Dawson function

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: $$D_+(x) = e^{-x^2} \int_0^x e^{t^2}\,dt,$$ also denoted as $$F(x)$$ or $$D(x),$$ or alternatively $$D_-(x) = e^{x^2} \int_0^x e^{-t^2}\,dt.\!$$

The Dawson function is the one-sided Fourier–Laplace sine transform of the Gaussian function, $$D_+(x) = \frac12 \int_0^\infty e^{-t^2/4}\,\sin(xt)\,dt.$$

It is closely related to the error function erf, as

where erfi is the imaginary error function, erfi(x) = −i erf(ix). Similarly, $$D_-(x) = \frac{\sqrt{\pi}}{2} e^{x^2} \operatorname{erf}(x)$$ 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: $$F(z) = {\sqrt{\pi} \over 2} e^{-z^2} \operatorname{erfi} (z) = \frac{i\sqrt{\pi}}{2} \left[ e^{-z^2} - w(z) \right],$$ which simplifies to $$D_+(x) = F(x) = \frac{\sqrt{\pi}}{2} \operatorname{Im}[w(x)]$$ $$D_-(x) = i F(-ix) = -\frac{\sqrt{\pi}}{2} \left[ e^{x^2} - w(-ix) \right]$$ 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 $$F(x) = \sum_{k=0}^\infty \frac{(-1)^k \, 2^k}{(2k+1)!!} \, x^{2k+1} = x - \frac{2}{3} x^3 + \frac{4}{15} x^5 - \cdots,$$ while for large $$x$$ it has the asymptotic expansion $$F(x) = \frac{1}{2 x} + \frac{1}{4 x^3} + \frac{3}{8 x^5} + \cdots.$$

More precisely $$\left|F(x) - \sum_{k=0}^{N} \frac{(2k-1)!!}{2^{k+1} x^{2k+1}}\right| \leq \frac{C_N}{x^{2N+3}}.$$ where $$n!!$$ is the double factorial.

$$F(x)$$ satisfies the differential equation $$\frac{dF}{dx} + 2xF = 1\,\!$$ with the initial condition $$F(0) = 0.$$ Consequently, it has extrema for $$F(x) = \frac{1}{2 x},$$ resulting in x = ±0.92413887..., F(x) = ±0.54104422....

Inflection points follow for $$F(x) = \frac{x}{2 x^2 - 1},$$ 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 $$H(y) = \pi^{-1} \operatorname{P.V.} \int_{-\infty}^\infty \frac{e^{-x^2}}{y-x} \, dx$$

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 $${1 \over u} = \int_0^\infty dk \, \sin ku = \int_0^\infty dk \, \operatorname{Im} e^{iku}.$$

With $$1/u = 1/(y-x),$$ we use the exponential representation of $$\sin(ku)$$ and complete the square with respect to $$x$$ to find $$\pi H(y) = \operatorname{Im} \int_0^\infty dk \,\exp[-k^2/4+iky] \int_{-\infty}^\infty dx \, \exp[-(x+ik/2)^2].$$

We can shift the integral over $$x$$ to the real axis, and it gives $$\pi^{1/2}.$$ Thus $$\pi^{1/2} H(y) = \operatorname{Im} \int_0^\infty dk \, \exp[-k^2/4+iky].$$

We complete the square with respect to $$k$$ and obtain $$\pi^{1/2}H(y) = e^{-y^2} \operatorname{Im} \int_0^\infty dk \, \exp[-(k/2-iy)^2].$$

We change variables to $$u = ik/2+y:$$ $$\pi^{1/2}H(y) = -2e^{-y^2} \operatorname{Im} i \int_y^{i\infty+y} du\ e^{u^2}.$$

The integral can be performed as a contour integral around a rectangle in the complex plane. Taking the imaginary part of the result gives $$H(y) = 2\pi^{-1/2} F(y)$$ where $$F(y)$$ is the Dawson function as defined above.

The Hilbert transform of $$x^{2n}e^{-x^2}$$ is also related to the Dawson function. We see this with the technique of differentiating inside the integral sign. Let $$H_n = \pi^{-1} \operatorname{P.V.} \int_{-\infty}^\infty \frac{x^{2n}e^{-x^2}}{y-x} \, dx.$$

Introduce $$H_a = \pi^{-1} \operatorname{P.V.} \int_{-\infty}^\infty {e^{-ax^2} \over y-x} \, dx.$$

The $$n$$th derivative is $${\partial^nH_a \over \partial a^n} = (-1)^n\pi^{-1} \operatorname{P.V.} \int_{-\infty}^\infty \frac{x^{2n}e^{-ax^2}}{y-x} \, dx.$$

We thus find $$\left. H_n = (-1)^n \frac{\partial^nH_a}{\partial a^n} \right|_{a=1}.$$

The derivatives are performed first, then the result evaluated at $$a = 1.$$ A change of variable also gives $$H_a = 2\pi^{-1/2}F(y\sqrt a).$$ Since $$F'(y) = 1-2yF(y),$$ we can write $$H_n = P_1(y)+P_2(y)F(y)$$ where $$P_1$$ and $$P_2$$ are polynomials. For example, $$H_1 = -\pi^{-1/2}y + 2\pi^{-1/2}y^2F(y).$$ Alternatively, $$H_n$$ can be calculated using the recurrence relation (for $$n \geq 0$$) $$H_{n+1}(y) = y^2 H_n(y) - \frac{(2n-1)!!}{\sqrt{\pi} 2^n} y.$$