Integrating the gaussian kernel on unbounded curves in $\Bbb C$

Question

It is well known (e.g. using Fourier transform) that if $\beta\in\Bbb C$ with $\Re\beta>0$, then $$ \int_{\Bbb R}e^{-bu^2}\,du=\sqrt{\frac{\pi}{\beta}}\;. $$ Once chosen a branch for the square root, by the variable change $\sqrt\beta u=w$ we get $$ \frac1{\sqrt\pi}\int_{\sqrt\beta\Bbb R}e^{-w^2}\,dw=1\;. $$ Now, $\sqrt\beta\Bbb R$ is a straight line in $\Bbb C$. Does this holds true on more generic curves in $\Bbb C$ which are diffeomorphic to $\Bbb R$? For example, if $h\colon\Bbb R\to\Bbb R$ is $\mathscr C^1$, bounded with $h'(x)\to0$ as $x\to\infty$, if we consider the curve $$ \Gamma:=\{x+ih(x)\;:\;x\in\Bbb R\} $$ Does $$ \frac1{\sqrt\pi}\int_{\Gamma}e^{-z^2}\;dz=1 $$ hold true?