# Complex Analysis: #3 Path Integrals

_{0}< t

_{1}be two real numbers. Then a continuous mapping γ : [t

_{0}, t

_{1}] → G ⊂ ℂ is a

*path*in the region G of ℂ. In the analysis lecture we learned that γ is

*rectiﬁable*if a number L exists such that for all ∈ > 0, a δ > 0 exists such that for every partition t

_{0}= a

_{0}< a

_{1}< · · · < a

_{n}= t

_{1}which is such that a

_{j+1}− a

_{j}< δ for all j, we have

Let γ(t) = γ

_{r}(t) + iγ

_{i}(t), where γ

_{r}, γ

_{i}: [t

_{0}, t

_{1}] → ℜ are real-valued functions. Then we say that the path is

*continuously differentiable*if both the functions γ

_{r}and γ

_{i}are continuously differentiable. In this case, γ' = γ

_{r}' + iγ

_{i}' is also a path in ℂ. [Thinking in terms of 2-dimensional real geometry, we can say that γ'(t) is the "tangent vector" to γ(t).] We also learned that continuously differential paths are always rectiﬁable, and we have

All of this has already been dealt with in the analysis lecture. For us now, the interesting thing is to think about path integrals through a region where a complex-valued function is given.

**Deﬁnition 3**

Let G ⊂ ℂ be a region, and let f : G → ℂ be a function. Furthermore, let γ : [t

_{0}, t

_{1}] → G be a

*differentiable path*. Then the path integral of f along γ is

assuming it exists.

The integral here is simply the sum of the integrals over the real and the imaginary parts. It is not necessary to assume that γ is continuously differentiable, but we will assume that it is

*piecewise*continuously differentiable. That is, there is a partition of the interval [t

_{0}, t

_{1}] such that it is continuously differentiable along the pieces of the partition. So from now on, we will (almost) always assume that all paths considered are piecewise continuously differentiable.

As an exercise (using the substitution rule for integrals), one sees that the path integral does not depend on the way the path is parameterized. The simplest case is that, say γ(t) = t. Then (taking t from 0 to 1) we just have ∫

_{γ}f(z)dz = ∫

_{0}

^{1}f(t)dt. Almost equally simple is the case that γ(t) = it. Then we have ∫

_{γ}f(z)dz = i∫

_{0}

^{1}f(it)dt.

Increasing the complexity of our thoughts ever so slightly, we arrive at the ﬁrst version of Cauchy’s integral theorem.

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