In mathematics, the Dawson function or Dawson integral (named after H. G. Dawson ) is the one-sided Fourier–Laplace sine transform of the Gaussian function.

Definition

The Dawson function is defined as either: also denoted as F(x) or D(x), or alternatively The Dawson function is the one-sided Fourier–Laplace sine transform of the Gaussian function, It is closely related to the error function erf, as where erfi is the imaginary error function, erfi(x) = −i erf(ix). Similarly, in terms of the real error function, erf. In terms of either erfi or the Faddeeva function w(z), the Dawson function can be extended to the entire complex plane: which simplifies to for real x. For |x| near zero, F(x) ≈ x. For |x| large, F(x) ≈ 1/(2x). More specifically, near the origin it has the series expansion while for large x it has the asymptotic expansion More precisely where n!! is the double factorial. F(x) satisfies the differential equation with the initial condition F(0) = 0. Consequently, it has extrema for resulting in x = ±0.92413887..., F(x) = ±0.54104422.... Inflection points follow for resulting in x = ±1.50197526..., F(x) = ±0.42768661.... (Apart from the trivial inflection point at x = 0, F(x) = 0.)

Relation to Hilbert transform of Gaussian

The Hilbert transform of the Gaussian is defined as P.V. denotes the Cauchy principal value, and we restrict ourselves to real y. H(y) can be related to the Dawson function as follows. Inside a principal value integral, we can treat 1/u as a generalized function or distribution, and use the Fourier representation With we use the exponential representation of \sin(ku) and complete the square with respect to x to find We can shift the integral over x to the real axis, and it gives \pi^{1/2}. Thus We complete the square with respect to k and obtain We change variables to u = ik/2+y: The integral can be performed as a contour integral around a rectangle in the complex plane. Taking the imaginary part of the result gives where F(y) is the Dawson function as defined above. The Hilbert transform of is also related to the Dawson function. We see this with the technique of differentiating inside the integral sign. Let Introduce The nth derivative is We thus find The derivatives are performed first, then the result evaluated at a = 1. A change of variable also gives Since we can write where P_1 and P_2 are polynomials. For example, Alternatively, H_n can be calculated using the recurrence relation (for n \geq 0)