5.1 Contour Integration

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Fix a domain DC and f:DC continuous. We are going to define γf where γ:[a,b]D is any smooth path. Such an integral will be called a contour integral.

Definition (Path Integral)

Given a smooth path γ:[a,b]D the path integral of f along γ is γf=abf(γ(t))γ(t)dt where the right-hand side is a Riemann integral. This means that the integral is formally defined as a limit abf(γ(t))γ(t)dt=limNbaNn=1Nf(γ(a+nbaN))γ(a+nbaN) of Riemann sums. As such, we can calculate these integrals using the rules of Calculus, treating i like any other constant.

Example

Define f:CC by f(z)=z2 and γ:[0,1]C by γ(t)=t2+it. Calculate the contour integral of f along γ.

Solution:

We first compute f(γ(t))=(t2+it)2=t4t2+2it3 and γ(t)=2t+i so that γf=01(t4t2+2it3)(2t+i)dt=012t54t3dt+i015t4t2dt=23+i23 using the usual rules of integration.

Definition (Contour Integral)

Fix DC a domain and f:DC continuous. Given a contour Γ=(γ1,,γn) in D we define the contour integral of f along Γ to be Γf=γ1f+γ2f++γnf which is permitted because each of the γi is a smooth path.

Contour integration in linear in the sense that both relationships Γf+g=Γf+ΓgΓcf=cΓf are true for any f,g:DC continuous and any cC.

It many not be clear why we should define path integrals the way we have. The definition we have given turns out to be correct for various reasons. First of all, we show how our definition is independent of our speed along γ and only depends on our route.

Definition (Reparameterization)

Fix a smooth path γ:[a,b]C. A reparameterization of γ is any smooth, increasing bijection ψ:[c,d][a,b].

If ψ is a reparameterization of γ:[a,b]C then γψ:[c,d]C defined by γψ(t)=γ(ψ(t)) is a smooth path satisfying γψ(c)=γ(a) and γψ(d)=γ(b). Thus γψ and γ begin and end in the same places. Moreover, they trace out the same route. They only differ in how quickly or slowly they traverse the route.

Example

Define γ:[0,1]C by γ(t)=eiπt. It is a smooth path from 1 to 1 travelling uniformly around the top half of the unit circle. Define ψ:[0,1][0,1] by ψ(t)=t5. It is a reparameterization of γ because it is a smooth, increasing bijection of the unit interval. The path γψ has the formula γψ(t)=eiπt5 and also travels from 1 to 1 by travelling around the top half of the unit circle. However, it does not do so uniformly: it is much slower in the beginning (near 1) and much faster at the end (near 1) than γ is.

Theorem

Fix DC a domain and f:DC continuous. Let γ:[a,b]D be a smooth path. Then γf=γψf for every reparameterization ψ of γ.

Proof:

We calculate γψf=cd(f(γψ(t))γψ(t)dt=cd(f(γ(ψ(t)))γ(ψ(t))ψ(t)dt and then apply the substitution s=ψ(t) to get cd(f(γ(ψ(t)))γ(ψ(t))ψ(t)dt=abf(γ(s))γ(s)ds=γf as desired.

The real and imaginary parts of our contour integral have interesting physical meanings. Writing f(x+iy)=u(x,y)+iv(x,y) and γ(t)=α(t)+iβ(t) we can calculate γf=ab(u(α(t),β(t))+iv(α(t),β(t)))(α(t)+iβ(t))dt=abu(α(t),β(t))α(t)+(v(α(t),β(t)))β(t)dt+iabu(α(t),β(t))β(t)+(v(α(t),β(t)))(α(t))dt and try to understand its real and imaginary parts. Write γ(t)=α(t),β(t). Let H be the vector field H(x,y)=u(x,y),v(x,y) on D. We see that Re(γf)=abu(α(t),β(t))α(t)+(v(α(t),β(t)))β(t)dt=Hdγ is the work done against H in moving along γ and that Im(γf)=abu(α(t),β(t))β(t)+(v(α(t),β(t)))(α(t))dt=Hdn is the flux of H across γ. Here n is the unit normal to γ defined by n(x,y)=β(t),α(t).

The vector field H associated to f:DC is called the Polya vector field of f.