Trick to solve an improper integral

Question

$$ \int_{-\infty}^{\infty} e^{-i a x} e^{-b x^2} \,\mathrm d x$$

Is there any good trick to solve this type integral? I guess there's a way to make that form like $\int e^{-x^2} \,\mathrm d x$. So I could get answer in terms of $\pi$. Is there any good trick that I can solve this easily?

This is really not precalc; that said, my first instinct would be to substitute $u = \frac{e^{-iax}}{-ia}$ — user361424, Jun 14 '18 at 08:48
In Probability Theory this integral is essentially the 'characteristic function ' of the normal distribution. There are many ways of evaluating this integral but they all require some knowledge of complex analysis. For example, you can evaluate this when $a$ is a purely imaginary number by basic Calculus and use theorems in Complex Analysis to say what the value is when $a$ is real. — Kavi Rama Murthy, Jun 14 '18 at 09:18
Notice that the integral is real and equals $\int\cos ax,e^{-bx^2}dx$. — , Jun 14 '18 at 09:39
Your integral can be seen as the Fourier transform of $e^{-bx^2}$;it can be shown the Fourier transform of the gaussian is still a gaussian. — edmz, Jun 14 '18 at 12:23
Related: How to calculate the Fourier transform of a Gaussian function? — Rodrigo de Azevedo, Jun 15 '18 at 06:12

score 3 · Accepted Answer · 2018-06-14T09:41:56.917

3

If we can differentiate under the integration sign,

$$\frac{\partial I}{\partial a}=-i\int x\,e^{-iax-bx^2/2}dx.$$ (Notice the $/2$ for convenience.)

Then we can write $$-iaI+\frac bi\frac{\partial I}{\partial a}=\int(-ia-bx)e^{-iax-bx^2/2}dx=\left.e^{-iax-bx^2/2}dx\right|_{-\infty}^\infty=0.$$

Hence the differential equation

$$\frac{\partial I}{\partial a}+\frac abI=0$$

that is separable and solved by

$$I=Ce^{-a^2/2b}.$$

The integration constant can be determined by setting $a=0$,

$$C=\int_{-\infty}^\infty e^{-bx^2/2}dx=\frac1{\sqrt b}\int_{-\infty}^\infty e^{-x^2/2}dx.$$

edited Jun 14 '18 at 09:41

answered Jun 14 '18 at 09:36

How did you know which to use as a parameter between a and b? I really like it. – Rumplestillskin Jun 14 '18 at 09:56
Using $b$ leads nowhere because it will introduce a factor $x^2$ when what you need is $x$ to form the derivative of $iax+bx^2/2$. – Jun 14 '18 at 10:35
if for example the exponent of the $e^{-bx^2/2}$ was something different that required you to include an $x$ term in the differential equation for $I$. Is this legal? So for example if the DE was $-iaI + bx/i \frac{\partial I}{\partial a} = \cdots = 0$. – Rumplestillskin Jul 17 '18 at 06:14

score -1 · Answer 2 · answered Jun 14 '18 at 12:00

Trick, is that what you want? $$ F(z) = \int_{-\infty}^\infty e^{zx}e^{-bx^2} $$ is an analytic function of $z$, so if you know it on an infinite set (such as $z \in \mathbb R$, where Rhys's change of variables works without extra effort), then you get it for free on the rest of the values of $z$.

score -3 · Answer 3 · answered Jun 14 '18 at 09:26

-3

You have $$\int_{-\infty}^{\infty}{\frac{1}{e^{bx^2+axi}}dx}$$ Try substituting out $u=bx^2+axi$

answered Jun 14 '18 at 09:26

Rhys Hughes

12,842

2

This is hopeless. – Jun 14 '18 at 09:26
I cannot see this working at all. Can you take it further? – Rumplestillskin Jun 14 '18 at 12:16

Trick to solve an improper integral

3 Answers3