Special Functions – The gamma and the beta function

The gamma function \(\Gamma(z)\) and the beta function \(B(u,v)\) play an important role in the theory of Special Functions. The gamma function is essential for the definition of other special functions such as hypergeometric functions and Bessel functions.

Sometimes we will use Cauchy's residue theorem.

Cauchy's residue theorem is a consequence of Cauchy's integral formula

\[f(z_0)=\frac{1}{2\pi\,i}\oint_{\mathcal{C}}\frac{f(z)}{z-z_0}\,dz,\]

where \(f\) is an analytic function and \(\mathcal{C}\) is a simple closed contour in the complex plane enclosing the point \(z_0\) with positive orientation which means that it is traversed counterclockwise.

Since the integrand is analytic except for \(z=z_0\), the integral is equal to the same integral with \(\mathcal{C}\) replaced by a small circle inside the contour \(\mathcal{C}\) with center \(z_0\). This implies that we have with \(z=z_0+r\,e^{i\theta}\) and \(0\leq\theta\leq2\pi\)

\[\oint_{\mathcal{C}}\frac{f(z)}{z-z_0}\,dz=\lim\limits_{r\downarrow 0}\int_0^{2\pi}\frac{f(z_0+r\,e^{i\theta})}{r\,e^{i\theta}}\,i\,r\,e^{i\theta}\,d\theta =i\,f(z_0)\int_0^{2\pi}\,d\theta=2\pi\,i\,f(z_0),\]

which proves Cauchy's integral formula.

This formula can be iterated to

\[f^{(n)}(z_0)=\frac{n!}{2\pi\,i}\oint_{\mathcal{C}}\frac{f(z)}{(z-z_0)^{n+1}}\,dz,\quad n=0,1,2,\ldots.\]

If \(f\) is an analytic function except for an isolated singularity at \(z=z_0\), then \(f\) has a Laurent series representation of the form

\[f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n.\]

Then the coefficient \(a_{-1}\) is called the residue of \(f\) at \(z=z_0\). We use the notation \(\text{Res}_f(z_0)=a_{-1}\). Now we have Cauchy's residue theorem:

Theorem: If \(\mathcal{C}\) is a simple closed, positively oriented contour in the complex plane and \(f\) is analytic except for some points \(z_1,\;z_2,\;\ldots,\;z_n\) inside the contour \(\mathcal{C}\), then

\[\oint_{\mathcal{C}}f(z)\,dz=2\pi\,i\sum_{k=1}^n\mathrm{Res}_f(z_k).\]

If \(f\) has a removable singularity at \(z=z_0\), then the residue is equal to zero. If \(f\) has a single pole at \(z=z_0\), then

\[\text{Res}_f(z_0)=\lim\limits_{z\to z_0}(z-z_0)f(z)\]

and if \(f\) has a pole of order \(k\) at \(z=z_0\), then

\[\text{Res}_f(z_0)=\frac{1}{(k-1)!}\lim\limits_{z\to z_0}\frac{d^k}{dz^k}\left\{(z-z_0)^kf(z)\right\},\quad k\in\{1,2,3,\ldots\}.\]

Cauchy's residue theorem can be used to evaluate real integrals by applying an appropriate contour in the complex plane.

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Last modified on May 15, 2021