Let \(f: [a, b] \rightarrow \mathbb{R} \) be continuous. Then

\[ \begin{aligned} \int_a^b f(t)\,dt &= \lim_{n \rightarrow \infty} \sum_{j=0}^{n-1} f(t_j)(t_{j+1} - t_j) \\ F(x) &= \int_a^x f(t)\,dt \\ F'(x) &= f(x) \,\,\, \text{ for } x \in [a, b] \\ \end{aligned} \]

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.

# posted by rot13(Unafba Pune) @ 8:11 AM