Integration by parts
by Hidenori
Proposition
Prove the following integration by parts statement: Let $f$ and $g$ be holomorphic in $G$ and suppose $\gamma \subset G$ is a piecewise smooth path from $\gamma(a)$ to $\gamma(b)$. Then
\[\begin{align*} \int_{\gamma} fg' = f(\gamma(b))g(\gamma(b)) - f(\gamma(a))g(\gamma(a)) - \int_{\gamma} f'g. \end{align*}\]Solution
\[\begin{align*} \int_{\gamma} (fg)' &= \int_{\gamma} f'g + fg' \\ &= \int_{a}^{b} (f'g + fg')(\gamma(t))\gamma'(t) dt \\ &= \int_{a}^{b} (f'g)(\gamma(t))\gamma'(t) dt + \int_{a}^{b} (fg')(\gamma(t))\gamma'(t) dt \\ &= \int_{\gamma} f'g + \int_{\gamma} fg'. \end{align*}\]On the other hand, $(fg) \circ \gamma$ is a function whose derivative is $((fg)’ \circ \gamma) \cdot \gamma’$.
Therefore, $\int_{\gamma} (fg)’ = \int_{a}^{b} (fg)’(\gamma(t))\gamma’(t) dt = ((fg) \circ \gamma)(b) - ((fg) \circ \gamma)(a) = f(\gamma(b))g(\gamma(b)) - f(\gamma(a))g(\gamma(a))$.
By putting them together, we obtain \(\begin{align*} \int_{\gamma} fg' = f(\gamma(b))g(\gamma(b)) - f(\gamma(a))g(\gamma(a)) - \int_{\gamma} f'g. \end{align*}\)
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