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I'm reading a proof of the Cauchy formula for the derivatives of an holomorphic function from some lecture notes. The author first proves that $$f^{(n)}(z)=\frac{1}{2\pi i}\int_{C}\frac{f^{(n)}(\zeta)}{\zeta-z}\;d\zeta$$ where $C$ is a circumference enclosing $z$. Then he says: "... integrating this by parts $n$ times gives the required formula..." I don't understand what he means with integrating by parts; this is a technique used in real analysis! In complex analysis, what is the integration by parts? Practically from the above formula I have problems to get:

$$f^{(n)}(z)=\frac{n!}{2\pi i}\int_{C}\frac{f(\zeta)}{{(\zeta-z)}^{n+1}}\;d\zeta$$

Thanks in advance.

user93089
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Dubious
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2 Answers2

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It's the same thing. The fundamental theorem of calculus here says: Let $\gamma \colon [0,1] \to \mathbb{C}$ a differentiable path (I restrict to differentiable paths for simplicity, piecewise differentiable paths can be handled with minimal additional considerations), and $f$ a holomorphic function defined in a neighbourhood of the trace of $\gamma$. Then

$$f(\gamma(1)) - f(\gamma(0)) = \int_\gamma f'(z)\,dz.$$

Inserting the definitions, we get

$$\int_\gamma f'(z)\,dz = \int_0^1 f'(\gamma(t))\cdot \gamma'(t)\,dt = \int_0^1 \bigl(f\circ\gamma\bigr)'(t)\,dt,$$

which is just the real version.

So if we have two holomorphic functions $f,g$, then by the real theory

$$\begin{align} \int_\gamma f'(z)g(z)\,dz &= \int_0^1 f'(\gamma(t))g(\gamma(t))\gamma'(t)\,dt\\ &= \int_0^1 \bigl(f\circ\gamma\bigr)'(t)(g\circ\gamma)(t)\,dt\\ &= \left[(f\circ\gamma)\cdot(g\circ\gamma)\right]_0^1 - \int_0^1 (f\circ\gamma)(t)\bigl(g\circ\gamma)'(t)\,dt\\ &= - \int_\gamma f(z)g'(z)\,dz \end{align}$$

for a closed path $\gamma$. The boundary term remains of course if $\gamma$ is not closed.

Daniel Fischer
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I think it is better to differentiate the Cauchy-theorem equality $f(z)={1\over 2\pi i}\int_\gamma {f(\zeta)\,d\zeta\over \zeta-z}$ with respect to $z$ to get the formula for the derivative.

paul garrett
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    And why one can exchange the derivative respect to $z$ with the line integral? – Dubious Sep 26 '13 at 15:28
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    The derivative can pass inside the integral because $f$ is uniformly continuous on the path (it is compact), and the very definition of derivative can be invoked to see that the limit of difference quotients exists, and is equal to the expected expression. (No necessity to invoke general facts about interchange of such limits.) – paul garrett Sep 26 '13 at 17:59