Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\gamma\colon[a,b]\to \mathbb{C}$ denote a piecewise differentiable path , and let $\varphi:$ Image $\gamma\colon \to \mathbb{C}$ be a continuous function.

Define $g: D = \mathbb{C}$-Image$\varphi \to \mathbb{C}$ by: $$ g\left( z \right) = \int\limits_\gamma {\frac{{\varphi \left( u \right)}} {{u - z}}du}. $$

Prove that: $$ g^{\left( n \right)} \left( z \right) = n!\int\limits_\gamma {\frac{{\varphi \left( u \right)}} {{\left( {u - z} \right)^{n + 1} }}du}. $$

Well , it's obvious that is enough to prove only the case $n=1$. Computing $$ \eqalign{ & \frac{{g\left( {z + h} \right) - g\left( z \right)}} {h} = \int\limits_\gamma {\frac{{\varphi \left( u \right)}} {{\left( {u - z - h} \right)}} - \frac{{\,\varphi \left( u \right)}} {{u - z}}du = } \cr & \int\limits_\gamma {\varphi \left( u \right)\frac{h} {{\left( {u - z - h} \right)\left( {u - z} \right)}}du} \cr} $$ Well I have no idea how to continue.

share|cite|improve this question
Your $h$ in the denominator somehow disappeared, it should cancel out the $h$ in the numerator in the last line. Then you have to let $h\to 0$, and justify that you can interchange limit and integral. Alternatively, there are some theorems that tell you when you can differentiate under an integral. And for the higher derivatives you just keep on going. (Think of induction proofs...) – Lukas Geyer Oct 6 '12 at 17:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.