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In mathematics, the Dawson function or Dawson integral^[1] (named after H. G. Dawson^[2]) is the one-sided Fourier–Laplace sine transform of the Gaussian function.

The Dawson function is defined as either: $D_{+}\left(x\right)=e^{-x^2}{\int }_0^x{e}^{t^2}dt,$ also denoted as $F\left(x\right)$ or $D\left(x\right),$ or alternatively $D_{-}\left(x\right)=e^{x^2}{\int }_0^x{e}^{-t^2}dt.$

The Dawson function is the one-sided Fourier–Laplace sine transform of the Gaussian function, $D_{+}\left(x\right)=\frac{1}{2}{\int }_0^{\infty }{e}^{-t^2/4}\sin \left(xt\right)dt.$

It is closely related to the error function erf, as

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

where erfi is the imaginary error function, erfi(x) = −i erf(ix).
Similarly, $D_{-}\left(x\right)=\frac{\sqrt{\pi }}{2}{e}^{x^2}\mathrm{erf}\left(x\right)$ in terms of the real error function, erf.

In terms of either erfi or the Faddeeva function $w\left(z\right),$ the Dawson function can be extended to the entire complex plane:^[3] $F\left(z\right)=\frac{\sqrt{\pi }}{2}{e}^{-z^2}\mathrm{erfi}\left(z\right)=\frac{i\sqrt{\pi }}{2}[e^{-z^2}-w(z\left)\right],$ which simplifies to $D_{+}\left(x\right)=F\left(x\right)=\frac{\sqrt{\pi }}{2}\mathrm{Im}\left[w\right(x\left)\right]$ $D_{-}\left(x\right)=iF(-ix)=-\frac{\sqrt{\pi }}{2}[e^{x^2}-w(-ix\left)\right]$ for real $x.$

For $\left|x\right|$ near zero, F(x) ≈ x. For $\left|x\right|$ large, F(x) ≈ 1/(2x). More specifically, near the origin it has the series expansion $F\left(x\right)=\sum _{k=0}^{\infty }\frac{(-1{)}^k{2}^k}{(2k+1)!!}{x}^{2k+1}=x-\frac{2}{3}{x}^3+\frac{4}{15}{x}^5-\cdots ,$ while for large $x$ it has the asymptotic expansion $F\left(x\right)=\frac{1}{2x}+\frac{1}{4x^3}+\frac{3}{8x^5}+\cdots .$

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

$F\left(x\right)$ satisfies the differential equation $\frac{dF}{dx}+2xF=1$ with the initial condition $F\left(0\right)=0.$ Consequently, it has extrema for $F\left(x\right)=\frac{1}{2x},$ resulting in x = ±0.92413887... (OEIS: A133841), F(x) = ±0.54104422... (OEIS: A133842).

Inflection points follow for $F\left(x\right)=\frac{x}{2x^2-1},$ resulting in x = ±1.50197526... (OEIS: A133843), F(x) = ±0.42768661... (OEIS: A245262). (Apart from the trivial inflection point at $x=0,$ $F\left(x\right)=0.$)

The Hilbert transform of the Gaussian is defined as $H\left(y\right)={\pi }^{-1}\mathrm{P.V.}{\int }_{-\infty }^{\infty }\frac{e^{-x^2}}{y-x}dx$

P.V. denotes the Cauchy principal value, and we restrict ourselves to real $y.$ $H\left(y\right)$ 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 $\frac{1}{u}={\int }_0^{\infty }dk\sin ku={\int }_0^{\infty }dk\mathrm{Im}{e}^{iku}.$

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

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

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

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

The integral can be performed as a contour integral around a rectangle in the complex plane. Taking the imaginary part of the result gives $H\left(y\right)=2{\pi }^{-1/2}F\left(y\right)$ where $F\left(y\right)$ 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 $H_n={\pi }^{-1}\mathrm{P.V.}{\int }_{-\infty }^{\infty }\frac{x^{2n}{e}^{-x^2}}{y-x}dx.$

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

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

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

• List of mathematical functions

• 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 Archived 2019-11-01 at the Wayback Machine