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Let $B_1(0) \subset \mathbb{C}$ be the unit ball in the complex plane and $f:B_1(0) \to \mathbb{C}, z \mapsto z^n$. I want to know how one would calculate the integral $$\int_{B_1(0)}f(z) \text{d}z$$ I expect the integral to be 0 because of the orthogonality in $L^2$ of the normed monomials. But I want to calculate the integral "by hand". Would one use the lebesgue measure $\lambda^2$? It seems that something like polar coordinates seem to work better but it is not that obious how to proceed. I also thought about splitting $f$ into its real and imaginary part. But using the binomial theorem for $z^n=(a+ib)^n$ does not get me any further.

Flynn Fehre
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    Polar coordinates are a good idea. Write $z^n = e^{i \theta n} \cdot r^n$. How does the integral look now? This should now be computable by hand and will in fact give 0. – Sven-Ole Behrend Nov 08 '23 at 14:30
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    @Sven-OleBehrend Thanks. Is it right that the point then will be that $e^{i\theta n} = \cos(n \theta) + i \sin(n \theta)$ which splits the integral and both real and imaginary part become 0? – Flynn Fehre Nov 08 '23 at 14:45
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    @FlynnFehre More simply : the antiderivative of an exponential is itself an exponential. – Abezhiko Nov 08 '23 at 17:19
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    The integral doesn't make sense. $B_1(0)$ is a disk, so to integrate over it is to use an area integral. But $dz$ is not an area differential. You use $dz$ for integrating along curves. Do you mean $\int_{B_1(0)} f(z)|dz|$? or $\int_{\partial B_1(0)} f(z)dz$? – Paul Sinclair Nov 09 '23 at 17:05
  • @PaulSinclair I am not quite sure really. The thing is that I have/want to show that $z^n$ is an orthogonal system in $L^2$ but I have not been given a scalar product. So I assumed it was gonna be the integral of the product where the second argument is conjugated. That is also why I was not sure to use the lebesgue measure $\lambda^2$ (which is kind of weird because it is $\mathbb{C}$ I am integrating over and not $\mathbb{R}^2$). But, yeah, that is why I do not really know. Do you perhaps know the convention for the scalar product of $L^2$ over $\mathbb{C}$? – Flynn Fehre Nov 09 '23 at 18:42
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    "$L^2(X,\Bbb C)$" for any measure space $(X, \mu)$ refers to the Hilbert space of all square-integrable complex-valued functions on $X$, with scalar product $\langle f, g\rangle = \int_X \overline f g,d\mu$. Apparently $X = B_1(0)$ in your case, so you need only figure out the measure to use on $B_1(0)$. $dz$ is not a measure (or the differential of one) on that set. The obvious choice is the area measure. – Paul Sinclair Nov 09 '23 at 18:54
  • @PaulSinclair I am not familiar with the "area measure". It sounds to me that this is just the two dimensional Lebesgue measure, isn't it? – Flynn Fehre Nov 09 '23 at 19:00
  • Yes. That is exactly what "area measure" means. – Paul Sinclair Nov 09 '23 at 19:41

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One could/would indeed use the lebesgue measure. With the substitution for multiple variables, fubini and the usual function for polar coordinates $\Phi(r,\phi) = (r \cos(\phi), r \sin(\phi))$, as such $\det(D \Phi) = r$, one would get: \begin{align*} \int_{B_1(0)} z^n d z &=^1 \int_{[0,1) \times [0, 2\pi)} (r e^{i \phi})^n r d \lambda^2 \\ &= \int_{[0,1) \times [0, 2\pi)} r^{n+1} e^{n i \phi} r d \lambda^2\\ &=^2 \int_{[0,1)} r^{n+1} \int _{[0,2\pi)} e^{n i \phi} d \phi d r\\ &= \int_{[0,1)} r^{n+1} \int _{[0,2\pi)} e^{n i \phi} d \phi d r\\ &=^3 \int_{[0,1)} r^{n+1} \left[\frac{1}{n i} e^{n i \phi} \right]_0 ^{2 \pi} d r\\ &= \int_{[0,1)} r^{n+m+1} \frac{1}{n i} \left[ e^{n i \phi} \right]_0 ^{2 \pi} d r\\ &= \int_{[0,1)} r^{n+m+1} \frac{1}{n i} 0 \; d r\\ &= 0 \end{align*} $^1$ Multivariate substitution
$^2$ Fubini
$^3$ Note that $(e^x)' = e^x$ even for $x \in \mathbb{C}$.

Which is $0$ as you guessed.

Raoul Luqué
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