219. Curvilinear integrals

Let $AB$ be an arc $C$ of a curve defined by the equations $$x=\varphi (t),y=\psi (t),$$ where $\varphi$ and $\psi$ are functions of $t$ with continuous differential coefficients ${\varphi }^{\prime }$ and ${\psi }^{\prime }$; and suppose that, as $t$ varies from $t_0$ to $t_1$, the point $(x,y)$ moves along the curve, in the same direction, from $A$ to $B$.

Then we define the curvilinear integral $$\begin{array}{}\text{(1)}& {\int }_C\{g(x,y)dx+h(x,y)dy\},\end{array}$$ where $g$ and $h$ are continuous functions of $x$ and $y$, as being equivalent to the ordinary integral obtained by effecting the formal substitutions $x=\varphi (t)$, $y=\psi (t)$,  to $${\int }_{t_0}^{t_1}\{g(\varphi ,\psi ){\varphi }^{\prime }+h(\varphi ,\psi ){\psi }^{\prime }\}dt.$$ We call $C$ the path of integration.

Let us suppose now that $$z=x+iy=\varphi (t)+i\psi (t),$$ so that $z$ describes the curve $C$ in Argand’s diagram as $t$ varies. Further let us suppose that $$f(z)=u+iv$$ is a polynomial in $z$ or rational function of $z$.

Then we define $$\begin{array}{}\text{(2)}& {\int }_{C}f(z)dz\end{array}$$ as meaning $${\int }_C(u+iv)(dx+idy),$$ which is itself defined as meaning $${\int }_C(udx–vdy)+i{\int }_C(vdx+udy).$$