Let f:[a,b]→R be continuous. Then
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}
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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.
# posted by rot13(Unafba Pune) @ 8:11 AM
