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## Proof of CIFD

Dear all,

Ken asked me about the proof of the Cauchy Integral Formula for Derivatives that uses the real analysis fact of differentiating under the integral sign.

Statement: If $f$ is holomorphic on a simply connected domain $D$ and $\alpha$ is a counterclockwise oriented Jordan rectifiable curve in $D$, then for every $z$ inside the interior of $\alpha$, we have

$f^{(n)}(z)=\frac{n!}{2\pi i} \int_{\alpha} \frac{f(w)}{(w-z)^{n+1}} dw$.

Proof: For fixed positive integers $k$ and $z$ in the interior of $\alpha$,  the differential form $\frac{f(w)}{(w-z)^{k}} dw$ has bounded and smooth complex coefficients on $\alpha$, and likewise all of its derivatives wrt $z$ are smooth and bounded. In particular, the functions mentioned are absolutely integrable over $\alpha$. Thus, by the theorem on differentiating under the integral sign, any partial derivative of the expression

$G(x,y)=\int_{\alpha} \frac{f(w)}{(w-(x+iy))^{k}} dw$.

may be computed by differentiating under the integral sign. For instance,

$\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right) G(x,y)=\frac{1}{2}\int_{\alpha} \left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\frac{f(w)}{(w-(x+iy))^{k}} dw$.

Therefore, by the Cauchy integral formula

$f(z)=\frac{1}{2\pi i} \int_{\alpha} \frac{f(w)}{(w-z)} dw$,

$f^{(n)}(z)=\frac{1}{2\pi i} \int_{\alpha} \frac{d^n}{dz^n}\frac{f(w)}{(w-z)} dw$

$=\frac{n!}{2\pi i} \int_{\alpha} \frac{f(w)}{(w-z)^{n+1}} dw$.

QED

Categories: Complex questions
1. June 15, 2011 at 5:12 pm

Another clever way is to use the power series expansion. Though now you’re switching integration with infinite summation.