Dawson function

From Wikipedia, the free encyclopedia
  (Redirected from Dawson's function)
Jump to: navigation, search
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 for 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) = \frac12 \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) = i F(-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}{2 x} + \frac{1}{4 x^3} + \frac{3}{8 x^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}{2 x},

resulting in x = ±0.92413887… (OEISA133841), F(x) = ±0.54104422… (OEISA133842).

Inflection points follow for

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

resulting in x = ±1.50197526… (OEISA133843), F(x) = ±0.42768661… (OEISA245262). (Apart from the trivial inflection point at x = 0, F(x) = 0.)

Relation to Hilbert transform of Gaussian[edit]

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 \Im e^{iku}

With 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) = \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) = \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} \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} \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

 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


H_a = \pi^{-1} P.V. \int_{-\infty}^\infty {e^{-ax^2} \over y-x} dx

The nth derivative is

{\partial^nH_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^nH_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^2F(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 .


  1. ^ 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. 
  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.

External links[edit]