Dawson function explained

In mathematics, the Dawson function or Dawson integral(named after H. G. Dawson^[1]) is the one-sided Fourier–Laplace sine transform of the Gaussian function.

Definition

The Dawson function is defined as either:$$D\_+(x)\; =\; e^\; \backslash int\_0^x\; e^\backslash ,dt,$$also denoted as

F(x) or D(x), or alternatively$$D\_-(x)\; =\; e^\; \backslash int\_0^x\; e^\backslash ,dt.\backslash !$$

The Dawson function is the one-sided Fourier–Laplace sine transform of the Gaussian function,$$D\_+(x)\; =\; \backslash frac12\; \backslash int\_0^\backslash infty\; e^\backslash ,\backslash sin(xt)\backslash ,dt.$$

It is closely related to the error function erf, as

$D\_+(x)\; =\; e^\; \backslash operatorname\; (x)\; =\; -\; e^\; \backslash operatorname\; (ix)$where erfi is the imaginary error function,
Similarly,$$D\_-(x)\; =\; \backslash frac\; e^\; \backslash operatorname(x)$$in terms of the real error function, erf.

w(z), the Dawson function can be extended to the entire complex plane:^[2] $$F(z)\; =\; e^\; \backslash operatorname\; (z)\; =\; \backslash frac\; \backslash left[e^\{-z^2\}\; -\; w(z)\; \backslash right],$$which simplifies to$$D\_+(x)\; =\; F(x)\; =\; \backslash frac\; \backslash operatorname[w(x)]$$$$D\_-(x)\; =\; i\; F(-ix)\; =\; -\backslash frac\; \backslash left[e^\{x^2\}\; -\; w(-ix)\; \backslash right]$$for real x.

For

|x| near zero, For |x| large, More specifically, near the origin it has the series expansion$$F(x)\; =\; \backslash sum\_^\backslash infty\; \backslash frac\; \backslash ,\; x^\; =\; x\; -\; \backslash frac\; x^3\; +\; \backslash frac\; x^5\; -\; \backslash cdots,$$while for large x it has the asymptotic expansion$$F(x)\; =\; \backslash frac\; +\; \backslash frac\; +\; \backslash frac\; +\; \backslash cdots.$$

More precisely $$\backslash left|F(x)\; -\; \backslash sum\_^\; \backslash frac\backslash right|\; \backslash leq\; \backslash frac.$$where

n!! is the double factorial. F(x) satisfies the differential equation$$\backslash frac\; +\; 2xF\; =\; 1\backslash ,\backslash !$$with the initial condition F( 0 ) = 0 . Consequently, it has extrema for$$F(x)\; =\; \backslash frac,$$resulting in x = ±0.92413887..., F(x) = ±0.54104422... .

Inflection points follow for$$F(x)\; =\; \backslash frac,$$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$$H(y)\; =\; \backslash pi^\; \backslash operatorname\; \backslash int\_^\backslash infty\; \backslash frac\; \backslash ,\; dx$$

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$$=\; \backslash int\_0^\backslash infty\; dk\; \backslash ,\; \backslash sin\; ku\; =\; \backslash int\_0^\backslash infty\; dk\; \backslash ,\; \backslash operatorname\; e^.$$

With

1 /u = 1 /(y - x), we use the exponential representation of \sin(ku) and complete the square with respect to x to find$$\backslash pi\; H(y)\; =\; \backslash operatorname\; \backslash int\_0^\backslash infty\; dk\; \backslash ,\backslash exp[-k^2/4+iky]\; \backslash int\_^\backslash infty\; dx\; \backslash ,\; \backslash exp[-(x+ik/2)^2].$$

We can shift the integral over

x to the real axis, and it gives \pi ^{1/2} . Thus$$\backslash pi^\; H(y)\; =\; \backslash operatorname\; \backslash int\_0^\backslash infty\; dk\; \backslash ,\; \backslash exp[-k^2/4+iky].$$

We complete the square with respect to

k and obtain$$\backslash pi^H(y)\; =\; e^\; \backslash operatorname\; \backslash int\_0^\backslash infty\; dk\; \backslash ,\; \backslash exp[-(k/2-iy)^2].$$

We change variables to

u = ik/ 2+ y: $$\backslash pi^H(y)\; =\; -2e^\; \backslash operatorname\; i\; \backslash int\_y^\; du\backslash \; e^.$$

The integral can be performed as a contour integral around a rectangle in the complex plane. Taking the imaginary part of the result gives$$H(y)\; =\; 2\backslash pi^\; F(y)$$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$$H\_n\; =\; \backslash pi^\; \backslash operatorname\; \backslash int\_^\backslash infty\; \backslash frac\; \backslash ,\; dx.$$

Introduce$$H\_a\; =\; \backslash pi^\; \backslash operatorname\; \backslash int\_^\backslash infty\; \backslash ,\; dx.$$

The

n th derivative is$$=\; (-1)^n\backslash pi^\; \backslash operatorname\; \backslash int\_^\backslash infty\; \backslash frac\; \backslash ,\; dx.$$

We thus find$$\backslash left\; .\; H\_n\; =\; (-1)^n\; \backslash frac\; \backslash right|\_.$$

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) = 1-2 yF(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 )$$H\_(y)\; =\; y^2\; H\_n(y)\; -\; \backslash frac\; y.$$

External links

• gsl_sf_dawson in the GNU Scientific Library
• libcerf, numeric C library for complex error functions, provides a function voigt(x, sigma, gamma) with approximately 13–14 digits precision. It is based on the Faddeeva function as implemented in the MIT Faddeeva Package
• Dawson's Integral (at Mathworld)
• Error functions

Notes and References

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