Special Functions: Gamma, Beta, and Zeta

Gamma Function

Let \(z \in \mathbb C\) and define, for \(\operatorname{Re}(z) > 0\),

\[\Gamma(z) = \int_0^\infty e^{-t} t^{z-1} \; dt\]

This function has a particularly interesting property - consider

\[\begin{align} \Gamma(z+1) &= \int_0^\infty e^{-t} t^{z} \; dt \\ &= \underbrace{\left[ -e^{-t} t^{z} \right]_0^\infty}_0 + \int_0^\infty e^{-t} \cdot z t^{z-1} \; dt &&\text{integrating by parts}\\ &= z \Gamma(z) \end{align}\]

Observe also that

\[\Gamma(1) = \int_0^\infty e^{-t} \; dt = 1\]

which means that, for \(n \in \mathbb N\), we have

\[\Gamma(n+1) = n \Gamma(n) = n \times (n-1) \times \dots \times 1 = n!\]

so that \(\Gamma(z)\) is one possible extension of the factorial to \(\operatorname{Re}(z) > 0\). Moreover, we can extend this to all \(z \in \mathbb C \setminus \lbrace 0, -1, -2, \dots \rbrace\) - to do so, observe that

\[\Gamma(z) = \frac{\Gamma(z+1)}{z}\]

holds for \(\operatorname{Re}(z) > 0\), but that in fact, this relationship provides an analytic continuation into the region \(\operatorname{Re}(z) > -1\) (excluding \(z = 0\)), and as such, defines a meromorphic function on \(\operatorname{Re}(z) > -1\). This can then be repeated to define \(\Gamma(z)\) at all points in \(\mathbb C\) except the non-positive integers.

Beta Function

Another useful function is the Beta function, defined for \(\operatorname{Re}(z), \operatorname{Re}(w) > 0\) by

\[B(z, w) = \int_0^1 t^{z-1} (1-t)^{w-1} \; dt\]

One can derive a useful identity by considering the following:

\[\begin{align} \Gamma(z) \Gamma(w) &= \int_0^\infty e^{-t} t^{z-1} \; dt \cdot \int_0^\infty e^{-s} s^{w-1} \; ds \\ &= \int_0^\infty \int_0^\infty e^{-t} t^{z-1} e^{-s} s^{w-1} \; dt \, ds \end{align}\]

Now introduce two new variables \(\lambda, \eta\) and define

\[t = \lambda \eta, \quad s = \lambda (1 - \eta)\]

The Jacobian for this change of coordinates is given by

\[J = \frac{\partial(t, s)}{\partial(\lambda, \eta)} = \begin{pmatrix} \eta & \lambda \\ 1 - \eta & -\lambda \end{pmatrix}\]

which gives

\[\det J = -\eta \lambda - \lambda (1-\eta) = -\lambda \implies dt \, ds = \lvert \lambda \rvert \, d\lambda \, d\eta\]

Our region of integration \((t, s) \in [0, \infty)^2\) becomes \((\lambda, \eta) \in [0, \infty) \times [0, 1]\), so that

\[\begin{align} \Gamma(z) \Gamma(w) &= \int_0^\infty \int_0^1 \left[ e^{-\lambda \eta} \lambda^{z-1} \eta^{z-1} e^{\lambda(1-\eta)} \lambda^{w-1} (1-\eta)^{w-1} \right] \cdot \lambda \; d \eta \, d\lambda \\ &= \int_0^\infty \int_0^1 e^\lambda \lambda^{z+w-1} \eta^{z-1} (1-\eta)^{w-1} \; d\eta \, d\lambda \\ &= \int_0^\infty e^\lambda \lambda^{z+w-1} d\lambda \cdot \int_0^1 \eta^{z-1} (1-\eta)^{w-1} \; d\eta \\ &= \Gamma(z+w) B(z, w) \end{align}\]

and hence

\[B(z,w) = \frac{\Gamma(z) \Gamma(w)}{\Gamma(z+w)}\]

In this form, it is clear that the Beta function is symmetric in its arguments - this fact can also be deduced straightforwardly by making the transformation \(t \mapsto 1-t\) in the integral definition.

One final identity can be obtained by the substitution \(t = \sin^2 \theta\) in the definition of \(B(z, w)\). With this, we have

\[B(z, w) = \int_0^{\frac{\pi}{2}} \sin^{2z-2} \theta \cdot \cos^{2w-2} \theta \cdot 2 \sin \theta \cos \theta \, d\theta = 2\int_0^{\frac{\pi}{2}} \sin^{2z-1} \theta \cos^{2w-1} \theta \; d\theta\]

This allows one to quickly conclude that

\[B\left( \frac{1}{2}, \frac{1}{2} \right) = \pi\]

which, using the relation with \(\Gamma\) and using \(\Gamma(1) = 1\) gives

\[\Gamma\left(\frac{1}{2}\right)^2 = \pi \implies \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}\]

where we’ve determined the sign by going back to the integral definition and noting that necessarily \(\Gamma(x) > 0\) for \(x > 0\).

We can also use this relation to express a trigonometric integral in terms of \(\Gamma\) - let \(z = \frac{n}{2}+\frac{1}{2}\) and \(w = \frac{1}{2}\), giving

\[B(z,w) = 2 \int_0^{\frac{\pi}{2}} \sin^n \theta \; d\theta = \frac{\Gamma\left(\frac{n}{2}+\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{n}{2}+1\right)}\]

so that

\[\int_0^{\frac{\pi}{2}} \sin^n \theta \; d\theta = \frac{\sqrt{\pi}}{2} \cdot \frac{\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{\Gamma\left(\frac{n}{2}+1\right)}\]

Moreover, since Beta is symmetric in \(z\) and \(w\), considering \(B(w, z)\) with the above values of \(z\) and \(w\) gives

\[\int_0^{\frac{\pi}{2}} \cos^n \theta \; d\theta = \int_0^{\frac{\pi}{2}} \sin^n \theta \; d\theta\]

which can also be seen straightforwardly by substituting \(\theta \mapsto \frac{\pi}{2} - \theta\).

Zeta Function

Let \(s \in \mathbb C\) and define, for \(\operatorname{Re} s > 1\),

\[\zeta(s) = \sum_{n = 1}^\infty \frac{1}{n^s}\]

As shown in the post about Fourier series, the value of \(\zeta(2)\) is given by

\[\zeta(2) = \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}\]