Dawson function

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) = \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} \operatorname{erfi} (x) = - {i \sqrt{\pi} \over 2 }e^{-x^2} \operatorname{erf} (ix) where erfi is the imaginary error function, 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) = i F(-ix) = -\frac{\sqrt{\pi}}{2} \left[ e^{x^2} - w(-ix) \right] for real x. For |x| near zero, For |x| large, 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}{2 x} + \frac{1}{4 x^3} + \frac{3}{8 x^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}{2 x}, resulting in x = ±0.92413887... (), F(x) = ±0.54104422... (). Inflection points follow for F(x) = \frac{x}{2 x^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 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 nth derivative is {\partial^nH_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^nH_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^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. ==See also==