I obtain the following integral after doing a FT of a function
$$\int_{-\infty}^{\infty} e^{-\pi(x + i\xi)^2}dx$$
I am not sure how to evaluate it. I tried change of variable $y = x+ i\xi$. but what is the limit of this as $x \to + \infty$ or $-\infty$? I know the Gaussian integral but I cant see how it is useful for this
Help is much appreciated
note: $\xi$ is a real constant
On
Upon enforcing the substitution $x\to x-i\xi$, we obtain
$$\int_{-\infty}^\infty e^{-\pi(x+i\xi)^2}\,dx=\int_{-\infty-i\xi}^{\infty-i\xi} e^{-\pi x^2}\,dx \tag 1$$
Using Cauchy's Integral Theorem, we assert that since $e^{-\pi z^2}$ is an entire function then
$$\oint_C e^{-\pi z^2}\,dz=0 \tag 2$$
for any closed contour $C$.
Take $C$ as comprised of the line segments (i) from $-R$ to $R$, (ii) from $R$ to $R-i\xi$, (iii) from $R-i\xi$ to $-R-i\xi$, and (iv) from $-R-i\xi$ to $-R$.
It is easy to show that the contributions to the right-hand side of $(2)$ from (ii) and (iv) vanish as $R\to \infty$. Therefore, we find that
$$\int_{-\infty-i\xi}^{\infty-i\xi} e^{-\pi x^2}\,dx=\int_{-\infty}^\infty e^{-\pi x^2}\,dx \tag 3$$
Using $(3)$ in $(1)$ yields
$$\begin{align} \int_{-\infty}^\infty e^{-\pi(x+i\xi)^2}\,dx&=\int_{-\infty}^\infty e^{-\pi x^2}\,dx\\\\ &=1 \end{align}$$
NOTE:
In this note, we show that the contributions to the integral on the left-hand side of $(2)$ from integrations over the vertical strips approach zero. To that end, we note that
$$\begin{align} \left|\int_0^{\xi}e^{-\pi (\pm R+iy)^2}\,i\,dy\right| & \le e^{-\pi (R^2-\xi^2)}\to 0\,\,\text{as}\,\,R\to \infty \end{align}$$
as was to be shown!
You may re-write the integral as:
$$ e^{\pi \xi^2}\int_{-\infty}^{+\infty}\exp\left(-\pi x^2 - 2\pi i \xi x\right)\,dx.\tag{1}$$ Consider the function: $$ J(\xi) = \int_{-\infty}^{+\infty}\exp(-\pi x^2-2\pi i \xi x)\,dx \tag{2}$$ and check through differentiation under the integral sign and integration by parts that it fulfills: $$ J'(\xi) = -2\pi \xi\, J(\xi)\tag{3} $$ so: $$ J(\xi) = K\cdot e^{-\pi \xi^2}\tag{4} $$ and since $J(0)=1$, $K=1$. By plugging in $(4)$ into $(1)$, it follows that the original integral just equals $\color{red}{\large 1}$ (it does not really depend on $\xi$).