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Dawson function

Mathematical function

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Source updated February 22, 2026

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.

01Definition

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

It is closely related to the error function erf, as

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

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 $D_{+}(x)=F(x)={\frac {\sqrt {\pi }}{2}}\operatorname {Im} [w(x)]$ $D_{-}(x)=iF(-ix)=-{\frac {\sqrt {\pi }}{2}}\left[e^{x^{2}}-w(-ix)\right]$ 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 $F(x)={\frac {1}{2x}}+{\frac {1}{4x^{3}}}+{\frac {3}{8x^{5}}}+\cdots .$

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 $F(x)={\frac {1}{2x}},$ 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 $x=0,$ $F(x)=0.$ )

02Relation 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 ${\partial ^{n}H_{a} \over \partial a^{n}}=(-1)^{n}\pi ^{-1}\operatorname {P.V.} \int _{-\infty }^{\infty }{\frac {x^{2n}e^{-ax^{2}}}{y-x}}\,dx.$

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

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, $H_{1}=-\pi ^{-1/2}y+2\pi ^{-1/2}y^{2}F(y).$ Alternatively, $H_{n}$ can be calculated using the recurrence relation (for $n\geq 0$ ) $H_{n+1}(y)=y^{2}H_{n}(y)-{\frac {(2n-1)!!}{{\sqrt {\pi }}2^{n}}}y.$

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Sources and credits

This article is adapted from the Wikipedia article “Dawson function”, written by its contributors and licensed under CC BY-SA 4.0. Fathomly has changed the layout, removed citation markers, navigation and maintenance notices, and adjusted punctuation. This adapted version is shared under the same license. For references, see the original article.

Images, from Wikimedia Commons:

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.svg by WalkingRadiance, CC BY-SA 4.0
DawsonDp.svg by LSzilard, CC BY-SA 3.0
DawsonDm.svg by LSzilard, CC BY-SA 3.0

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Dawson function

01Definition

02Relation to Hilbert transform of Gaussian

Sources and credits

Related topics

List of mathematical functions