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In mathematics, the Dawson function or Dawson integral (named after H. G. Dawson) is the one-sided Fourier–Laplace sine transform of the Gaussian function. Definition The Dawson function is defined as either: also denoted as F ( x ) {\displaystyle F(x)} or D ( x ) , {\displaystyle D(x),} or alternatively The Dawson function is the one-sided Fourier–Laplace sine transform of the Gaussian function, It is closely related to the error function erf, as D + ( x ) = π 2 e − x 2 erfi ⁡ ( x ) = − i π 2 e − x 2 erf ⁡ ( i x ) {\displaystyle 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, erfi(x) = −i erf(ix). Similarly, in terms of the real error function, erf. In terms of either erfi or the Faddeeva function w ( z ) , {\displaystyle w(z),} the Dawson function can be extended to the entire complex plane: which simplifies to for real x . {\displaystyle x.} For | x | {\displaystyle |x|} near zero, F(x) ≈ x. For | x | {\displaystyle |x|} large, F(x) ≈ 1/(2x). More specifically, near the origin it has the series expansion while for large x {\displaystyle x} it has the asymptotic expansion More precisely where n ! ! {\displaystyle n!!} is the double factorial. F ( x ) {\displaystyle F(x)} satisfies the differential equation with the initial condition F ( 0 ) = 0. {\displaystyle F(0)=0.} Consequently, it has extrema for resulting in x = ±0.92413887... (OEIS: A133841), F(x) = ±0.54104422... (OEIS: A133842). Inflection points follow for resulting in x = ±1.50197526... (OEIS: A133843), F(x) = ±0.42768661... (OEIS: A245262). (Apart from the trivial inflection point at x = 0 , {\displaystyle x=0,} F ( x ) = 0. {\displaystyle F(x)=0.} ) Relation to Hilbert transform of Gaussian The Hilbert transform of the Gaussian is defined as P.V. denotes the Cauchy principal value, and we restrict ourselves to real y . {\displaystyle y.} H ( y ) {\displaystyle H(y)} can be related to the Dawson function as follows. Inside a principal value integral, we can treat 1 / u {\displaystyle 1/u} as a generalized function or distribution, and use the Fourier representation With 1 / u = 1 / ( y − x ) , {\displaystyle 1/u=1/(y-x),} we use the exponential representation of sin ⁡ ( k u ) {\displaystyle \sin(ku)} and complete the square with respect to x {\displaystyle x} to find We can shift the integral over x {\displaystyle x} to the real axis, and it gives π 1 / 2 . {\displaystyle \pi ^{1/2}.} Thus We complete the square with respect to k {\displaystyle k} and obtain We change variables to u = i k / 2 + y : {\displaystyle u=ik/2+y:} The integral can be performed as a contour integral around a rectangle in the complex plane. Taking the imaginary part of the result gives where F ( y ) {\displaystyle F(y)} is the Dawson function as defined above. The Hilbert transform of x 2 n e − x 2 {\displaystyle 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 Introduce The n {\displaystyle n} th derivative is We thus find The derivatives are performed first, then the result evaluated at a = 1. {\displaystyle a=1.} A change of variable also gives H a = 2 π − 1 / 2 F ( y a ) . {\displaystyle H_{a}=2\pi ^{-1/2}F(y{\sqrt {a}}).} Since F ′ ( y ) = 1 − 2 y F ( y ) , {\displaystyle F'(y)=1-2yF(y),} we can write .... Discover the H A Dawson popular books. Find the top 100 most popular H A Dawson books.