Home > Complex questions > Proof of CIFD

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

 

Advertisements
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.

  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: