On the integral $\int_{-\infty}^\infty e^{-(x-ti)^2} dx$

Note that $$ \frac{d}{dt}\int_{-\infty}^\infty e^{-(x-it)^2}\,dx = -i\int_{-\infty}^\infty -2(x-it)e^{-(x-it)^2}\,dx = -i e^{-(x-it)^2}\big|_{x=-\infty}^\infty = 0. $$ So the integral is constant as a function of $t$, and you can set $t=0$ to find the constant.

We start with $$ \int_{-\infty}^\infty \mathrm{e}^{-(x-ti)^2} \mathrm{d}x = \mathrm{e}^{t^2} \cdot \int_{-\infty}^\infty \mathrm{e}^{-x^2} \cdot \cos(2 t x) \mathrm{d}x $$ Now, let $\mathcal{I}(t) = \int_{-\infty}^\infty \mathrm{e}^{-x^2} \cdot \cos(2 t x) \mathrm{d}x$. Then $$ \mathcal{I}^{\prime}(t) = -2\int_{-\infty}^\infty x \cdot \mathrm{e}^{-x^2} \cdot \sin(2 t x) \mathrm{d}x = \int_{-\infty}^\infty \sin(2 t x) \cdot \mathrm{d} \left( \mathrm{e}^{-x^2} \right) \stackrel{\text{by parts}}{=}\\ -2t \int_{-\infty}^\infty \cos(2 t x) \cdot \mathrm{e}^{-x^2} \mathrm{d} x = -2 t \mathcal{I}(t) $$ Thus $$ \mathcal{I}(t) = \mathcal{I}(0) \cdot \mathrm{e}^{-t^2} = \sqrt{\pi} \cdot \mathrm{e}^{-t^2} $$

Hence $$ \int_{-\infty}^\infty \mathrm{e}^{-(x-ti)^2} \mathrm{d}x = \sqrt{\pi} $$

First show that $$ F(z) := \int_{-\infty}^\infty e^{-(x-z)^2} dx $$ exists for all $z \in \mathbb C$, and that it is analytic in $z$. Then evaluate it for some convenient values (say $z$ real) that have a limit point. Then you may conclude the value for all $z$.