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?

Answer 1

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 $a=0$,

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

Answer 2

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

Answer 3

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 $z$.