Cauchy's Integral Formula/General Result

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Theorem

Let $D = \set {z \in \C: \cmod z \le r}$ be the closed disk of radius $r$ in $\C$.

Let $f: U \to \C$ be holomorphic on some open set $U$ such that $D \subseteq U$.

Let $n \in \N$ be a natural number.


Then for each $a$ in the interior of $D$, the $n$-th derivative of $f$ at $a$ exists and can be written as:

$\ds \map {f^{\paren n} } a = \frac {n!} {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$

where $\partial D$ is the boundary of $D$, and is traversed anticlockwise.


Corollary

Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.

Let the coefficient of $z^n$ extracted from $\map G z$ be denoted:

$\sqbrk {z^n} \map G z := a_n$

Let $\map G z$ be convergent for $z = z_0$ and $0 < r < \cmod {z_0}$.


Then:

$\sqbrk {z^n} \map G z = \ds \frac 1 {2 \pi i} \oint_{\cmod z \mathop = r} \dfrac {\map G z \d z} {z^{n + 1} }$


Proof

Proof by induction:

For all $n \in \N$, let $\map P n$ be the proposition:

$\ds \map {f^{\paren n} } a = \frac {n!} {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$


Basis for the Induction

$\map P 0$ holds, as this is:

$\ds \map f a = \frac 1 {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}} \rd z$

which is Cauchy's Integral Formula.


This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$\ds \map {f^{\paren k} } a = \frac {k!} {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}^{k + 1} } \rd z$


Then we need to show:

$\ds \map {f^{\paren {k + 1} } } a = \frac {\paren {k + 1}!} {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}^{k + 2} } \rd z$


Induction Step

This is our induction step:

\(\ds \frac {\rd} {\rd a} \map {f^{\paren k} } a\) \(=\) \(\ds \frac {k!} {2 \pi i} \oint_{\partial D} \frac {\rd} {\rd a} \frac {\map f z} {\paren {z - a}^{k + 1} } \rd z\)
\(\ds \) \(=\) \(\ds \frac {k!} {2 \pi i}\oint_{\partial D} \frac {\paren {k + 1} \map f z} {\paren {z - a}^{k + 2} } \rd z\)
\(\ds \) \(=\) \(\ds \frac {\paren {k + 1}!} {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}^{k + 2} } \rd z\)


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To justify the passage of the derivative under the integral sign in line 1, note that for $\alpha$ near $a$, we have the following bound for all $z \in \partial D$:

$\ds \cmod {\frac {\paren {k + 1} \map f z} {\paren {z - \alpha}^{k + 2} } } = \frac {\paren {k + 1} \cmod {\map f z} }{r^{k + 2} }$

Since $f$ is continuous on the compact set $\partial D$, it follows that $\ds \sup_{z \mathop \in \partial D} \cmod {\map f z} < \infty$.

Thus the arc length integral:

$\ds \oint_{\partial D} \frac {\paren {k + 1} \cmod {\map f z} }{r^{k + 2}} ds(z) \le \paren {k + 1} \map \ell {\partial D} \sup_{z \mathop \in \partial D}\frac{\cmod {\map f z} }{r^{k + 2} } < \infty$


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Hence an argument involving the Fundamental Theorem of Calculus followed by the dominated convergence theorem justifies the swap.

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\ds \forall n \in \N: \map {f^{\paren n} } a = \frac {n!} {2 \pi i} \oint_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$

$\blacksquare$


Also known as

The General Form of Cauchy's Integral Formula can also be referred to as Cauchy's Integral Formula for Derivatives, or just Cauchy's Formula for Derivatives.


Source of Name

This entry was named for Augustin Louis Cauchy.


Sources

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