How do I compute $\int_{-\infty}^\infty e^{-\frac{x^2}{2t}} e^{-ikx} \, \mathrm dx$ for $t \in \mathbb{R}_{>0}$ and $k \in \mathbb{R}$? Let $t \in \mathbb{R}_{>0}$ and $k \in \mathbb{R}$. I want to find
$$\int_{-\infty}^\infty e^{-\frac{x^2}{2t}} e^{-ikx} \, \mathrm dx.$$
A hint told me to first determine $\int_{-\infty}^\infty e^{-\frac{x^2}{2}} \, \mathrm dx$ which I found to be equal to $\sqrt{2 \pi}$. Now I am told to compute the integral using Cauchy's formula for a convenient Cauchy contour. As I have not yet practically applied Cauchy's formula and I have no idea how it would be helpful in this case, I ask for a little help, a hint would be enough really. Thanks in advance.

Here is a method which circumvents complex analysis. As DonAntonio pointed out, we have 
$$ \exp\left\{ -\frac{x^2}{2t} - ikx \right\} = \exp\left\{ -\frac{1}{2t}\left( x + ikt \right)^2 - \frac{k^2t}{2} \right\} $$
Since 
$$ \frac{d}{du} \exp\left\{ -\frac{1}{2t}\left( x + iut \right)^2 \right\} = -i(x+iut) \exp\left\{ -\frac{1}{2t}\left( x + iut \right)^2 \right\}, $$
we have
$$ \begin{align*}
& \int_{-\infty}^{\infty} \exp\left\{ -\frac{1}{2t}\left( x + ikt \right)^2 \right\} \, dx
- \int_{-\infty}^{\infty} \exp\left\{ -\frac{x^2}{2t} \right\} \, dx \\
&= \int_{-\infty}^{\infty} \left[ \frac{d}{du} \exp\left\{ -\frac{1}{2t}\left( x + iut \right)^2 \right\} \right]_{u=0}^{u=k} \, dx \\
&= -i \int_{-\infty}^{\infty} \int_{0}^{k} (x+iut) \exp\left\{ -\frac{1}{2t}\left( x + iut \right)^2 \right\} \, dudx.
\end{align*}$$
Since the integrand is Lebesgue integrable on $(x,u) \in \Bbb{R} \times [0, k]$, we can apply Fubini's theorem and we have
$$ \begin{align*}
& \int_{-\infty}^{\infty} \exp\left\{ -\frac{1}{2t}\left( x + ikt \right)^2 \right\} \, dx
- \int_{-\infty}^{\infty} \exp\left\{ -\frac{x^2}{2t} \right\} \, dx \\
&= -i \int_{0}^{k} \int_{-\infty}^{\infty} (x+iut) \exp\left\{ -\frac{1}{2t}\left( x + iut \right)^2 \right\} \, dxdu.  \\
&= \int_{0}^{k} \left[ it \exp\left\{ -\frac{1}{2t}\left( x + iut \right)^2 \right\} \right]_{x=-\infty}^{x=\infty} \, du
= 0.
\end{align*}$$
Therefore
$$ \begin{align*}
\int_{-\infty}^{\infty} \exp\left\{ -\frac{x^2}{2t} - ikx \right\} \, dx
&= \int_{-\infty}^{\infty} \exp\left\{ -\frac{1}{2t}\left( x + ikt \right)^2 - \frac{k^2t}{2} \right\} \, dx \\
&= \exp\left\{  - \frac{k^2t}{2} \right\} \int_{-\infty}^{\infty} \exp\left\{ -\frac{x^2}{2t} \right\} \, dx \\
&= \sqrt{2\pi t} \exp\left\{  - \frac{k^2t}{2} \right\}.
\end{align*}$$
