Substitution and integration by parts for functions of a complex variable

This term I am teaching “Funktionentheorie”, i.e. functions of a complex variable. It turned out that it would be useful to have the rule of partial integration and a substitution rule for line integrals in this context. Well, these rules hold true and are not difficult to prove, but for some reason they are not treated in the textbooks we used. I state them here for convenience.

Proposition 1 (Partial integration for line integrals of holomorphic functions) Let ${f}$ and ${g}$ be holomorphic functions on a domain ${G\subset {\mathbb C}}$ and let ${\gamma:[a,b]\rightarrow G}$ be continuous and piecewise differentiable. Then it holds that

$\displaystyle \begin{array}{rcl} \int_\gamma f(z)g'(z) dz & =& -\int_\gamma f'(z) g(z)dz\\ & & \quad+ \Big[ f(\gamma(b))g(\gamma(b)) - f(\gamma(a))g(\gamma(a))\Big]. \end{array}$

Proof: It holds that

$\displaystyle (f\,g)'(z) = f'(z)g(z) + f(z) g'(z).$

Integration this along ${\gamma}$ and using that ${f\, g}$ is obviously an antiderivative of ${(f\,g)'}$ we obtain

$\displaystyle \Big[ f(\gamma(b))g(\gamma(b)) - f(\gamma(a))g(\gamma(a))\Big] = \int_\gamma f'(z)g(z)dz + \int_\gamma f(z)g'(z)dz.$

$\Box$

Proposition 2 (Substitution rule for line integrals of functions of a complex variable) Let ${\gamma:[a,b]\rightarrow {\mathbb C}}$ be continuous and piecewise differentiable, ${f}$ be continuous on a neighborhood of ${g([a,b])}$ and let ${\phi:{\mathbb C}\rightarrow{\mathbb C}}$ be bi-holomorphic. Then it holds that

$\displaystyle \int_\gamma f(z) dz = \int_{\phi^{-1}\circ \gamma} f(\phi(\zeta))\phi'(\zeta) d\zeta.$

Proof: We simply calculate by substituting the parameterization and using the rule for the derivative of the inverse function:

$\displaystyle \begin{array}{rcl} \int_{\phi^{-1}\circ \gamma} f(\phi(\zeta))\phi'(\zeta)d\zeta & = & \int_a^b f(\phi(\phi^{-1}(\gamma(t))))\phi'(\phi^{-1}(\gamma(t))) (\phi^{-1}\circ\gamma)'(t)dt\\ & = & \int_a^b f(\gamma(t)) \phi'(\phi^{-1}(\gamma(t))) \frac{\gamma'(t)}{\phi'(\phi^{-1}(\gamma(t)))} dt\\ & = & \int_a^b f(\gamma(t))\gamma'(t)dt\\ & = & \int_\gamma f(z) dz. \end{array}$

$\Box$

Remark: It is enough that the transformation ${\phi}$ is biholomorphic from a neighborhood ${U}$ of ${\gamma([a,b])}$ onto its image ${\phi(U)}$ .