Compute complex integral resulting from FT

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

$$\begin{align} \int_{-\infty}^\infty e^{-\pi(x+i\xi)^2}\,dx&=\int_{-\infty}^\infty e^{-\pi x^2}\,dx\\\\ &=1 \end{align}$$

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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 $\xi$).