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Let $\sum_{n\geq 0} a_n z^n$ be a power series with radius of convergence greater or equal to 1. Let $0<r<1$, show that for all $n \geq 1$, $$a_{n}=\frac{1}{\pi r^n}\int_{0}^{2\pi} Re(f(re^{i\theta}))e^{-ni\theta}d\theta$$.

I know that :

$$\int_{0}^{2\pi}f(re^{i\theta})e^{-ni\theta}d\theta=\int_{0}^{2\pi}(\sum_{p=0}^{+\infty}a_p r^p e^{i(p-n)\theta})d\theta=\sum_{p=0}^{+\infty}a_p r^p \int_{0}^{2\pi}e^{i(p-n)\theta}d\theta$$

as we can exchange the sum and the integral because $\sum_{p=0}^{+\infty}a_p r^p e^{i(p-n)\theta}$ is converges normally. Noticing that $\int_{0}^{2\pi}e^{i(p-n)\theta}d\theta =0 $ if $p\neq n$ and $2\pi$ if $p=n$, we obtain :

$$2\pi r^n a_n = \int_{0}^{2\pi}f(re^{i\theta})e^{-ni\theta}d\theta$$

Then I think i should make a change of variable $\alpha = 2\pi - \theta$ in this integral but I wasn't able to conclude.

Kieran McShane
  • 398

1 Answers1

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$$\int_{0}^{2\pi}Re(f(re^{i\theta}))e^{-ni\theta}d\theta = \frac{1}{2} \int_{0}^{2\pi}(f(re^{i\theta})+\overline{f(re^{i\theta})})e^{-ni\theta}d\theta$$

Now : $$\int_{0}^{2\pi}\overline{f(re^{i\theta})}e^{-ni\theta}d\theta = \int_{0}^{2\pi} \sum_{p\geq0} \overline{a_p}r^pe^{-ip\theta}e^{-ni\theta} d\theta = \sum_{p\geq0} \overline{a_p}r^p \int_{0}^{2\pi}e^{-(p+n)i\theta}d\theta$$

we can exchange the integral and the sum for the same reason as above.

Notice that for all $n \geq 1$ and $p\ge 0$, $\int_{0}^{2\pi}e^{-(p+n)i\theta}d\theta=0$.

Hence : $$\int_{0}^{2\pi}Re(f(re^{i\theta}))e^{-ni\theta}d\theta = \frac{1}{2} \int_{0}^{2\pi}f(re^{i\theta})e^{-ni\theta}d\theta = \pi r^n a_n$$

Kieran McShane
  • 398