Dawson function
# Dawson function

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:

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

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 $x=0,$ $F(x)=0.$)

## 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 $n$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 $n\geq 0$)

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

