Dawson function

The Dawson function, $F(x)=D_{+}(x)$, around the origin

The Dawson function, $D_{-}(x)$, around the origin

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:

$D_{+}(x)=e^{-x^2}\underset{0}{\overset{x}{\int }}{e}^{t^2}dt,$

also denoted as F(x) or D(x), or alternatively

$D_{-}(x)=e^{x^2}\underset{0}{\overset{x}{\int }}{e}^{-t^2}dt.$

The Dawson function is the one-sided Fourier–Laplace sine transform of the Gaussian function,

$D_{+}(x)=\frac{1}{2}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-t^2/4}\sin (xt)dt.$

It is closely related to the error function erf, as

$D_{+}(x)=\frac{\sqrt{\pi }}{2}{e}^{-x^2}\mathrm{erfi}(x)=-\frac{i\sqrt{\pi }}{2}{e}^{-x^2}\mathrm{erf}(ix)$

where erfi is the imaginary error function, erfi(x) = −i erf(ix). Similarly,

$D_{-}(x)=\frac{\sqrt{\pi }}{2}{e}^{x^2}\mathrm{erf}(x)$

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:

$F(z)=\frac{\sqrt{\pi }}{2}{e}^{-z^2}\mathrm{erfi}(z)=\frac{i\sqrt{\pi }}{2}[e^{-z^2}-w(z)],$

which simplifies to

$D_{+}(x)=F(x)=\frac{\sqrt{\pi }}{2}\mathrm{Im}[w(x)]$

$D_{-}(x)=iF(-ix)=-\frac{\sqrt{\pi }}{2}[e^{x^2}-w(-ix)]$

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

$F(x)=\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k{2}^k}{(2k+1)!!}{x}^{2k+1}=x-\frac{2}{3}{x}^3+\frac{4}{15}{x}^5-\cdots ,$

while for large x it has the asymptotic expansion

$F(x)=\sum _{k=0}^{\mathrm{\infty }}\frac{(2k-1)!!}{2^{k+1}{x}^{2k+1}}=\frac{1}{2x}+\frac{1}{4x^3}+\frac{3}{8x^5}+\cdots ,$

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

$F(x)=\frac{1}{2x},$

resulting in x = ±0.92413887... (OEIS: A133841), F(x) = ±0.54104422... (OEIS: A133842).

Inflection points follow for

$F(x)=\frac{x}{2x^2-1},$

resulting in x = ±1.50197526... (OEIS: A133843), F(x) = ±0.42768661... (OEIS: A245262). (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

$H(y)={\pi }^{-1}\mathrm{P.V.}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-x^2}}{y-x}dx$

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

$\frac{1}{u}=\underset{0}{\overset{\mathrm{\infty }}{\int }}dk\sin ku=\underset{0}{\overset{\mathrm{\infty }}{\int }}dk\mathrm{Im}{e}^{iku}$

With $1/u=1/(y-x)$, we use the exponential representation of $\sin (ku)$ and complete the square with respect to $x$ to find

$\pi H(y)=\mathrm{Im}\underset{0}{\overset{\mathrm{\infty }}{\int }}dk\exp [-k^2/4+iky]\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}dx\exp [-(x+ik/2{)}^2]$

We can shift the integral over $x$ to the real axis, and it gives ${\pi }^{1/2}$. Thus

${\pi }^{1/2}H(y)=\mathrm{Im}\underset{0}{\overset{\mathrm{\infty }}{\int }}dk\exp [-k^2/4+iky]$

We complete the square with respect to $k$ and obtain

${\pi }^{1/2}H(y)=e^{-y^2}\mathrm{Im}\underset{0}{\overset{\mathrm{\infty }}{\int }}dk\exp [-(k/2-iy{)}^2]$

We change variables to $u=ik/2+y$:

${\pi }^{1/2}H(y)=-2e^{-y^2}\mathrm{Im}i\underset{y}{\overset{i\mathrm{\infty }+y}{\int }}du{e}^{u^2}$

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

$H_n={\pi }^{-1}\mathrm{P.V.}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{x^{2n}{e}^{-x^2}}{y-x}dx$

Introduce

$H_a={\pi }^{-1}\mathrm{P.V.}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-a{x}^2}}{y-x}dx$

The nth derivative is

$\frac{{\partial }^n{H}_a}{\partial a^n}=(-1{)}^n{\pi }^{-1}\mathrm{P.V.}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{x^{2n}{e}^{-a{x}^2}}{y-x}dx$

We thus find

${H_n=(-1{)}^n\frac{{\partial }^n{H}_a}{\partial a^n}|}_{a=1}$

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 $F^{\prime }(y)=1-2yF(y)$, we can write $H_n=P_1(y)+P_2(y)F(y)$ where $P_1$ and $P_2$ are polynomials. For example, $H_1=-{\pi }^{-1/2}y+2{\pi }^{-1/2}{y}^{2}F(y)$. Alternatively, $H_n$ can be calculated using the recurrence relation (for $n\ge 0$)

$H_{n+1}(y)=y^2{H}_n(y)-\frac{(2n-1)!!}{\sqrt{\pi }{2}^n}y.$

External links

• gsl_sf_dawson in the GNU Scientific Library
• libcerf, numeric C library for complex error functions, provides a function voigt(x, sigma, gamma) with approximately 13–14 digits precision. It is based on the Faddeeva function as implemented in the MIT Faddeeva Package
• Dawson's Integral (at Mathworld)
• Error functions