To obtain the inverse fourier transform of the gaussian integral presented consider
$$\begin{array}{rcl}f(x) &=& \int_{-\infty}^{\infty} F(\omega) \textrm{e}^{-i\omega x} d\omega\\ &=& \int_{-\infty}^{\infty} \textrm{e}^{-\alpha \omega^2} \textrm{e}^{-i\omega x} d\omega\end{array}.$$
By completing the square,
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$$\begin{array}{rcl}f(x) &=& \int_{-\infty}^{\infty} \textrm{e}^{-\alpha \left( \omega ^2 + i \frac{x \omega}{\alpha} \right)} \textrm{d}\omega\\&=& \int_{-\infty}^{\infty} \textrm{e}^{ -\alpha \left( \omega + i \frac{x}{2\alpha} \right) ^2} \textrm{e}^{-\frac{x^2}{4\alpha}} \textrm{d}\omega\\ &=& \textrm{e}^{-\frac{x^2}{4\alpha}} \int_{-\infty + i \frac{x}{2\alpha}}^{\infty + i \frac{x}{2\alpha}} \textrm{e}^{-\alpha z^2} \textrm{d}z\\ &=& \textrm{e}^{-\frac{x^2}{4\alpha}} \int_{-\infty}^{\infty} \textrm{e}^{-\alpha z^2} \textrm{d}z.\end{array}$$ |
(A.1) |
The last simplifications were obtained by substituting
$$\begin{array}{ccc} z = \omega + i\frac{x}{2\alpha} & \textrm{and} & \textrm{d}z = \textrm{d}\omega \end{array}.$$
By virtue of Cauchy's Integral Theorem[1] this function, integrated on the complex plane, yields an equality of the last two expressions in (Eq. A.1). (i.e. Integration on the infinite imaginary space is equal to the infinite integration on the real axis.) Hence a real solution is attained.
Finally, with the substitution of $y = \sqrt{\alpha}z$ and $\textrm{d}y = \sqrt{\alpha}z$,
$$f(x) = \frac{\textrm{e}^{-x^2/4\alpha}}{\sqrt{\alpha}} \int_{-\infty}^{\infty} \textrm{e}^{-y^2} dy.$$
The Gaussian integral is carried out in Appendix A.3.
References
- , Mathematical Methods for Physicists. Academic Press, 1995.