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Monday, November 18, 2013

 

Integration in C

Let f:[a,b]R be continuous. Then
  baf(t)dt=lim
Let \gamma be a curve ie a smooth or piecewise smooth function in \mathbb{C}:
  \begin{aligned} &\gamma: [a, b] \rightarrow \mathbb{C} \\ &\gamma(t) = x(t) + iy(t) \\ \end{aligned}
If f is complex-valued on \gamma,
  \begin{aligned} \int_\gamma f(z)\,dz &= \lim_{n \rightarrow \infty} \sum_{j=0}^{n-1} f(z_j)(z_{j+1} - z_j) \\ \end{aligned}
where z_j = \gamma(t_j).

But
  \begin{aligned} \int_\gamma f(z)\,dz &= \int_a^b f(\gamma(t)) \, \gamma'(t)\,dt \\ \end{aligned}
Why ? This is easy to see, as
  \begin{aligned} z &= \gamma(t) \\ dz &= \gamma'(t)\,dt \\ \end{aligned}
Such integration over the curve is called the Path Integral. Furthermore, if g(t) = u(t) + iv(t) , then
  \begin{aligned} \int_a^b g(t)\,dt &= \int_a^b u(t) \,dt + i\int_a^b v(t) \,dt \\ \end{aligned}
What is \displaystyle \int_{\left| z \right| = 1} \frac{1}{z}\,dz ? Kind of cute.

Turns out, in general, \displaystyle \int_{\left| z \right| = 1} z^m \,dz = 2\pi i when m = -1 but zero otherwise !

More at Analysis of a Complex Kind.


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