# Why does $\int_{\Omega} f\, dA = 2\pi \,f(0)$ and not zero?

I have been experimenting with quadrature domains. The most obvious one is a circle. Let $f(z)$ be holomorphic on a large enough region and $\Omega= \{|z| < 1 \}$ then:

$$\int_{\Omega} f\, dA = \pi \,f(0)$$

I almost reasoned the answer should be zero. We could change the area to polar coordinates:

$$\int_{\Omega} f\, dA = \int_0^1 \int_{|z| = r } f(x+iy) \,dx\, dy = \int_0^1 \int_{|z| = r } f( r \, e^{i\theta}) \,r \, dr \, d\theta = 0$$

since the integral around the circle should be zero. Assuming there is no pole in side the circle $\{ |z| < 1\}$:

$$\oint_{|z| = 1} f(z) \, dz = \int_{|z| = 1} f ( r \, e^{i\theta})\, r e^{i\theta} d\theta = 0 \cdot f(0) = 0$$

Short Version: $dz\ne d\theta$.

In Detail:

Forget $r$ for the moment - it's irrelevant to the problem.

The notation $\int_{|z|=1}f(e^{i\theta})\,d\theta$ doesn't really make sense, because there's no $z$ in the integrand. Assumng $f$ is hoolomorphic in a neighborhood of the closed unit disk both of the following hold: $$\int_{|z|=1}f(z)\,dz=0$$(Cauchy's Theorem) $$\frac1{2\pi}\int_0^{2\pi}f(e^{i\theta})\,d\theta=f(0)$$ (mean-value property for holmorphic (hence harmonic) functions).

The two integrals are simply not the same at all. If we parametrize the unit circle via $z=e^{i\theta}$, $0\le\theta\le2\pi$ we get $$dz=ie^{i\theta}\,d\theta,$$hence $$\int_{|z|=1}f(z)\,dz=i\int_0^{2\pi}f(e^{i\theta})e^{i\theta}\,d\theta \ne\int_0^{2\pi}f(e^{i\theta})\,d\theta.$$

Now when you do the original integral in polar coordinates you get $d\theta$, not $dz$. A correct version of your polar-coordinate calculation is this: $$\int_{\Omega} f\, dA = \int_0^1 \int_0^{2\pi} f(x+iy) \,dx\, dy = \int_0^1 \int_0^{2\pi} f( r \, e^{i\theta}) \,r \, dr \, d\theta = 2\pi\int_0^1 f(0)r\,dr=\pi f(0).$$The integral $\int_{|z|=r}f(z)\,dz$ simply doesn't come up there.

(Look up the polar coordinates formula. You don't see $\int_{|z|=r}$ there.)

$\int_{|z|=1} f(r e^{i \theta}) e^{i \theta}\,d\theta$ is zero as you say, but $\int_{|z|=1} f(r e^{i \theta}) \,d\theta$ doesn't have to be. You could think about the constant function $f=1$, for instance.

• $\int_{|z|=1} f(r e^{i \theta}) e^{i \theta}\,d\theta$ doesn't really make sense, or at least not the sense you intend - if the integral is wrt $\theta$ then the limits should say something about $\theta$, not $z$. Not to be pedantic, but it seems to me that being careless about this sort of thing is what's leading to the OP's confusion. – David C. Ullrich Dec 8 '17 at 15:54

Note that $$\frac{1}{2\pi}\int_0^{2\pi} f(re^{it}) dt = \frac{1}{2\pi i}\int_{\partial B_r(0)} \frac{f(z)}zdz$$ and use Cauchy's theorem.