Special functions · Complex analysis
The Gamma function began as an answer to "what is \(\frac12!\)?". Viewed as a function of a complex variable it becomes much more: a function analytic everywhere except at \(0, -1, -2, \dots\), with an infinite product formula, a reflection formula linking it to the sine, and a whole family of identities that make it one of the most useful special functions in mathematics and physics. This post takes the complex point of view: how the integral definition is extended to the whole plane, and what the main formulas say and why they hold.
Here \( t^{z-1} = e^{(z-1)\ln t} \) with the real logarithm of \(t > 0\), so \( |t^{z-1}| = t^{\operatorname{Re}z - 1} \). This explains the condition \( \operatorname{Re}z > 0 \): near \(t = 0\) the integrand behaves like \(t^{\operatorname{Re}z - 1}\), which is integrable exactly when \( \operatorname{Re}z > 0 \); at infinity \(e^{-t}\) wins over any power. The integral converges uniformly on every strip \( \delta \le \operatorname{Re}z \le M \), and differentiation under the integral sign (see the posts on parametric integrals) shows that \(\Gamma\) is analytic in the right half-plane.
Integrating by parts gives \( \Gamma(z + 1) = z\Gamma(z) \), and since \(\Gamma(1) = 1\), \( \Gamma(n + 1) = n! \): the Gamma function interpolates the factorials. The real-variable story, including \( \Gamma(\frac12) = \sqrt\pi \) and many integrals evaluated with \(\Gamma\) and \(\mathrm B\), is in the earlier post on Euler integrals; here we go into the complex plane.
How it works. Apply (2) repeatedly: \( \Gamma(z + n + 1) = (z + n)(z + n - 1)\cdots z\,\Gamma(z) \). Now read this backwards. The right-hand side of the boxed formula makes sense whenever \( \operatorname{Re}(z + n + 1) > 0 \), i.e. \( \operatorname{Re}z > -n - 1 \), provided the denominator is not zero. So it defines \(\Gamma\) in the bigger half-plane \( \operatorname{Re}z > -n - 1 \), and it agrees with (1) where both make sense. Taking \(n = 0, 1, 2, \dots\) pushes the boundary as far left as we like.
By the uniqueness of analytic continuation, the different \(n\) give consistent values: there is only one way to extend an analytic function to a larger domain. The result is a function analytic in \( \mathbb C\setminus\{0, -1, -2, \dots\} \), i.e. a meromorphic function.
Near \(z = -n\), only the factor \(z + n\) in the denominator vanishes, so \(\Gamma\) has a simple pole there, with residue
\[ \operatorname*{Res}_{z=-n}\Gamma(z) = \lim_{z\to-n}(z + n)\Gamma(z) = \frac{\Gamma(1)}{(-n)(-n + 1)\cdots(-1)} = \frac{(-1)^n}{n!}. \]Another fact, less obvious but very important: \(\Gamma(z)\) is never zero. So \(\frac{1}{\Gamma(z)}\) is analytic in the whole plane (an entire function), with zeros exactly at \(0, -1, -2, \dots\). That is why the product formula below is written for \(\frac1\Gamma\) rather than for \(\Gamma\).
Where it comes from. For \( \operatorname{Re}z > 0 \), replace \(e^{-t}\) in (1) by its classical approximation \( \big(1 - \frac tk\big)^k \) on \([0, k]\):
\[ \int_0^k t^{z-1}\Big(1 - \frac tk\Big)^k dt = k^z\int_0^1 s^{z-1}(1 - s)^k\,ds = k^z\,\mathrm B(z, k + 1) = \frac{k!\,k^z}{z(z + 1)\cdots(z + k)}, \]and as \(k \to \infty\) the left side tends to \(\Gamma(z)\). The right side, however, makes sense for all \(z\) except the poles, which gives a formula valid in the whole plane. The intuition: \(\Gamma(z + k + 1) \approx k!\,k^z\) for large \(k\) (the factorial grows by a factor of about \(k\) per unit step), and the formula of Section 2 does the rest.
This is the factorization of the entire function \(\frac1\Gamma\) by its zeros \(0, -1, -2, \dots\), in the same spirit as factoring a polynomial by its roots. The factors \(e^{-z/k}\) are needed because \(\prod(1 + \frac zk)\) alone diverges (the harmonic series again); they make the infinite product converge (see the post on infinite products). It follows from Gauss's formula: write \( k^z = e^{z\ln k} = e^{z(\ln k - 1 - \frac12 - \cdots - \frac1k)}\,e^{z(1 + \frac12 + \cdots + \frac1k)} \) and let \(k \to \infty\); the first exponential tends to \(e^{-\gamma z}\).
Why. From Weierstrass's product, \( \frac{1}{\Gamma(z)\Gamma(-z)} = -z^2\prod_{k\ge1}\big(1 - \frac{z^2}{k^2}\big) \) (the exponential factors cancel in pairs). By Euler's product for the sine, \( \sin\pi z = \pi z\prod_{k\ge1}\big(1 - \frac{z^2}{k^2}\big) \). Since \( \Gamma(1 - z) = -z\Gamma(-z) \), the formula follows. Both sides are meromorphic with the same poles, at all the integers.
Consequences:
The most used special case is \(n = 2\), Legendre's duplication formula:
\[ \Gamma(z)\,\Gamma\Big(z + \frac12\Big) = 2^{1 - 2z}\sqrt\pi\,\Gamma(2z). \]It can be proved with Gauss's limit: write all the Gamma values as limits and compare. For integer arguments it says something about factorials; e.g. \(z = 1\): \( 1\cdot\Gamma(\frac32) = \frac12\sqrt\pi\cdot 1! \), true since \( \Gamma(\frac32) = \frac12\Gamma(\frac12) \).
Why: take the logarithm of Weierstrass's product and differentiate term by term. A product turns into a sum, which is exactly why one prefers to work with \(\ln\Gamma\):
\[ -\ln\Gamma(z) = \ln z + \gamma z + \sum_{k\ge1}\Big[\ln\Big(1 + \frac zk\Big) - \frac zk\Big] \ \Rightarrow\ -\psi(z) = \frac1z + \gamma + \sum_{k\ge1}\Big(\frac{1}{z + k} - \frac1k\Big), \]which, after moving the term \(\frac1z\) into the sum, is the boxed formula.
Useful values: \( \psi(1) = -\gamma \), i.e. \( \Gamma'(1) = -\gamma \approx -0.5772 \), and from \( \psi(z + 1) = \psi(z) + \frac1z \) (the logarithmic derivative of (2)),
\[ \psi(n + 1) = -\gamma + 1 + \frac12 + \cdots + \frac1n. \]So the digamma function interpolates the harmonic numbers, just as \(\Gamma\) interpolates the factorials. (Check: \( \psi(4) = -\gamma + \frac{11}{6} \approx 1.2561 \).)
The symmetry follows from the substitution \(t \to 1 - t\). The second identity, proved in the post on Euler integrals by writing \(\Gamma(p)\Gamma(q)\) as a double integral, does more than evaluate integrals: it shows that \(\mathrm B\) extends to all complex \(p, q\) (as a meromorphic function of each variable), since the right side does. It also gives the common trigonometric form
\[ \int_0^{\pi/2}\sin^{2p-1}\theta\,\cos^{2q-1}\theta\,d\theta = \frac12\mathrm B(p, q). \]| Formula | Meaning |
|---|---|
| \( \Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt \), \(\operatorname{Re}z > 0\) | definition |
| \( \Gamma(z + 1) = z\Gamma(z) \) | functional equation; continuation to \(\mathbb C\setminus\{0, -1, \dots\}\) |
| \( \operatorname{Res}_{-n}\Gamma = \frac{(-1)^n}{n!} \) | simple poles; \(\frac1\Gamma\) is entire |
| \( \Gamma(z) = \lim\frac{k!\,k^z}{z(z + 1)\cdots(z + k)} \) | Gauss's limit |
| \( \frac1\Gamma = ze^{\gamma z}\prod(1 + \frac zk)e^{-z/k} \) | Weierstrass's product |
| \( \Gamma(z)\Gamma(1 - z) = \frac{\pi}{\sin\pi z} \) | reflection |
| \( \prod_{j=0}^{n-1}\Gamma(z + \frac jn) = n^{\frac12 - nz}(2\pi)^{\frac{n-1}{2}}\Gamma(nz) \) | multiplication (duplication for \(n = 2\)) |
| \( \psi = \frac{\Gamma'}{\Gamma} = -\gamma + \sum(\frac1n - \frac{1}{z + n - 1}) \) | digamma; \(\psi(n + 1) = -\gamma + H_n\) |
| \( \mathrm B(p, q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p + q)} \) | Beta via Gamma |
Try each one before opening the solution. More solved problems from this topic are in Complex Numbers.
\( \Gamma(-\frac52) = \frac{\Gamma(\frac12)}{(-\frac52)(-\frac32)(-\frac12)} = -\frac{8}{15}\sqrt\pi \approx -0.945 \).
\( \frac{(-1)^3}{3!} = -\frac16 \).
\( \frac{\pi}{\sin\frac\pi4} = \pi\sqrt2 \).
With \( z = \frac12 + iy \): \( 1 - z = \frac12 - iy = \bar z \), so \( \Gamma(z)\Gamma(\bar z) = |\Gamma(z)|^2 = \frac{\pi}{\sin(\frac\pi2 + i\pi y)} = \frac{\pi}{\cos(i\pi y)} = \frac{\pi}{\cosh\pi y} \).
\( \Gamma'(2) = \Gamma(2)\psi(2) = 1\cdot(-\gamma + 1) \approx 0.4228 \).
\( 2p - 1 = 4 \), \( 2q - 1 = 2 \): \( p = \frac52 \), \( q = \frac32 \). \( \frac12\mathrm B(\frac52, \frac32) = \frac12\cdot\frac{\frac34\sqrt\pi\cdot\frac12\sqrt\pi}{\Gamma(4)} = \frac12\cdot\frac{3\pi/8}{6} = \frac{\pi}{32} \).
In the complex plane, the Gamma function shows its full structure. The functional equation \(\Gamma(z + 1) = z\Gamma(z)\), read backwards, carries Euler's integral across the whole plane, leaving simple poles at \(0, -1, -2, \dots\). Gauss's limit and Weierstrass's product describe \(\Gamma\) without any integral and display the zeros of \(\frac1\Gamma\); the reflection formula ties \(\Gamma\) to the sine and shows that \(\Gamma\) never vanishes; the multiplication and digamma formulas follow from the product. And the Beta function, through \( \mathrm B = \frac{\Gamma\Gamma}{\Gamma} \), comes along for free.
Related posts: Euler integrals (the real-variable introduction), infinite products, and integrals depending on a parameter.