# Proving the following integral formula

In proving $$f^{(n)}(w) = \frac{n!}{\rho^n \pi} \int_{0}^{2\pi} \Re ( f(w+\rho e^{i\theta})) e^{-in\theta} d\theta$$ where $$f$$ is analytic in $$D(0;R)$$, $$0, and let $$w$$ be an arbitary point in $$D(0;r)$$, I have some confusion.

The first idea is the Cauchy Integral Formula for derivatives, where we have $$f^{(n)}(w) = \frac{n!}{2\pi i} \int_{\partial D(w;\rho)} \frac{f(z)}{(z-w)^{n+1}} dz.$$

Plug in $$z = w + \rho e^{i\theta}$$ we have $$f^{(n)}(w) = \frac{n!}{2\pi i} \int_{0}^{2\pi} \frac{f(w+\rho e^{i\theta})}{\rho^n e^{in\theta}} d\theta = \frac{n!}{2 \rho^n \pi i} \int_{0}^{2\pi} \frac{\Re(f(w+\rho e^{i\theta}))e^{i\theta}}{e^{in\theta}} d\theta,$$ and that's where I can go no further. I didn't know how to cancel the $$2i$$ in the denominator, or did the integrand has wrong power?

What is the trick?

## 2 Answers

There are two problems in your calculation. First, as Kavi said, with the parametrization $$\gamma(t) = w+ \rho e^{i\theta}$$ we have $$\gamma'(t) = i\rho e^{i\theta}$$, and therefore $$\tag{1} f^{(n)}(w) = \frac{n!}{2\pi i} \int_\gamma \frac{f(z)}{(z-w)^{n+1}} \, dz = \frac{n!}{2 \pi \rho^n} \int_0^{2 \pi} f( w+ \rho e^{i\theta}) e^{-in\theta} \, d\theta \, .$$

The second problem is that you cannot simply replace $$f( w+ \rho e^{i\theta})$$ by its real part in the integral. Instead we use Cauchy's integral theorem $$\tag{2} 0 = \frac{n!}{2\pi i} \int_\gamma f(z)(z-w)^{n-1} \, dz = \frac{n!\rho^n}{2 \pi } \int_0^{2 \pi} f( w+ \rho e^{i\theta}) e^{in\theta} \, d\theta$$ for $$n \ge 1$$. Dividing this by $$\rho^{2n}$$ and taking the complex conjugate gives $$\tag{3} 0 = \frac{n!}{2 \pi \rho^n} \int_0^{2 \pi} \overline{f( w+ \rho e^{i\theta})} e^{-in\theta} \, d\theta \, .$$ Now we add $$(1)$$ and $$(3)$$ and note that $$f( w+ \rho e^{i\theta}) + \overline{f( w+ \rho e^{i\theta})} = 2 \operatorname{Re}\left(f( w+ \rho e^{i\theta}) \right) \, ,$$ so that the factor $$2$$ cancels in the result: $$f^{(n)}(w) = \frac{n!}{ \pi \rho^n} \int_0^{2 \pi} \operatorname{Re}\left(f( w+ \rho e^{i\theta}) \right) e^{-in\theta} \, d\theta \, .$$

Remark: We have assumed that $$n \ge 1$$. The formula does not hold for $$n=0$$.

• My initial thought is to rewrite $f((w+ρe^{i\theta})$ as $R(f(w+ρe^{i\theta}))e^{i\theta}$, since we can rewrite a complex number as a product of its real part and its argument. Is this a valid apporach? – Dinoman May 11 at 11:32
• @Dinoman: That does not work. What you perhaps mean is that a complex number is the product of its absolute value and $e^{i\theta}$ where $\theta$ is the argument. Also, the argument of $f(w+ρe^{i\theta})$ is not $\theta$ or $e^{i\theta}$. – Martin R May 11 at 11:35

With $$z = w + \rho e^{i\theta}$$ we have $$f^{n}(w) = \frac{n!}{2\pi i} \int_{0}^{2\pi} \frac{f(w+\rho e^{i\theta})}{\rho^{n+1} e^{i(n+1)\theta}} i\rho e^{i\theta} d\theta$$ = $$\frac{n!}{2 \rho^n \pi } \int_{0}^{2\pi} {\{f(w+\rho e^{i\theta})\}e^{-in\theta}} d\theta$$.

• I think I have replaced it and it got cancelled off. The original denominator is to the power (n+1) – Dinoman May 11 at 9:18
• How does $i$ still remain in the final expression? @Dinoman – Kavi Rama Murthy May 11 at 9:24
• You made several mistakes in your calculations. How did you get $(z-w)^{n}$ instead of $(z-w)^{n+1}$ in the denominator? Please check each step carefully. – Kavi Rama Murthy May 11 at 9:27