where \(C\) is a positive contour(anti-clockwise), and there are \(m\) poles inside the contour. Each \(d_k\) is the residual of each pole.
Residual
For a \(n\)-th order pole at \(s_0\), its residual
\[d = \frac{1}{(n-1)!} \frac{d^{n-1}}{ds^{n-1}}(s-s_0)^n f(s)\Bigg|_{s=s_0}\]for example, for a first order pole
\[d = (s-s_0)f(s)\Big|_{s=s_0}\]Example 01
\[\begin{aligned} \int_{C} \frac{ds}{a^2 - s^2} \end{aligned}\]where \(C\) is a positive contour around \(s_0 = a\). Since it is a first order pole
\[\begin{aligned} d = \frac{s - a}{a^2 - s^2} \Bigg|_{s=a} = -\frac{1}{a+s}\Bigg|_{s=a} = -\frac{1}{2a} \end{aligned}\]thus
\[\begin{aligned} \int_{C} \frac{ds}{a^2 - s^2} = 2\pi j \Big(-\frac{1}{2a}\Big) = -j\frac{\pi}{a} \end{aligned}\]side note
using Jordan’s Lemma, we can calculate the below real integral, using the above result
\[\int_{-\infty}^{\infty}\frac{dy}{a^2 + y^2} = -j (2 \pi j) (-\frac{1}{2a}) (-1) = \frac{\pi}{a}\]