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existence of derivatives of all orders

Another application is to deliberately introduce a singularity and then to evaluate the new integral. Consider

\begin{eqnarray*}f(z_0) & = & \frac{1}{2 \pi i} \oint \frac{f(z)}{z-z_0}dz

which has a simple pole at z0 and residue f(z0).

This is Cauchy's integral representationCauchy's integral of an analytic function, which is nothing more than an integral version of the observation that an analytic function averages its surrounding values. Here each value is given the weight $dz/(z-z_0) = d \ln(z-z_0)$. If z-z0 is written in polar form, the weight is a combination of the reciprocal of the distance of an element of arc and its angular aperture. That normalizes all arcs to an equivalent distance (that's the factor r), and performs the average.

Exploiting the fact that the derivative of an integral is the integral of the derivative of its integrand,

\begin{eqnarray*}f'(z_0) & = & \frac{1}{2\pi i} \frac{d}{dz_0} \oint \frac{f(z)}...
... dz \\
& = & \frac{1}{2\pi i} \oint \frac{f(z)}{(z-z_0)^2} dz

which surely ought to converge as well as the original integral. Note that the minus sign arising from the negative power is cancelled by the sign of -z0 in the denominator, which is the variable for this differentiation. Repeating the process turns up a whole infinite series of derivatives, which stands in contrast to what may happen for a real variable, that derivatives may eventually suffer discontinuities and other faults. In particular, there is no analytic function mimicking the peaks in the sawtooth function.

For convenience of reference, we could just collect the formulas for the first few derivatives:

f(z0) = $\displaystyle \frac{1}{2 \pi i} \oint \frac{f(z)}{z-z_0}dz$ (146)
f'(z0) = $\displaystyle \frac{1}{2\pi i} \oint \frac{f(z)}{(z-z_0)^2} dz$ (147)
f''(z0) = $\displaystyle \frac{2}{2\pi i} \oint \frac{f(z)}{(z-z_0)^3} dz$ (148)
    $\displaystyle \ldots$  
f(n)(z0) = $\displaystyle \frac{n!}{2\pi i} \oint \frac{f(z)}{(z-z_0)^{n+1}} dz$ (149)

Still another useful induced singularity makes use of the logarithmic derivative, supposing that there is a region containing only a finite number of zeroes and poles of f(z). Then.

$\displaystyle \frac{1}{2 \pi i} \oint \frac{f'(z)}{f(z)}dz$ = $\displaystyle {\rm (number\ of\ zeroes)} - {\rm (number\ of\ poles)},$ (150)

both being counted with their respective multiplicities. Such a function could be the quotient of two polynomials. In fact, if f were a polynomial and the contour were a sufficiently large circle, the highest power would dominate the logarithmic derivative, provoking the conclusion that there were just as many zeroes inside, as the degree of the polynomial.

This result, being true for any polynomial with complex coefficients, lays to rest the question of whether anything new can be gotten from taking the square root of -i (bearing in mind the humble origin of i itself). Neither x2+i nor any other combination of powers and complex numbers needs anything beside complex numbers as roots. An interesting point here is the way that the contour integral transforms the result that a power has n roots into the fact that any polynomial of that degree also has roots, although they are not necessarily (real multiples of) roots of unity.

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Next: Liouville's theorem: a bounded Up: Contour Integrals Previous: evaluation of integrals by