# A complex variable function integrated over an infinitesimal disk [closed]

Let $0<\delta\in \mathbb R$ and let $f$ be a complex function analytic inside a disk with a radius $\delta$. Prove that $$\lim_{\varepsilon\to 0}\int_{\gamma_\varepsilon}f(z)dz=0$$ where $\gamma_\varepsilon$ is the upper half circle of radius $\varepsilon$ counterclockwise.

It is very obvious that such an integral would be zero because the radius is going to zero, but I need some help proving it.

Well for $\varepsilon < \delta$ the function $f$ will be analytic inside $\gamma_\varepsilon$. Thus $$\int_{\gamma_\varepsilon} f(z) dz = 0.$$

Taking the limit for $\varepsilon \to 0$ will result in $$\lim\limits_{\varepsilon \to 0}\int_{\gamma_\varepsilon} f(z) dz = 0$$ because for $\varepsilon$ close enough to $0$, the inequality $\varepsilon < \delta$ will hold.

• This looks coorect. Thanks.
– E Be
Commented May 12, 2015 at 19:34

Since $f$ is analytic for $|z|<\delta$ there is an $r<\delta$ and an $M>0$ such that $$\bigl|f(z)\bigr|\leq M\qquad\bigl(|z|\leq r\bigr)\ .$$ Let $\gamma_\epsilon$ be the upper half circle of radius $\epsilon\leq r$. Then $$\left|\int_{\gamma_\epsilon}f(z)\>dz\right|\leq M\cdot {\rm length}(\gamma_\epsilon)=M\>\pi\epsilon\ .$$ It follows that $$\lim_{\epsilon\to0+}\int_{\gamma_\epsilon}f(z)\>dz=0\ .$$

• The proof really only uses the fact that $f$ is continuous in the disc...
– DVD
Commented May 23, 2015 at 23:10