Use integration along $\partial Q$ of $Q=[-R,R]+\mathrm i[0,Y]$ to show that for all $Y\geq 0$ it holds that
$$\int_{-\infty}^\infty \exp(-(x+\mathrm iY)^2)~\mathrm dx = \int_{-\infty}^\infty \exp(-x^2)~\mathrm dx.$$
Similiar to one of my other questions referring to a rectangle I was going to rewrite $\partial Q$ as four curves but the integrands became really complicated. Using the bottom curve of the rectangle I got $\require{cancel}\cancel{\gamma(t)=(1-t)(-R)+tR=R(2t-1)}$ hence
$$\cancel{\int_\gamma\exp(-(x+\mathrm iY)^2)~\mathrm dx =\int_0^1 \exp(-(R(2t-1)+\mathrm iY)^2)\cdot 2R~\mathrm dt}$$
which seems like a very tough integral. I was thinking of a similiar (and more simple) curve $\hat{\gamma}$ using another parametrisation to get easier integrals.
What would be your suggested approach for this problem?
Results of some attempts after getting some help
After much help of πr8 I have the following results for all four curves thus far
$$ \begin{align*} \gamma_1(t) &= (1-t)(-R) + tR = R(2t-1)\\ \gamma_1'(t) &= 2R\\ \int_{\gamma_1}\exp(-(x+\mathrm iY)^2)\,\mathrm dx &= \int_0^1 \exp(-(\underbrace{R(2t-1)}_{u}+\mathrm iY)^2)\cdot 2R\,\mathrm dt\\ &= \int_{-R}^R \exp(-(u+\mathrm iY)^2)\,\mathrm du \end{align*} $$
$$ \begin{align*} \gamma_2(t) &= (1-t)R + t(R+\mathrm iY) = R+t\mathrm iY\\ \gamma_2'(t) &= \mathrm iY\\ \int_{\gamma_2}\exp(-(x+\mathrm iY)^2)\,\mathrm dx &= \int_0^1 \exp(-(\underbrace{R+t\mathrm iY}_{u}+\mathrm iY)^2)\cdot \mathrm iY\,\mathrm dt\\ &= \int_{R}^{R+\mathrm iY} \exp(-(u+\mathrm iY)^2)\,\mathrm du \end{align*} $$
$$ \begin{align*} \gamma_3(t) &= (1-t)(R+\mathrm iY) + t(-R+\mathrm iY) = R(1-2t)+\mathrm iY\\ \gamma_3'(t) &= -2R\\ \int_{\gamma_3}\exp(-(x+\mathrm iY)^2)\,\mathrm dx &= \int_0^1 \exp(-(\underbrace{R(1-2t)+\mathrm iY}_{u}+\mathrm iY)^2)\cdot (-2R)\,\mathrm dt\\ &= \int_{R+\mathrm iY}^{-R+\mathrm iY} \exp(-(u+\mathrm iY)^2)\,\mathrm du \end{align*} $$
So far I could rewrite all integrals to have the same boundaries as the curves used to deduce them. For the last curve however this wasn't as straight-forward as for the other ones (I was eager to have four identical integrands) and I had to use another substitution yielding
$$ \begin{align*} \gamma_4(t) &= (1-t)(-R+\mathrm iY) + t(-R) = -R-(t-1)\mathrm iY\\ \gamma_4'(t) &= -\mathrm iY\\ \int_{\gamma_4}\exp(-(x+\mathrm iY)^2)\,\mathrm dx &= \int_0^1 \exp(-(\underbrace{-R-t\mathrm iY+\mathrm iY}_{u})^2)\cdot \mathrm iY\,\mathrm dt\\ &= \int_{-R+\mathrm iY}^{-R} \exp(-u^2)\,\mathrm du. \end{align*} $$
Is there anything salvageable in the above equations?
Since $z\mapsto \exp(-z^2)$ is an entire function, by Cauchy's theorem the integral $\int_{\partial Q} \exp(-z^2)\, dz$ is zero. On the other hand,
$$\int_{\partial Q} \exp(-z^2)\, dz = \left(\int_{-R}^R + \int_R^{R + iY} - \int_{-R + iY}^{R + iY} - \int_{-R}^{-R + iY}\right)\exp(-z^2)\, dz$$ Thus
$$\int_{-R + iY}^{R + iY} \exp(-z^2)\, dz = \int_{-R}^R \exp(-x^2)\, dx + \int_R^{R + iY} \exp(-z^2)\, dz - \int_{-R}^{-R + iY} \exp(-z^2)\, dz\tag{1}$$
Now, the integral along the vertical edges of $Q$ are $O\!\left(\exp(-R^2)\right)$ as $R\to \infty$. Indeed, consider the vertical edge from $R$ to $R + iY$, parametrized by $z = R + it, 0 \le t \le Y$.
$$\left\lvert \int_R^{R + iY} \exp(-z^2)\, dz\right\rvert = \left\lvert\int_0^Y \exp\{-(R + it)^2\}i\, dt\right\rvert \le \int_0^Y \exp\{-(R^2 - t^2)\}\, dt = Ce^{-R^2}$$
where $C = \int_0^Y \exp(t^2)\, dt$. Similarly, $\left\lvert\int_{-R}^{-R + iY} \exp(-z^2)\, dz\right\rvert \le Ce^{-R^2}$. Hence, by $(1)$,
$$\int_{-R+iY}^{R + iY} \exp(-z^2)\, dz = \int_{-R}^R \exp(-x^2)\, dx + O\!\left(\exp(-R^2)\right) \tag{2} $$
Parametrizing the horizontal edge $[-R + iY,R + iY]$ via $z = x + iY$, $-R\le x \le R$, we have $\int_{-R + iY}^{R + iY} \exp(-z^2)\, dz = \int_{-R}^R \exp(-(x + iY)^2)\, dx$. Therefore, $(2)$ becomes
$$\int_{-R}^R \exp(-(x + iY)^2)\, dx = \int_{-R}^R \exp(-x^2)\, dx + O\!\left(\exp(-R^2)\right)\tag{3}$$ Letting $R \to \infty$ in $(3)$, the result is established.