# Integration around circular arcs

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Prove the following

If $$\displaystyle f$$ has a simple pole at $$\displaystyle z=c$$ and $$\displaystyle C_r$$ is any circular arc bounded by $$\displaystyle \theta_1 , \theta_2$$ and centered at $$\displaystyle c$$ with radius $$\displaystyle r$$

$$\displaystyle \lim_{r \to 0^+} \int_{C_r} f(z) \, dz = i ( \theta_2 - \theta_1 ) \text{Res} (f;c)$$​

#### GJA

##### Well-known member
MHB Math Scholar
Hi ZaidAlyafey,

I'm assuming we are taking $$\displaystyle f$$ to be meromorphic on an open subset, say $$\displaystyle U$$, of $$\displaystyle \mathbb{C}$$ that contains $$\displaystyle c.$$ I also assume the circular arcs we are considering contain the endpoints of the arc (i.e. are compact). That said, we can prove the claim by looking at the Laurent series expansion for $$\displaystyle f$$.

Since $$\displaystyle z=c$$ is a simple pole of $$\displaystyle f,$$ we know that the Laurent series expansion for $$\displaystyle f$$ is of the form

$$\displaystyle f(z)=\frac{Res(f;c)}{z-c}+\sum_{n=0}^{\infty}a_{n}(z-c)^{n}\qquad (*),$$

where the above holds on an annulus of inner radius $$\displaystyle \rho$$ and outer radius $$\displaystyle R.$$ Now

$$\displaystyle \rho = \limsup_{n\rightarrow\infty}|a_{-n}|^{1/n}=0,$$

and since $$\displaystyle f$$ is meromorphic on $$\displaystyle U,$$ $$\displaystyle R>0;$$ the fact that $$\displaystyle \rho=0$$ is what allows us to let $$\displaystyle r\rightarrow 0^{+}.$$ We must note that $$\displaystyle (*)$$ converges uniformly on compact subsets of our annular region (Laurent series - Wikipedia, the free encyclopedia).

Since, for small enough $$\displaystyle r,$$ $$\displaystyle C_{r}$$ is a compact subset of our annular region, $$\displaystyle (*)$$ converges uniformly on $$\displaystyle C_{r}.$$ Using the uniform convergence to justify interchanging integral and sum we have

$$\displaystyle \int_{C_{r}}f(z)dz=Res(f;c)\int_{C_{r}}\frac{1}{z-c}dz+\sum_{n=0}^{\infty}a_{n}\int_{C_{r}}(z-c)^{n}dz.$$

Parameterizing $$\displaystyle C_{r}$$ via $$\displaystyle z=c+re^{i\theta}$$ and integrating gives

$$\displaystyle \int_{C_{r}}f(z)dz=i(\theta_{2}-\theta_{1})Res(f;c)+ir\sum_{n=0}^{\infty}a_{n}\int_{\theta_{1}}^{\theta_{2}}r^{n}e^{i(n+1)\theta}d \theta$$

The sum on the right exists, because the integral on the left exists. Hence, taking the limit we obtain

$$\displaystyle \lim_{r\rightarrow 0^{+}}\int_{C_{r}}f(z)dz=i(\theta_{2}-\theta_{1})Res(f;c),$$

as desired