Gauss transform

The linear functional transform $W ( \zeta ) [ x]$ of a function $x( t)$ defined by the integral

$$W ( \zeta ) [ x] = \ \frac{1}{\sqrt {\pi \zeta } } \int\limits _ {- \infty } ^ \infty \mathop{\rm exp} \left ( - \frac{u ^ {2} } \zeta \right ) x ( t + u) du,$$

$$\mathop{\rm Re} \zeta > 0.$$

If $x( t) \in L _ {2} ( - \infty , \infty )$, then $W ( \zeta ) [ x] \in L _ {2} ( - \infty , \infty )$; for real values $\zeta = \overline \zeta \;$, the operator $W ( \zeta )$ is self-adjoint [1]. The inversion formula for the Gauss transform is

$$x ( t) = \mathop{\rm exp} \left \{ - { \frac \zeta {4} } \frac{d ^ {2} }{dt ^ {2} } \right \} W ( \zeta ) [ x ( t)].$$

If $\zeta = 4$, the Gauss transform is known as the Weierstrass transform.

References

 [1] E. Hille, R.S. Phillips, "Functional analysis and semi-groups" , Amer. Math. Soc. (1957) [2] V.A. Ditkin, A.P. Prudnikov, "Integral transforms" Itogi Nauk. Ser. Mat. Mat. Anal. (1966) pp. 7–82 (In Russian)

The above inversion formula can be interpreted in terms of semi-groups. Another way to invert the Gauss transform is to write in the first equation $t + u = v$, from which substitution a double-sided Laplace transform results. Then the inversion formula follows from well-known Laplace-transform techniques.