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)

Comments

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.

References

[a1]	I.N. Sneddon, "The use of integral transforms" , McGraw-Hill (1972)