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).

Please look http://math.stackexchange.com/questions/331283/fourier-transform in particular the answer — J L, Apr 27 '13 at 16:44

score 7 · Accepted Answer · answered Apr 27 '13 at 17:57

7

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 $w \in \Bbb{C}$.

answered Apr 27 '13 at 17:57

Sangchul Lee

167,468

i guess typo $z > 0$ before (1) – hasExams Apr 27 '13 at 18:09
@Crocodile, Why do you think that $z > 0$ is a typo? – Sangchul Lee Apr 27 '13 at 18:13
isn't $z$ a complex number? – hasExams Apr 27 '13 at 18:22
@Crocodile, It is tacitly assumed that only real cases are in consideration when we write $z > 0$. Indeed, we first restrict our attention to the real positive $z$ where we know that $(1)$ holds, and then extend the validity of $(1)$ to complex $z$ with $\Re z > 0$ with the identity theorem. – Sangchul Lee Apr 27 '13 at 18:27
1

ok ok ... i everytime i see $z$ i assume complex. sorry for bugging unnecessarily. – hasExams Apr 27 '13 at 18:28
Thank you for your comment. It solved the question :). – yarnamc Apr 27 '13 at 20:47

score 2 · Answer 2 · answered Apr 27 '13 at 16:18

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 $y=\sqrt{a}x + \frac{d}{2\sqrt{a}}$.

answered Apr 27 '13 at 16:18

Matt L.

10,636

1

He wants to solve when $a$ is complex, not real. – Integral Apr 27 '13 at 16:23
Read the comment in parentheses at the end of the question. Or did I get that wrong? – Matt L. Apr 27 '13 at 17:12
I meant indeed for $a$ complex, but thanks for your answer. – yarnamc Apr 27 '13 at 20:47

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

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