- #1
Fingolfin_Noldo
- 2
- 0
Homework Statement
Given [tex]f(x) = e^{-ax^2/2} [/tex] with a > 0 then show that [tex]\^{f} = \int_{-\infty}^{\infty} e^{-i \xi x - ax^2/2} \, \mathrm{d}x = \surd\frac{2}{a} = e^{-\xi^2/2a}[/tex] by completing the square in the exponent, using Cauchy's theorem to shift the path of integration from the real axis (Im x = 0) to the horizontal line I am x = -[tex]\xi/a[/tex] and finally use the relation [tex] \int_{-\infty}^{\infty} e^{- ax^2/2} \, \mathrm{d}x = \surd\pi[/tex]
Homework Equations
Show [tex]\^{f} = \int_{-\infty}^{\infty} e^{-i \xi x - ax^2/2} \, \mathrm{d}x = \surd\frac{2}{a} e^{-\xi^2/2a}[/tex]
The Attempt at a Solution
I completed the squares and set x = x' + i [tex]\xi/a[/tex]. But now I have an integral from [tex]-\infty + i\xi/a[/tex] to [tex]\infty + i\xi/a[/tex]. Is there something I am missing here? Should I construct some contour? Constructing a semi-circle won't work since the arc does not converge. Thanks