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.

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One Response to Complex integration

  1. Pingback: Cauchy Integral Theorem « Andy Soffer

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