \right).

Now the Euler expansion of $\sin z$ (VI, p. 287, th. 2) shows that

z \prod_{n=1}^{\infty} \left( 1 - \frac{z^2}{n^2} \right) = \frac{1}{\pi} \sin \pi z;

taking account of the functional equation (4) of VII, p. 315, one then sees that:

Proposition 1

For every complex $z$ one has

\frac{1}{\Gamma(z)\Gamma(1-z)} = \frac{1}{\pi} \sin \pi z

(the complements' relation).

COROLLARY — For every real $t$ one has

|\Gamma(it)| = \sqrt{\frac{\pi}{t \sinh \pi t}} \quad (t \neq 0)

|\Gamma(\frac{1}{2} + it)| = \sqrt{\frac{\pi}{\cosh \pi t}}.

Indeed, one deduces from (8) that $\Gamma(it)\Gamma(-it) = \frac{i\pi}{t \sin \pi it} = \frac{\pi}{t \sinh \pi t}$ , and one has $\Gamma(-it) = \overline{\Gamma(it)}$ ; similarly, (8) gives

\Gamma\left(\frac{1}{2} + it\right)\Gamma\left(\frac{1}{2} - it\right) = \frac{\pi}{\sin (\frac{\pi}{2} + \pi it)} = \frac{\pi}{\cos \pi it} = \frac{\pi}{\cosh \pi t},

and one has

\Gamma\left(\frac{1}{2} - it\right) = \overline{\Gamma\left(\frac{1}{2} + it\right)}.

Now let $p$ be any integer $> 0$ and consider the product

f(z) = \Gamma\left(\frac{z+1}{p}\right) \Gamma\left(\frac{z+2}{p}\right) \ldots \Gamma\left(\frac{z+p}{p}\right).

By (3) (VII, p. 315), for every $z \neq -n \ (n \in \mathbf{N})$ , $f(z)$ is the limit of the product

\frac{n^{(z+1)/p} n!}{\left(\frac{z+1}{p}\right) \left(\frac{z+1}{p} + 1\right) \ldots \left(\frac{z+1}{p} + n\right)}

\frac{n^{(z+2)/p} n!}{\left(\frac{z+2}{p}\right) \left(\frac{z+2}{p} + 1\right) \ldots \left(\frac{z+2}{p} + n\right)} \cdots

\ldots \frac{n^{(z+p)/p} n!}{\left(\frac{z+p}{p}\right) \left(\frac{z+p}{p} + 1\right) \ldots \left(\frac{z+p}{p} + n\right)}

= \frac{n^{z+(p+1)/2} p^{(n+1)p} (n!)^p}{(z+1)(z+2) \ldots (z+(n+1)p)}

and in particular $f(0)$ is the limit of the product

\frac{n^{(p+1)/2} p^{(n+1)p} (n!)^p}{((n+1)p)!}

from which it follows that $f(z)/f(0)$ is the limit of

\frac{n^z ((n+1)p)!}{(z+1)(z+2)\ldots(z+(n+1)p)} = z\ p^{-z} \left( \frac{n}{n+1} \right)^z \cdot \frac{((n+1)p)^z ((n+1)p)!}{z(z+1)(z+2)\ldots(z+(n+1)p)}

which, by (3) (VII, p. 315), gives

f(z) = f(0)\ z\ p^{-z}\ \Gamma(z).

But one can write

f(0) = \prod_{k=1}^{p-1} \Gamma \left( \frac{k}{p} \right) = \prod_{k=1}^{p-1} \Gamma \left( 1 - \frac{k}{p} \right) = \sqrt{\prod_{k=1}^{p-1} \Gamma \left( \frac{k}{p} \right) \Gamma \left( 1 - \frac{k}{p} \right)}

since

f(0) > 0

; the complements’ relation then gives

f(0) = \sqrt{\pi^{p-1} / \prod_{k=1}^{p-1} \sin \frac{k\pi}{p}}

and since the product on the right-hand side is equal to

p/2^{p-1}

(VI, p. 284, cor. 1) one sees finally that:

Proposition 2

For every complex number* $z$ not an integer $\leqslant 0$ and for every integer $p > 0$ one has

\Gamma \left( \frac{z}{p} \right) \Gamma \left( \frac{z+1}{p} \right) \ldots \Gamma \left( \frac{z+p-1}{p} \right) = (2\pi)^{(p-1)/2}\ p^{\frac{1}{2}-z}\ \Gamma(z)

(Legendre-Gauss multiplication formula).

Proposition 3

For every real* $x > 0$ one has

\int_1^{x+1} \log \Gamma(t)\ dt = x(\log x - 1) + \frac{1}{2} \log 2\pi

(Raabe’s integral).

First we establish formula (13) for $x = 0$ . Since $\log \Gamma(x) \sim \log \frac{1}{x}$ as $x$ tends to 0, the integral $\int_0^1 \log \Gamma(x)\ dx$ converges. Further, the function $\log \Gamma(x)$ decreases on $]0, 1]$ (VII, p. 310); for every $\alpha > 0$ one thus has

\frac{1}{n} \sum_{k=1}^q \log \Gamma \left( \frac{k}{n} \right) \leqslant \int_0^\alpha \log \Gamma(x)\ dx,

$q$ being the largest integer such that $q/n \leqslant \alpha$ . Since $\int_0^\alpha \log \Gamma(x) dx$ tends to 0 with $\alpha$ and also $\frac{1}{n} \sum_{k=q+1}^n \log \Gamma \left( \frac{k}{n} \right)$ tends to $\int_\alpha^1 \log \Gamma(x) dx$ as $n$ tends to $+\infty$ (II, p. 57, prop. 5) one has

\int_0^1 \log \Gamma(x) dx = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \log \Gamma \left( \frac{k}{n} \right).

But, by (12), the right-hand side of this formula is the limit of

\frac{n-1}{2n} \log 2\pi - \frac{1}{2} \frac{\log n}{n},

whence

\int_0^1 \log \Gamma(x) dx = \frac{1}{2} \log 2\pi.

(14)

Next we remark that from the identity

\log \Gamma(x+1) = \log \Gamma(x) + \log x

one deduces, on integrating, that for $x > 0$

\int_0^x \log \Gamma(t+1) dt = \int_0^x \log \Gamma(t) dt + \int_0^x \log t dt.

But the integral on the left-hand side is also equal to $\int_1^{x+1} \log \Gamma(t) dt$ . Thus, by (14),

\int_1^{x+1} \log \Gamma(t) dt = \int_0^x \log t dt + \frac{1}{2} \log 2\pi = x (\log x - 1) + \frac{1}{2} \log 2\pi.

3. Stirling's expansion

Let $x$ and $y$ be two complex numbers not lying on the real negative half-axis; by formula (3) of VII, p. 315, with the conventions of VII, p. 316, concerning the logarithms, $\log \Gamma(x) - \log \Gamma(y)$ is congruent modulo $2\pi i$ to the limit of the expression

(x-y) \log n + \sum_{k=0}^n \left( \log(y+k) - \log(x+k) \right).

(15)

Let us put $f(t) = \log(y+t) - \log(x+t)$ ; we apply the Euler-Maclaurin summation formula (VI, p. 288) to the function $f$ :

f(0) + f(1) + \cdots + f(n) = \int_0^{n+1} f(t) dt - \frac{1}{2} (f(n+1) - f(0))

+ \sum_{k=1}^p \frac{b_{2k}}{(2k)!} \left( f^{(2k-1)}(n+1) - f^{(2k-1)}(0) \right) + T_p(n)

with

|T_p(n)| \leq \frac{4 e^{2\pi}}{(2\pi)^{2p+1}} \int_0^{n+1} |f^{(2p+1)}(u)|\ du.

(16)

Since

f^{(m)}(t) = (-1)^{m-1} (m-1)! \left( \frac{1}{(y+t)^m} - \frac{1}{(x+t)^m} \right),

f^{(2k-1)}(n+1)

tends to 0 as

n

tends to

+\infty

, for all

k \geq 1

; it is the same for

f(n+1) = \log \left( 1 + \frac{y}{n+1} \right) - \log \left( 1 + \frac{x}{n+1} \right).

Moreover, one has

\int_0^{n+1} \log(x+t)\ dt = (x+n+1)(\log(x+n+1)-1) - x(\log x - 1);

n

tends to

+\infty

, one has the asymptotic expansion

(x+n)(\log(x+n)-1) = n \log n - n + x \log n + O \left( \frac{1}{n} \right).

Substituting in the expression (15) one sees finally that, as

n

tends to

+\infty

T_p(n)

has a limit

R_p(x,\ y)

and that one can write

\log \Gamma(x) - g(x) \equiv \log \Gamma(y) - g(y) + R_p(x,\ y) \quad (\text{mod. } 2\pi i)

on putting

g(x) = x \log x - x - \frac{1}{2} \log x + \sum_{k=1}^p \frac{b_{2k}}{2k(2k-1)} \frac{1}{x^{2k-1}}.

(17)

We shall now evaluate a bound for $R_p(x,\ y)$ with the help of (16), assuming that $x$ and $y$ are both in the subset $\mathbf{H}_A$ of $\mathbf{C}$ defined by the relation " $\mathscr{R}(z) \geq A$ or $|\mathscr{I}(z)| \geq A$ ", where $A$ is an arbitrary number $> 0$ (fig. 2). To this end we remark that if $x = s + it$ with $s > A$ one has $|x+u| \geq A+u$ for every $u > 0$ , and consequently

\int_0^{n+1} \frac{du}{|x+u|^{2p+1}} \leq \int_0^\infty \frac{du}{(A+u)^{2p+1}} = \frac{1}{2pA^{2p}}.

Similarly, if

|t| \geq A

one has

|x+u| = |s+u+it| \geq \sqrt{A^2 + (s+u)^2}

for all real

u

, whence

\int_0^{n+1} \frac{du}{|x+u|^{2p+1}} \leq \int_{-\infty}^{+\infty} \frac{du}{(A^2 + u^2)^{p+1/2}} = \frac{2}{A^{2p}} \int_0^\infty \frac{dv}{(1+v^2)^{p+1/2}}.

Thus one sees that, when

x

and

y

are in

\mathbf{H}_A

, one has

|R_p(x,\ y)| \leq \frac{C_p}{A^{2p}}

![A complex plane diagram showing axes labeled t, s, and regions H_A, A, -A, 0](Fig. 2)

where C_p depends only on p. Now let $\mathfrak{F}$ be the filter having the sets H_A as basis; the Cauchy criterion shows that, along the filter $\mathfrak{F}$ , the function $\log \Gamma(z) - g(z)$ has a finite limit $\delta$ (modulo $2\pi i$ ) and that, if one puts $\omega(z) = \max(\mathscr{R}(z), |\mathscr{I}(z)|)$ , one has

\log \Gamma(z) - g(z) - \delta \equiv O \left( \frac{1}{(\omega(z))^{2p}} \right) \quad (\text{mod. } 2\pi i).

For x real and > 0 one has $\Gamma(x) > 0$ , and $g(x)$ is real, so one can assume that $\delta$ is real and one has

\log \Gamma(x) = g(x) + \delta + O \left( \frac{1}{x^{2p}} \right).

Now we shall deduce the value of the constant $\delta$ : by prop. 2 of VII, p. 318, applied for $p = 2$ , one has, for real $x$ tending to $+\infty$

\begin{align*} &\frac{x-1}{2} \log \frac{x}{2} - \frac{x}{2} + \frac{x}{2} \log \frac{x+1}{2} - \frac{x+1}{2} + 2\delta \\ &\qquad = x \log x - x - \frac{1}{2} \log x + (\frac{1}{2} - x) \log 2 + \frac{1}{2} \log 2\pi + \delta + o(1) \end{align*}

from which one deduces easily that $\delta = \frac{1}{2} \log 2\pi$ . Finally one has the following result:

Proposition 4

Along the filter $\mathfrak{F}$ one has (for every integer $p \geqslant 1$ ) the asymptotic expansion

\log \Gamma(z) \equiv z \log z - z - \frac{1}{2} \log z + \frac{1}{2} \log 2\pi + \sum_{k=1}^{p} \frac{b_{2k}}{2k(2k-1)} \frac{1}{z^{2k-1}} + O \left( \frac{1}{(\omega(z))^{2p}} \right) \quad (\text{mod. } 2\pi i)

(Stirling’s expansion).

Corollary

Along the filter $\mathfrak{F}$ one has

\Gamma(z) \sim \sqrt{2\pi} \exp(z \log z - z - \frac{1}{2} \log z).

In particular, for $x$ real tending to $+\infty$ the formula (20) can be written

\Gamma(x) \sim \sqrt{2\pi} \, x^{x-1/2} e^{-x},

whence, as the integer $n$ tends to $+\infty$ ,

n! \sim \sqrt{2\pi} \, n^{n+1/2} \, e^{-n}

(cf. V, p. 244).

One can deduce numerous formulae from this. For example, for every complex number $\alpha$ and every integer $n$ one has, as $n$ tends to $+\infty$ ,

\frac{\Gamma(n+\alpha)}{\Gamma(n)} \sim n^{\alpha} \quad (= e^{\alpha \log n}).

Similarly, for every complex number $a$ not an integer $\leqslant 0$ one has

a(a+1)(a+2)\ldots(a+n) = \frac{\Gamma(n+a+1)}{\Gamma(a)} \sim \frac{\sqrt{2\pi}}{\Gamma(a)} \, n^{n+a+1/2} \, e^n

and for every complex number $a$ not an integer $\geqslant 0$

\binom{a}{n} = \frac{(-1)^n}{\Gamma(-a)} \frac{\Gamma(n-a)}{\Gamma(n+1)} \sim \frac{(-1)^n}{\Gamma(-a)} \, n^{-a-1}.

Finally, for every real constant $k > 1$ one has

\binom{kn}{n} = \frac{\Gamma(kn+1)}{\Gamma(n+1) \Gamma((k-1)n+1)} \sim \sqrt{\frac{k}{2\pi(k-1)n}} \left( \frac{k^k}{(k-1)^{k-1}} \right)^n.

The same argument leads to the following analogous proposition:

Proposition 5

Along the filter $\mathfrak{F}$ one has (for every integer $p \geqslant 1$ ), the asymptotic expansion

\frac{\Gamma'(z)}{\Gamma(z)} = \log z - \frac{1}{2z} - \sum_{k=1}^{p} \frac{b_{2k}}{2k} \frac{1}{z^{2k}} + O \left( \frac{1}{(\omega(z))^{2p+1}} \right).

Instead of prop. 2 of VII, p. 318, one uses the formula

\int_{x}^{x+1} \frac{\Gamma'(t)}{\Gamma(t)} \, dt = \log \Gamma(x+1) - \log \Gamma(x) = \log x

in order to determine the constant.

Exercises

See the exercises for § 2.