Calculating $\int_{-\infty}^\infty e^{-ax^2}e^{ibx}dx$

Question

In my syllabus about quantum mechanics, they state that the following integral can be easily calculated:

$$\int_{-\infty}^\infty e^{-ax^2}e^{ibx}dx = \sqrt{\frac{\pi}{a}}e^{-b^2/4a}$$

if it is known that

$$\int_{-\infty}^{\infty}e^{-x^2}dx = \sqrt{\pi}.$$

This isn't indeed that hard, if $a$ is real. But they use it in a derivation, where $a$ has an imaginary part. I hope someone can show me, how this is also valid when $a$ and $b$ are complex. (If needed you may assume that it is known that the integral is valid for $a$ real and $b$ complex, because I can prove that already).

Answer 1

Let

$$f(z, w) = \int_{-\infty}^{\infty} e^{-zx^2} e^{iwx} \, dx. $$

It is clear that the integral converges if $z$ and $w$ are complex numbers such that $\Re z > 0$. It is also possible to show that on this region $f(z, w)$ actually defines an analytic function. Finally, we also know that for $z > 0$ and $w \in \Bbb{R}$ we have

$$ f(z, w) = \sqrt{\frac{\pi}{z}} e^{-w^2 / 4z}. \tag{1}$$

Since the right-hand side of $(1)$ also defines an analytic function on $\Re z > 0$, in view of the identity theorem, we find that $(1)$ actually holds for any $\Re z > 0$ and $w \in \Bbb{C}$.

Answer 2

Let's assume that $a$ is real and $b=c+id$ is complex. Let's furthermore assume $a>0$. Then we can write the integrand as

$$e^{-ax^2}e^{ibx}=e^{-ax^2}e^{i(c+id)x}=e^{-(ax^2+dx)}e^{icx}= e^{\frac{d^2}{4a}}e^{-(\sqrt{a}x + \frac{d}{2\sqrt{a}})^2}e^{icx}$$ Then you can substitute $y=\sqrt{a}x + \frac{d}{2\sqrt{a}}$.