On the integral $\int_{-\infty}^\infty e^{-(x-ti)^2} dx$ It is known that
$$ \int_{-\infty}^\infty e^{-x^2} dx=\sqrt{\pi}.$$ 
What about $$\int_{-\infty}^\infty e^{-(x-ti)^2} dx, $$ where $t \in \mathbb{R}$, $i=
\sqrt{-1}$.
Thanks.
 A: 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 $z$.
A: 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.
A: 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
$$
  \int_{-\infty}^\infty \mathrm{e}^{-(x-ti)^2} \mathrm{d}x = \sqrt{\pi}
$$
