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 either

$F(x)=D_{+}(x)=e^{-x^{2}}\int _{0}^{x}e^{t^{2}}\,dt,$ 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, and 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

$F(x)=\sum _{k=0}^{\infty }{\frac {(2k-1)!!}{2^{k+1}x^{2k+1}}}={\frac {1}{2x}}+{\frac {1}{4x^{3}}}+{\frac {3}{8x^{5}}}+\cdots ,$ 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… (), F(x) = ±0.54104422… ().

$F(x)={\frac {x}{2x^{2}-1}},$ 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 nth 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 $n\geq 0$ )

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