# Complex integration

It’s now time to introduce the integral. Again, I’ll assume we already understand real integrals. As you might expect, if ${h:[a,b]\rightarrow{\mathbb C}}$, then we will say that ${h}$ is integrable if ${{\mbox{Re}}(h)}$ and ${{\mbox{Im}}(h)}$ are both integrable. For the pedantic, here I suppose I mean Riemann integrable. Anyway, if so, then we can define

$\displaystyle \int_a^bh(t)dt=\int_a^b({\mbox{Re}} h)(t)dt+i\int_a^b({\mbox{Im}} h)(t)dt.$

This is nice, but not exactly what we want. We want path integrals. If ${\gamma:[a,b]\rightarrow{\mathbb C}}$ is a path, and ${f}$ is defined on some open set containing the image of ${\gamma}$, then

$\displaystyle \int_\gamma f(z)dz=\int_a^bf(\gamma(t))\gamma'(t)dt.$

This may look mysterious, but I assure you, it isn’t. In fact, there is pretty much no difference between path integrals over ${{\mathbb C}}$, and path integrals over ${{\mathbb R}^2}$. They’re certainly defined similarly. Anyway, let’s do an example. If ${\gamma}$ is the unit circle ${\gamma(t)=e^{i\cdot t}}$ for ${t\in[0,2\pi]}$, then

$\displaystyle \begin{array}{rcl} \displaystyle\int_\gamma z^ndz &=& \displaystyle\int_0^{2\pi}\gamma(t)^n\gamma'(t)dt\\ &=& \displaystyle\int_0^{2\pi}e^{i\cdot nt}\cdot ie^{i\cdot t}\\ &=& i\displaystyle\int_0^{2\pi}e^{i\cdot (n+1)t}dt \end{array}$

If ${n=-1}$, we have ${i\displaystyle\int_0^{2\pi}dt=2\pi i}$. Otherwise, the integral is

$\displaystyle \left[\frac i{i\cdot(n+1)}\cdot e^{i\cdot (n+1)t}\right|_0^{2\pi}=0.$

Also, as one might expect, the “speed” at which you traverse a curve is irrelevant. If you reparameterize the curve, the integral computed is identical.

Theorem 1 If ${f}$ is integrable along a path ${\gamma}$, and ${\beta}$ is a reparameterization of ${\gamma}$, then

$\displaystyle \displaystyle\int_\gamma f(z)dz=\displaystyle\int_\beta f(z)dz.$

Proof: Write ${\beta=\gamma\circ\phi}$. We’ll have ${\beta:[a,b]\rightarrow{\mathbb C}}$, ${\gamma:[c,d]\rightarrow{\mathbb C}}$, and ${\phi:[a,b]\rightarrow[c,d]}$, all continuous. Then

$\displaystyle \begin{array}{rcl} \displaystyle\int_\beta f(z)dz &=& \displaystyle\int_a^bf(\beta(t))\beta'(t)dt\\ &=& \displaystyle\int_a^b f(\gamma(\phi(t)))\gamma'(\phi(t))\phi'(t)dt \end{array}$

Using the “${u}$-substitution” ${u=\phi(t)}$, so ${du=\phi'(t)dt}$, we get

$\displaystyle \begin{array}{rcl} \displaystyle\int_\beta f(z)dz &=& \displaystyle\int_a^b f(\gamma(\phi(t)))\gamma'(\phi(t))\phi'(t)dt\\ &=& \displaystyle\int_c^d f(\gamma(u))\gamma'(u)du\\ &=& \displaystyle\int_\gamma f(z)dz \end{array}$

$\Box$

There are a few more facts I’m going to take for granted. First, if I integrate along a loop ${\gamma}$, but I traverse it backwards, then the value of the integral is the opposite. If I traverse two paths, first ${\beta}$, then ${\gamma}$, The integral is the sum of the integrals for each piece separately.