Editing Dawson function (section)

==Relation to Hilbert transform of Gaussian==

The [[Hilbert transform]] of the Gaussian is defined as
<math display=block>H(y) = \pi^{-1} \operatorname{P.V.} \int_{-\infty}^\infty \frac{e^{-x^2}}{y-x} \, dx</math>

P.V. denotes the [[Cauchy principal value]], and we restrict ourselves to real <math>y.</math> <math>H(y)</math> can be related to the Dawson function as follows. Inside a principal value integral, we can treat <math>1/u</math> as a [[generalized function]] or distribution, and use the Fourier representation
<math display=block>{1 \over u} = \int_0^\infty dk \, \sin ku = \int_0^\infty dk \, \operatorname{Im} e^{iku}.</math>

With <math>1/u = 1/(y-x),</math> we use the exponential representation of <math>\sin(ku)</math> and complete the square with respect to <math>x</math> to find
<math display=block>\pi H(y) = \operatorname{Im} \int_0^\infty dk \,\exp[-k^2/4+iky] \int_{-\infty}^\infty dx \, \exp[-(x+ik/2)^2].</math>

We can shift the integral over <math>x</math> to the real axis, and it gives <math>\pi^{1/2}.</math>
Thus
<math display=block>\pi^{1/2} H(y) = \operatorname{Im} \int_0^\infty dk \, \exp[-k^2/4+iky].</math>

We complete the square with respect to <math>k</math> and obtain
<math display=block>\pi^{1/2}H(y) = e^{-y^2} \operatorname{Im} \int_0^\infty dk \, \exp[-(k/2-iy)^2].</math>

We change variables to <math>u = ik/2+y:</math>
<math display=block>\pi^{1/2}H(y) = -2e^{-y^2} \operatorname{Im} i \int_y^{i\infty+y} du\ e^{u^2}.</math>

The integral can be performed as a contour integral around a rectangle in the complex plane. Taking the imaginary part of the result gives
<math display=block>H(y) = 2\pi^{-1/2} F(y)</math>
where <math>F(y)</math> is the Dawson function as defined above.

The Hilbert transform of <math>x^{2n}e^{-x^2}</math> is also related to the Dawson function. We see this with the technique of differentiating inside the integral sign. Let
<math display=block>H_n = \pi^{-1} \operatorname{P.V.} \int_{-\infty}^\infty \frac{x^{2n}e^{-x^2}}{y-x} \, dx.</math>

Introduce
<math display=block>H_a = \pi^{-1} \operatorname{P.V.} \int_{-\infty}^\infty {e^{-ax^2} \over y-x} \, dx.</math>

The <math>n</math>th derivative is
<math display=block>{\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.</math>

We thus find
<math display=block>\left . H_n = (-1)^n \frac{\partial^nH_a}{\partial a^n} \right|_{a=1}.</math>

The derivatives are performed first, then the result evaluated at <math>a = 1.</math> A change of variable also gives <math>H_a = 2\pi^{-1/2}F(y\sqrt a).</math> Since <math>F'(y) = 1-2yF(y),</math> we can write <math>H_n = P_1(y)+P_2(y)F(y)</math> where <math>P_1</math> and <math>P_2</math> are polynomials. For example, <math>H_1 = -\pi^{-1/2}y + 2\pi^{-1/2}y^2F(y).</math> Alternatively, <math>H_n</math> can be calculated using the [[recurrence relation]] (for <math>n \geq 0</math>)
<math display=block>H_{n+1}(y) = y^2 H_n(y) - \frac{(2n-1)!!}{\sqrt{\pi} 2^n} y.</math>