Contour integration is most commonly used to calculate integrals along the real axis, by turning them into complex integrals.
Calculate the integral
|
{\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{−∞}^{∞} {1\over
1 + {x}^{2}}\kern 1.66702pt dx
| (A.11) |
We actually know this one: it is {\left [\mathop{arctan}\nolimits (x)\right ]}_{−∞}^{∞} = π. This is the simplest example of an integral doable by contour integration. Rewrite as a complex integral
As |z|→∞, the integral over the half circle z = R{e}^{iϕ} (R fixed) gives (dz = d\left (R{e}^{iϕ}\right ) = R{e}^{iϕ}idϕ)
This means we can close of the integral by adding a contour from ∞ to −∞ along a half circle. We easily see that {z}^{2} + 1 = (z + i)(z − i), and thus has poles in the upper and lower half plane. Only the one in the upper half plane is contained inside the contour, which goes around in positive sense.
The residue is − {i\over 2} and thus
as we know should be the case.
Note that in order to work, the ratio of denomiator over numerator should be at least {1\over { R}^{2}} for large radius.