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^[1]) 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}}\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,

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]

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

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… ( A133841), F(x) = ±0.54104422… ( A133842).

Inflection points follow for

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

resulting in x = ±1.50197526… ( A133843), F(x) = ±0.42768661… ( 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}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 $y$ . $H(y)$ 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}P.V.\int _{-\infty }^{\infty }{x^{2n}e^{-x^{2}} \over y-x}dx

Introduce

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

We thus find

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

The derivatives are performed first, then the result evaluated at $a=1$ . A change of variable also gives $H_{a}=2\pi ^{-1/2}F(y{\sqrt {a}})$ . 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

References

Press, WH; Teukolsky, SA; Vetterling, WT; Flannery, BP (2007), "Section 6.9. Dawson's Integral", Numerical Recipes: The Art of Scientific Computing (3rd ed.), New York: Cambridge University Press, ISBN 978-0-521-88068-8
Temme, N. M. (2010), "Error Functions, Dawson’s and Fresnel Integrals", in Olver, Frank W. J.; Lozier, Daniel M.; Boisvert, Ronald F.; Clark, Charles W., NIST Handbook of Mathematical Functions, Cambridge University Press, ISBN 978-0521192255, MR 2723248

↑ Dawson, H. G. (1897). "On the Numerical Value of $\int _{0}^{h}\exp(x^{2})dx$ ". Proceedings of the London Mathematical Society. s1-29 (1): 519–522. doi:10.1112/plms/s1-29.1.519.
↑ 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.

External links

gsl_sf_dawson in the GNU Scientific Library
Cephes – C and C++ language special functions math library
Faddeeva Package – C++ code for the Dawson function of both real and complex arguments, via the Faddeeva function
Dawson's Integral (at Mathworld)
Error functions

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