# $g\left( z \right) = \int\limits_\gamma {\frac{{\varphi \left( u \right)}} {{u - z}}du}$ computing the derivate

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.

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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