Dawson function

Plot of the Dawson integral function F(z) in the complex plane from -2-2i to 2+2i with colors created with Mathematica 13.1 function ComplexPlot3D

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

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

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, $$D_{+}(x)=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{e}^{-t^2/4}\sin (xt)dt.$$

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, Template:Nowrap
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, Template:Nowrap For $|x|$ large, Template:Nowrap 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 $$F(x)=\frac{1}{2x}+\frac{1}{4x^3}+\frac{3}{8x^5}+\cdots .$$

More precisely $$|F(x)-\sum _{k=0}^N\frac{(2k-1)!!}{2^{k+1}{x}^{2k+1}}|\le \frac{C_N}{x^{2N+3}}.$$ 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 $$F(x)=\frac{1}{2x},$$ resulting in x = ±0.92413887... (Template:OEIS2C), F(x) = ±0.54104422... (Template:OEIS2C).

Inflection points follow for $$F(x)=\frac{x}{2x^2-1},$$ resulting in x = ±1.50197526... (Template:OEIS2C), F(x) = ±0.42768661... (Template:OEIS2C). (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 $$H(y)={\pi }^{-1}\mathrm{P.V.}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{-x^2}}{y-x}dx$$

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 $$\frac{1}{u}={\int }_0^{\mathrm{\infty }}dk\sin ku={\int }_0^{\mathrm{\infty }}dk\mathrm{Im}{e}^{iku}.$$

With $1/u=1/(y-x),$ we use the exponential representation of $\sin (ku)$ and complete the square with respect to $x$ to find $$\pi H(y)=\mathrm{Im}{\int }_0^{\mathrm{\infty }}dk\exp [-k^2/4+iky]{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}dx\exp [-(x+ik/2{)}^2].$$

We can shift the integral over $x$ to the real axis, and it gives ${\pi }^{1/2}.$ Thus $${\pi }^{1/2}H(y)=\mathrm{Im}{\int }_0^{\mathrm{\infty }}dk\exp [-k^2/4+iky].$$

We complete the square with respect to $k$ and obtain $${\pi }^{1/2}H(y)=e^{-y^2}\mathrm{Im}{\int }_0^{\mathrm{\infty }}dk\exp [-(k/2-iy{)}^2].$$

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

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

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

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

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 $n\ge 0$) $$H_{n+1}(y)=y^2{H}_n(y)-\frac{(2n-1)!!}{\sqrt{\pi }{2}^n}y.$$

See also

• Template:Annotated link

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 Template:Webarchive