[proofplan] We prove the first identity by reducing the line integral to ordinary one-dimensional integrals on intervals where both $\gamma$ and $\phi$ are $C^1$. On each such interval, the chain rule gives \begin{align*} (\gamma \circ \phi)' = (\gamma' \circ \phi)\phi', \end{align*} and the one-dimensional substitution formula converts the integral over the new parameter back to the original parameter. The reversal identity is the same computation with the orientation-reversing affine change of variable $u = a+b-t$. [/proofplan]

[step:Choose partitions on which all paths are $C^1$] Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. Let \begin{align*} a = t_0 < t_1 < \cdots < t_m = b \end{align*} be a partition of $[a,b]$ such that the restriction \begin{align*} \gamma|_{[t_{k-1},t_k]}: [t_{k-1},t_k] \to \Omega \end{align*} is $C^1$ for every $k \in \{1,\dots,m\}$. Because $\phi: [c,d] \to [a,b]$ is increasing, it is monotone. A monotone map on an interval has one-sided limits at every interior point; if it had a jump discontinuity, every value strictly between the left and right limits at the jump would be omitted from its image. Since $\phi$ is bijective onto the entire interval $[a,b]$, no such jump can occur, so $\phi$ is continuous. Hence $\phi(c)=a$ and $\phi(d)=b$. For each $k \in \{0,\dots,m\}$, define $s_k := \phi^{-1}(t_k)$. Then \begin{align*} c = s_0 < s_1 < \cdots < s_m = d. \end{align*} Refine this partition further, if necessary, so that on every subinterval \begin{align*} [\sigma_{\ell-1},\sigma_\ell] \subset [c,d] \end{align*} the map \begin{align*} \phi|_{[\sigma_{\ell-1},\sigma_\ell]}: [\sigma_{\ell-1},\sigma_\ell] \to [a,b] \end{align*} is $C^1$, and the image $\phi([\sigma_{\ell-1},\sigma_\ell])$ lies inside some interval $[t_{k-1},t_k]$ on which $\gamma$ is $C^1$. Therefore \begin{align*} (\gamma \circ \phi)|_{[\sigma_{\ell-1},\sigma_\ell]}: [\sigma_{\ell-1},\sigma_\ell] \to \Omega \end{align*} is $C^1$ on each such subinterval. Hence $\gamma \circ \phi$ is piecewise $C^1$. [/step]

[step:Transform the integral on each smooth subinterval]Fix one subinterval $[\sigma_{\ell-1},\sigma_\ell]$ of the refined partition. Define \begin{align*} \alpha_\ell &:= \phi(\sigma_{\ell-1}), & \beta_\ell &:= \phi(\sigma_\ell). \end{align*} Since $\phi$ is increasing, $\alpha_\ell \leq \beta_\ell$. On this subinterval, the [Chain Rule](/page/Chain%20Rule) applies because $\phi$ is $C^1$ there and $\gamma$ is $C^1$ on the image interval. It gives \begin{align*} (\gamma \circ \phi)'(r) = \gamma'(\phi(r))\phi'(r) \end{align*} for every $r \in (\sigma_{\ell-1},\sigma_\ell)$ where the derivatives are defined. Define the continuous map \begin{align*} g_\ell: [\alpha_\ell,\beta_\ell] &\to \mathbb{C} \\ u &\mapsto f(\gamma(u))\gamma'(u). \end{align*} The [one-dimensional substitution formula](/page/Substitution%20Formula) applies to the increasing $C^1$ map $\phi|_{[\sigma_{\ell-1},\sigma_\ell]}$ and the continuous map $g_\ell$. It gives \begin{align*} \int_{\sigma_{\ell-1}}^{\sigma_\ell} f(\gamma(\phi(r)))\gamma'(\phi(r))\phi'(r)\, d\mathcal{L}^1(r) = \int_{\alpha_\ell}^{\beta_\ell} f(\gamma(u))\gamma'(u)\, d\mathcal{L}^1(u). \end{align*} Using the chain-rule identity, this is exactly \begin{align*} \int_{\sigma_{\ell-1}}^{\sigma_\ell} f((\gamma \circ \phi)(r))(\gamma \circ \phi)'(r)\, d\mathcal{L}^1(r) = \int_{\alpha_\ell}^{\beta_\ell} f(\gamma(u))\gamma'(u)\, d\mathcal{L}^1(u). \end{align*}[/step]

[guided]The goal on this subinterval is to compare the line integral written with the new parameter $r$ to the same curve written with the old parameter $u$. Since $\phi$ is $C^1$ on $[\sigma_{\ell-1},\sigma_\ell]$ and $\gamma$ is $C^1$ on the image interval $[\alpha_\ell,\beta_\ell]$, the composition \begin{align*} (\gamma \circ \phi)|_{[\sigma_{\ell-1},\sigma_\ell]}: [\sigma_{\ell-1},\sigma_\ell] \to \Omega \end{align*} is $C^1$. The [Chain Rule](/page/Chain%20Rule) applies to these two $C^1$ maps and gives \begin{align*} (\gamma \circ \phi)'(r) = \gamma'(\phi(r))\phi'(r). \end{align*} Now define \begin{align*} g_\ell: [\alpha_\ell,\beta_\ell] &\to \mathbb{C} \\ u &\mapsto f(\gamma(u))\gamma'(u). \end{align*} This map is continuous because $f$ is continuous, $\gamma$ is $C^1$ on $[\alpha_\ell,\beta_\ell]$, and $\gamma'$ is continuous there. The [one-dimensional substitution formula](/page/Substitution%20Formula) applies to the increasing $C^1$ map $\phi|_{[\sigma_{\ell-1},\sigma_\ell]}$ and the continuous integrand $g_\ell$. Thus \begin{align*} \int_{\sigma_{\ell-1}}^{\sigma_\ell} g_\ell(\phi(r))\phi'(r)\, d\mathcal{L}^1(r) = \int_{\phi(\sigma_{\ell-1})}^{\phi(\sigma_\ell)} g_\ell(u)\, d\mathcal{L}^1(u). \end{align*} Substituting the definition of $g_\ell$ and the definitions $\alpha_\ell=\phi(\sigma_{\ell-1})$ and $\beta_\ell=\phi(\sigma_\ell)$ gives \begin{align*} \int_{\sigma_{\ell-1}}^{\sigma_\ell} f(\gamma(\phi(r)))\gamma'(\phi(r))\phi'(r)\, d\mathcal{L}^1(r) = \int_{\alpha_\ell}^{\beta_\ell} f(\gamma(u))\gamma'(u)\, d\mathcal{L}^1(u). \end{align*} Finally, replacing $\gamma'(\phi(r))\phi'(r)$ by $(\gamma \circ \phi)'(r)$ identifies the left-hand side with the line-integral contribution of the reparametrised path on this subinterval.[/guided]

[step:Sum the subinterval identities to obtain reparametrisation invariance] By definition of the complex line integral for a piecewise $C^1$ path, \begin{align*} \int_{\gamma \circ \phi} f(z)\, dz = \sum_{\ell} \int_{\sigma_{\ell-1}}^{\sigma_\ell} f((\gamma \circ \phi)(r))(\gamma \circ \phi)'(r)\, d\mathcal{L}^1(r). \end{align*} Using the identity proved on each subinterval, \begin{align*} \int_{\gamma \circ \phi} f(z)\, dz = \sum_{\ell} \int_{\alpha_\ell}^{\beta_\ell} f(\gamma(u))\gamma'(u)\, d\mathcal{L}^1(u). \end{align*} The intervals $[\alpha_\ell,\beta_\ell]$ form a refinement of the original partition of $[a,b]$, in increasing order. Additivity of the one-dimensional integral over adjacent intervals gives \begin{align*} \sum_{\ell} \int_{\alpha_\ell}^{\beta_\ell} f(\gamma(u))\gamma'(u)\, d\mathcal{L}^1(u) = \int_a^b f(\gamma(u))\gamma'(u)\, d\mathcal{L}^1(u) = \int_\gamma f(z)\, dz. \end{align*} Therefore \begin{align*} \int_{\gamma \circ \phi} f(z)\, dz = \int_\gamma f(z)\, dz. \end{align*} [/step]

[step:Apply the orientation-reversing affine change of parameter] Define the affine map \begin{align*} \rho: [a,b] &\to [a,b] \\ t &\mapsto a+b-t. \end{align*} Then $-\gamma = \gamma \circ \rho$, and $\rho'(t)=-1$ for every $t \in [a,b]$. The reversed breakpoints \begin{align*} a+b-t_m < a+b-t_{m-1} < \cdots < a+b-t_0 \end{align*} form a partition of $[a,b]$, and on each of its subintervals the image under $\rho$ lies in one interval $[t_{k-1},t_k]$ on which $\gamma$ is $C^1$. Hence $-\gamma$ is piecewise $C^1$. On each such subinterval, the [Chain Rule](/page/Chain%20Rule) gives \begin{align*} (-\gamma)'(t) = -\gamma'(a+b-t). \end{align*} Hence, by the definition of the line integral, \begin{align*} \int_{-\gamma} f(z)\, dz &= \int_a^b f(\gamma(a+b-t))\bigl(-\gamma'(a+b-t)\bigr)\, d\mathcal{L}^1(t). \end{align*} Using the [one-dimensional substitution formula](/page/Substitution%20Formula) with the affine $C^1$ change of variable $u=a+b-t$, the interval $t \in [a,b]$ becomes $u \in [b,a]$, and the orientation is reversed. Therefore \begin{align*} \int_a^b f(\gamma(a+b-t))\bigl(-\gamma'(a+b-t)\bigr)\, d\mathcal{L}^1(t) &= -\int_a^b f(\gamma(u))\gamma'(u)\, d\mathcal{L}^1(u) \\ &= -\int_\gamma f(z)\, dz. \end{align*} Thus \begin{align*} \int_{-\gamma} f(z)\, dz = -\int_\gamma f(z)\, dz. \end{align*} [/step]