# Solving Gaussian Integral with Complex offset

1. Nov 10, 2007

### Fingolfin_Noldo

1. The problem statement, all variables and given/known data
Given $$f(x) = e^{-ax^2/2}$$ with a > 0 then show that $$\^{f} = \int_{-\infty}^{\infty} e^{-i \xi x - ax^2/2} \, \mathrm{d}x = \surd\frac{2}{a} = e^{-\xi^2/2a}$$ 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 Im x = -$$\xi/a$$ and finally use the relation $$\int_{-\infty}^{\infty} e^{- ax^2/2} \, \mathrm{d}x = \surd\pi$$

2. Relevant equations

Show $$\^{f} = \int_{-\infty}^{\infty} e^{-i \xi x - ax^2/2} \, \mathrm{d}x = \surd\frac{2}{a} e^{-\xi^2/2a}$$

3. The attempt at a solution
I completed the squares and set x = x' + i $$\xi/a$$. But now I have an integral from $$-\infty + i\xi/a$$ to $$\infty + i\xi/a$$. 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

2. Nov 10, 2007

### Dick

If you want an explicit contour your could draw a rectangle connecting the line with imaginary part $$i\xi/a$$ to the real axis. As the ends of the rectangle go to infinity their contribution vanishes. And there are no poles enclosed. So the integral over your line is the same as the integral over the real axis.

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