2

I'm trying to solve the following analytically:

$P(u) = {1\over 2\pi} \int^{+\infty}_{-\infty} e^{i ut} \int^{+\infty}_{-\infty} e^{-x^2\over2} e^{-i \alpha t x} dx dt $

Where $i$ is the imaginary unit, $\alpha$ is a real parameter. $x$, $u$ and $t$ are obviously variables. Does it have an analytical solution? Many thanks.

user2350366
  • 513
  • Further to @RonGordon's answer you use the following hints. hint 1: $$-\frac{x^{2}}{2}-iaxt= -\frac{\left(x+iat\right)^{2}+a^{2}t^{2}}{2}$$ hint 2: you can change variable and integrate over this as normal (even though it technical is a complex integral!) – Chinny84 Jan 24 '14 at 16:14
  • While this has already been answered: where did you come across this? What did you try? Do you know the Gaussian integral at all? This is so specifically a Fourier transform and IFT that it's hard to imagine you would hit it in any other context... – Steven Stadnicki Jan 24 '14 at 16:27

2 Answers2

1

The inner integral is a well-known Fourier transform of the Gaussian term and is

$$\sqrt{2 \pi} e^{-\alpha^2 t^2/2}$$

The outer integral is simply an inverse FT applied to the above result, scaled by the factor $\alpha$, or

$$P(u) = \frac1{\sqrt{\alpha}} e^{-u^2/(2 \alpha^2)}$$

Ron Gordon
  • 138,521
1

Since $\frac{x^2}{2} + i \alpha t x = \left(\frac{x}{\sqrt{2}} + i \alpha t \frac{\sqrt{2}}{2}\right )^2 + \frac{\alpha^2 t^2}{2}$, your first integral becomes:

$$\int^{\infty}_{-\infty} e^{-(x/\sqrt{2} + i \alpha t \sqrt{2}/2)^2} e^{-\alpha^2 t^2/2} \, dx = e^{-\alpha^2 t^2/2} \sqrt{2} \int^{\infty}_{-\infty}e^{-y^2}\,dy =\sqrt{2 \pi} e^{-\alpha^2 t^2/2} . $$

I hope this is useful for you to solve the other integral.

Dmoreno
  • 7,517