# Dawson function

The Dawson function, ${\displaystyle F(x)=D_{+}(x)}$, around the origin
The Dawson function, ${\displaystyle D_{-}(x)}$, around the origin

In mathematics, the Dawson function or Dawson integral (named after H. G. Dawson[1]) is either

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

also denoted as F(x) or D(x), or alternatively

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

The Dawson function is the one-sided Fourier-Laplace sine transform of the Gaussian function,

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

It is closely related to the error function erf, as

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

where erfi is the imaginary error function, erfi(x) = −i erf(ix). Similarly,

${\displaystyle D_{-}(x)={\frac {\sqrt {\pi }}{2}}e^{x^{2}}\mathrm {erf} (x)}$

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:[2]

${\displaystyle F(z)={{\sqrt {\pi }} \over 2}e^{-z^{2}}\mathrm {erfi} (z)={\frac {i{\sqrt {\pi }}}{2}}\left[e^{-z^{2}}-w(z)\right]}$,

which simplifies to

${\displaystyle D_{+}(x)=F(x)={\frac {\sqrt {\pi }}{2}}\operatorname {Im} [w(x)]}$
${\displaystyle D_{-}(x)=iF(-ix)=-{\frac {\sqrt {\pi }}{2}}\left[e^{x^{2}}-w(-ix)\right]}$

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

${\displaystyle F(x)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}\,2^{k}}{(2k+1)!!}}\,x^{2k+1}=x-{\frac {2}{3}}x^{3}+{\frac {4}{15}}x^{5}-\cdots }$,

while for large x it has the asymptotic expansion

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

where n!! is the double factorial.

F(x) satisfies the differential equation

${\displaystyle {\frac {dF}{dx}}+2xF=1\,\!}$

with the initial condition F(0) = 0. Consequently, it has extrema for

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

resulting in x = ±0.92413887… (), F(x) = ±0.54104422… ().

${\displaystyle F(x)={\frac {x}{2x^{2}-1}}}$,

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

${\displaystyle H(y)=\pi ^{-1}P.V.\int _{-\infty }^{\infty }{e^{-x^{2}} \over y-x}dx}$

P.V. denotes the Cauchy principal value, and we restrict ourselves to real ${\displaystyle y}$. ${\displaystyle H(y)}$ can be related to the Dawson function as follows. Inside a principal value integral, we can treat ${\displaystyle 1/u}$ as a generalized function or distribution, and use the Fourier representation

${\displaystyle {1 \over u}=\int _{0}^{\infty }dk\sin ku=\int _{0}^{\infty }dk\operatorname {Im} e^{iku}}$

With ${\displaystyle u=1/(y-x)}$, we use the exponential representation of ${\displaystyle \sin(ku)}$ and complete the square with respect to ${\displaystyle x}$ to find

${\displaystyle \pi H(y)=\operatorname {Im} \int _{0}^{\infty }dk\exp[-k^{2}/4+iky]\int _{-\infty }^{\infty }dx\exp[-(x+ik/2)^{2}]}$

We can shift the integral over ${\displaystyle x}$ to the real axis, and it gives ${\displaystyle \pi ^{1/2}}$. Thus

${\displaystyle \pi ^{1/2}H(y)=\operatorname {Im} \int _{0}^{\infty }dk\exp[-k^{2}/4+iky]}$

We complete the square with respect to ${\displaystyle k}$ and obtain

${\displaystyle \pi ^{1/2}H(y)=e^{-y^{2}}\operatorname {Im} \int _{0}^{\infty }dk\exp[-(k/2-iy)^{2}]}$

We change variables to ${\displaystyle u=ik/2+y}$:

${\displaystyle \pi ^{1/2}H(y)=-2e^{-y^{2}}\operatorname {Im} \int _{y}^{i\infty +y}du\ e^{u^{2}}}$

The integral can be performed as a contour integral around a rectangle in the complex plane. Taking the imaginary part of the result gives

${\displaystyle H(y)=2\pi ^{-1/2}F(y)}$

where ${\displaystyle F(y)}$ is the Dawson function as defined above.

The Hilbert transform of ${\displaystyle x^{2n}e^{-x^{2}}}$ is also related to the Dawson function. We see this with the technique of differentiating inside the integral sign. Let

${\displaystyle H_{n}=\pi ^{-1}P.V.\int _{-\infty }^{\infty }{x^{2n}e^{-x^{2}} \over y-x}dx}$

Introduce

${\displaystyle H_{a}=\pi ^{-1}P.V.\int _{-\infty }^{\infty }{e^{-ax^{2}} \over y-x}dx}$

The nth derivative is

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

We thus find

${\displaystyle H_{n}=(-1)^{n}{\partial ^{n}H_{a} \over \partial a^{n}}|_{a=1}}$

The derivatives are performed first, then the result evaluated at ${\displaystyle a=1}$. A change of variable also gives ${\displaystyle H_{a}=2\pi ^{-1/2}F(y{\sqrt {a}})}$. Since ${\displaystyle F'(y)=1-2yF(y)}$, we can write ${\displaystyle H_{n}=P_{1}(y)+P_{2}(y)F(y)}$ where ${\displaystyle P_{1}}$ and ${\displaystyle P_{2}}$ are polynomials. For example, ${\displaystyle H_{1}=-\pi ^{-1/2}y+2\pi ^{-1/2}y^{2}F(y)}$. Alternatively, ${\displaystyle H_{n}}$ can be calculated using the recurrence relation (for ${\displaystyle n\geq 0}$)

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

## References

1. ^ Dawson, H. G. (1897). "On the Numerical Value of ${\displaystyle \int _{0}^{h}\exp(x^{2})dx}$". Proceedings of the London Mathematical Society. s1-29 (1): 519–522. doi:10.1112/plms/s1-29.1.519.
2. ^ Mofreh R. Zaghloul and Ahmed N. Ali, "Algorithm 916: Computing the Faddeyeva and Voigt Functions," ACM Trans. Math. Soft. 38 (2), 15 (2011). Preprint available at arXiv:1106.0151.