How do I compute $\int_{-\infty}^\infty e^{-\frac{x^2}{2t}} e^{-ikx} \, \mathrm dx$ for $t \in \mathbb{R}_{>0}$ and $k \in \mathbb{R}$?

Question

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.

Answer 1

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

$$ \begin{align*} \int_{-\infty}^{\infty} \exp\left\{ -\frac{x^2}{2t} - ikx \right\} \, dx &= \int_{-\infty}^{\infty} \exp\left\{ -\frac{1}{2t}\left( x + ikt \right)^2 - \frac{k^2t}{2} \right\} \, dx \\ &= \exp\left\{ - \frac{k^2t}{2} \right\} \int_{-\infty}^{\infty} \exp\left\{ -\frac{x^2}{2t} \right\} \, dx \\ &= \sqrt{2\pi t} \exp\left\{ - \frac{k^2t}{2} \right\}. \end{align*}$$

Answer 2

After completing the square of the exponential argument, use the following contour to justify the change of variables,

Here is what $c$ is $$\displaystyle \int_{-\infty}^\infty e^{-p(t+c)^2}dt = \sqrt{\frac{\pi}{p}}, \quad p,c\in {\bf C},\;\mathrm{Re}\displaystyle \left\{p\right\}>0$$

Answer 3

Complete the square:

$$\frac{x^2}{2t}+ikx=\frac{1}{2t}(x^2+2tikx)=\frac{1}{2t}(x+tik)^2+\frac{t^2k^2}{2t}=\frac{1}{2t}\left[(x+tik)^2+t^2k^2\right]\Longrightarrow$$

$$\Longrightarrow e^{-\frac{x^2}{2t}-ikx}=e^{-\frac{1}{2t}(x+tik)^2}\,e^{-\frac{1}{2}tk^2}$$

Now, substituting

$$u:=\frac{1}{\sqrt{2t}}(x+tik)\Longrightarrow du=\frac{1}{\sqrt{2t}}dx\Longrightarrow$$

$$\Longrightarrow \int_{-\infty}^\infty e^{-\frac{1}{2t}(x+tik)^2}dx=\sqrt{2t}\int_{-\infty}^\infty e^{-u^2}du=\sqrt {2t\pi}$$

End now the exercise.