Contour Integrals Along Parametrised Paths
Integrate a complex function along a path by parametrising it, and bound the result by length times maximum modulus.
What a learner can do afterwards
- Evaluate a contour integral directly from a parametrisation
- Show the value is unchanged by reparametrising and reversed by reversing the path
- Apply the estimation lemma to bound an integral without evaluating it
1 · Read
A contour integral accumulates f along a parametrised path. Put gamma of t into f, multiply by the velocity gamma prime, and integrate in t. On gamma of t equal t over 0 to 3, the integral of z dz becomes the integral of t dt, which is 9 over 2, or 4.5.
Circles use gamma of t equal e to the it. Integrating 1 along the upper half from 1 to minus 1 gives end minus start, which is minus 2. Integrating 1 divided by z around the full circle gives 2 pi i, since the velocity cancels the integrand down to i.
Parametrising the same curve twice as fast changes nothing, because the velocity factor compensates exactly. Only the path plus its direction matter. Reversing direction negates the whole integral, so clockwise and anticlockwise runs are negatives of each other.
When exact evaluation is hopeless, bound instead. The estimation lemma says the modulus of the integral is at most the maximum of f times the path length. Length 3 with f at most 4 gives a bound of 12.
Feed the path into the function, weight by velocity, and watch direction while speed stays irrelevant.
2 · Watch
Take it off screen
Where it sits
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.