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The Cauchy integral formula is as follows:

$$f^{n}(a) = \dfrac{n!}{2 \pi i}\oint_C \dfrac{f(z)}{(z-a)^{n+1}}\,\mathrm{d}z.$$

However in every source I can find describing the Cauchy integral formula does not state the domain of $n$ and I have only ever seen it used for positive values of $n$. So my question is does the formula hold for negative values of $n$? If so what does $f^{n}$ mean when $n$ is negative?

Sriotchilism O'Zaic
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    For $n \in \mathbb{N}$ if $f$ is holomophic/analytic on a simply connected open contaning the closed contour $C$ then $\int_C f(z) (z-a)^n dz = 0$ by the Cauchy integral theorem – reuns Apr 02 '17 at 20:01
  • Define $f^{(n)}$ when $n$ is not a non-negative integer? – Mark Viola Apr 02 '17 at 20:01
  • @Dr.MV That is the crux of my question. What happens when $n$ is negative – Sriotchilism O'Zaic Apr 02 '17 at 20:03
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    See Cauchy's theorems. If $n$ is negative then the integrand is analytic on the interior of the closed contour $C$. However, the above formula is not well defined for negative $n$. – SamM Apr 02 '17 at 20:03
  • @SamM, that formula does give a well-defined result for negative $n$ and $f$ holomorphic on the disk, namely, $0$. As in Count Iblis' answer, this would give Laurent expansions coefficients for $f$ having an isolated singularity (e.g.), rather than being holomorphic, in the disk. – paul garrett Apr 02 '17 at 23:33

2 Answers2

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The following excerpt from Applied and Computational Complex Analysis, Vol. 1 by P. Henrici might be helpful since it explicitely states the domain.

Theorem 4.7b (Cauchy Integral formula)

Let $R$ be a simply connected domain, let $f$ be analytic everywhere in $R$, and let $z_0$ be a point of $R$. Then, if $\Gamma$ is a piecewise regular closed curve in $R$ not passing through $z_0$, \begin{align*} n(\Gamma,z_0)f(z_0)=\frac{1}{2\pi i}\int_{\Gamma}\frac{f(z)}{z-z_0}\,dz. \end{align*} In particular, if $\Gamma$ is a positively oriented Jordan curve that contains the point $z_0$ in its interior, then \begin{align*} f(z_0)=\frac{1}{2\pi i}\int_{\Gamma}\frac{f(z)}{z-z_0}\,dz. \end{align*}

and later on

Corollary 4.7c

Under the hypotheses of Theorem 4.7b, if $\color{blue}{k=0,1,2,\ldots}$, \begin{align*} n(\Gamma,z_0)f^{(k)}(z_0)=\frac{k!}{2\pi i}\int_{\Gamma}\frac{f(z)}{(z-z_0)^{k+1}}\,dz, \end{align*} and particularly, if $\Gamma$ is a positively oriented Jordan curve containing $z_0$ in its interior, \begin{align*} \frac{1}{k!}f^{(k)}(z_0)=\frac{1}{2\pi i}\int_{\Gamma}\frac{f(z)}{(z-z_0)^{k+1}}\,dz. \end{align*}

Conclusion: In the corollary 4.7c the domain of $k$ is explicitely stated to be non-negative integers. We conclude, that the Cauchy Integral formula stated in OPs question does not apply to negative $n$.

Markus Scheuer
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$\frac{f^{n}(a)}{n!}$ is the coefficient of $(z-a)^n$ in the series expansion of $f(z)$ around the point $z = a$. An analytic function does not have negative powers in the series expansion, therefore we can define this to be zero for an analytic function, while for a meromorphic function we can take such coefficients to be given by the Laurent expansion coefficients. Then the Cauchy integral is indeed zero for analytic functions when you take $n$ negative, while for meromorphic functions that have a pole at $z = a$ and no other poles inside the contour, you'll also get the correct answer.

So, the conclusion is that it does work for negative $n$ when you replace $\frac{f^{n}(a)}{n!}$ by the coefficient of $(z-a)^n$.

Count Iblis
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  • Indeed, the formula correctly computes Laurent coefficients in a reasonable context. Not that there's a notion of negative-order derivative, necessarily. – paul garrett Apr 02 '17 at 23:34
  • @paulgarrett: No doubt, the formula of course correctly computes Laurent coefficients in a reasonable context. I fully agree with your statement. But, with respect to the left-hand side of OPs expression, the validity is given only for non-negative integers. That's why I formulated in my conclusion section: ... the Cauchy Integral formula stated in OPs question. – Markus Scheuer Apr 03 '17 at 09:03