Editing Contour integration (section)

==Direct methods== 
Direct methods involve the calculation of the integral through methods similar to those in calculating line integrals in multivariate calculus. This means that we use the following method:
*parametrizing the contour
*: The contour is parametrized by a differentiable complex-valued function of real variables, or the contour is broken up into pieces and parametrized separately.
* substitution of the parametrization into the integrand
*: Substituting the parametrization into the integrand transforms the integral into an integral of one real variable.
* direct evaluation
*: The integral is evaluated in a method akin to a real-variable integral.

===Example===
A fundamental result in complex analysis is that the contour integral of {{math|{{sfrac|1|''z''}}}} is {{math|2π''i''}}, where the path of the contour is taken to be the unit circle traversed counterclockwise (or any positively oriented [[Jordan curve]] about 0). In the case of the unit circle there is a direct method to evaluate the integral
<math display=block>\oint_C \frac{1}{z}\,dz.</math>

In evaluating this integral, use the unit circle {{math|1={{abs|''z''}} = 1}} as a contour, parametrized by {{math|1=''z''(''t'') = ''e<sup>it</sup>''}}, with {{math|''t'' ∈ [0, 2π]}}, then {{math|1={{sfrac|''dz''|''dt''}} = ''ie<sup>it</sup>''}} and
<math display=block>\oint_C \frac{1}{z}\,dz = \int_0^{2\pi} \frac{1}{e^{it}} ie^{it}\,dt = i\int_0^{2\pi} 1 \, dt = i \, t\Big|_0^{2\pi} = \left(2\pi-0\right)i = 2\pi i</math>

which is the value of the integral. This result only applies to the case in which z is raised to power of -1. If the power is not equal to -1, then the result will always be zero.