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Dawson function

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Mathematical function
Plot of the Dawson integral function F(z) in the complex plane from -2-2i to 2+2i with colors created with Mathematica 13.1 function ComplexPlot3D
Plot of the Dawson integral function F(z) in the complex plane from -2-2i to 2+2i with colors created with Mathematica 13.1 function ComplexPlot3D

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, F ( x ) = D + ( x ) , {\displaystyle F(x)=D_{+}(x),} around the origin
The Dawson function, D ( x ) , {\displaystyle D_{-}(x),} around the origin

The Dawson function is defined as either: D + ( x ) = e x 2 0 x e t 2 d t , {\displaystyle D_{+}(x)=e^{-x^{2}}\int _{0}^{x}e^{t^{2}}\,dt,} also denoted as F ( x ) {\displaystyle F(x)} or D ( x ) , {\displaystyle D(x),} or alternatively D ( x ) = e x 2 0 x e t 2 d t . {\displaystyle 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 ) = 1 2 0 e t 2 / 4 sin ( x t ) d t . {\displaystyle D_{+}(x)={\frac {1}{2}}\int _{0}^{\infty }e^{-t^{2}/4}\,\sin(xt)\,dt.}

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, D ( x ) = π 2 e x 2 erf ( x ) {\displaystyle 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 ) , {\displaystyle w(z),} the Dawson function can be extended to the entire complex plane: F ( z ) = π 2 e z 2 erfi ( z ) = i π 2 [ e z 2 w ( z ) ] , {\displaystyle F(z)={{\sqrt {\pi }} \over 2}e^{-z^{2}}\operatorname {erfi} (z)={\frac {i{\sqrt {\pi }}}{2}}\left,} which simplifies to D + ( x ) = F ( x ) = π 2 Im [ w ( x ) ] {\displaystyle D_{+}(x)=F(x)={\frac {\sqrt {\pi }}{2}}\operatorname {Im} } D ( x ) = i F ( i x ) = π 2 [ e x 2 w ( i x ) ] {\displaystyle D_{-}(x)=iF(-ix)=-{\frac {\sqrt {\pi }}{2}}\left} 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 F ( x ) = k = 0 ( 1 ) k 2 k ( 2 k + 1 ) ! ! x 2 k + 1 = x 2 3 x 3 + 4 15 x 5 , {\displaystyle 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 {\displaystyle x} it has the asymptotic expansion F ( x ) = 1 2 x + 1 4 x 3 + 3 8 x 5 + . {\displaystyle F(x)={\frac {1}{2x}}+{\frac {1}{4x^{3}}}+{\frac {3}{8x^{5}}}+\cdots .}

More precisely | F ( x ) k = 0 N ( 2 k 1 ) ! ! 2 k + 1 x 2 k + 1 | C N x 2 N + 3 . {\displaystyle \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 ! ! {\displaystyle n!!} is the double factorial.

F ( x ) {\displaystyle F(x)} satisfies the differential equation d F d x + 2 x F = 1 {\displaystyle {\frac {dF}{dx}}+2xF=1\,\!} with the initial condition F ( 0 ) = 0. {\displaystyle F(0)=0.} Consequently, it has extrema for F ( x ) = 1 2 x , {\displaystyle F(x)={\frac {1}{2x}},} resulting in x = ±0.92413887... (OEISA133841), F(x) = ±0.54104422... (OEISA133842).

Inflection points follow for F ( x ) = x 2 x 2 1 , {\displaystyle F(x)={\frac {x}{2x^{2}-1}},} resulting in x = ±1.50197526... (OEISA133843), F(x) = ±0.42768661... (OEISA245262). (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 H ( y ) = π 1 P . V . e x 2 y x d x {\displaystyle 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 . {\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 1 u = 0 d k sin k u = 0 d k Im e i k u . {\displaystyle {1 \over u}=\int _{0}^{\infty }dk\,\sin ku=\int _{0}^{\infty }dk\,\operatorname {Im} e^{iku}.}

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 π H ( y ) = Im 0 d k exp [ k 2 / 4 + i k y ] d x exp [ ( x + i k / 2 ) 2 ] . {\displaystyle \pi H(y)=\operatorname {Im} \int _{0}^{\infty }dk\,\exp\int _{-\infty }^{\infty }dx\,\exp.}

We can shift the integral over x {\displaystyle x} to the real axis, and it gives π 1 / 2 . {\displaystyle \pi ^{1/2}.} Thus π 1 / 2 H ( y ) = Im 0 d k exp [ k 2 / 4 + i k y ] . {\displaystyle \pi ^{1/2}H(y)=\operatorname {Im} \int _{0}^{\infty }dk\,\exp.}

We complete the square with respect to k {\displaystyle k} and obtain π 1 / 2 H ( y ) = e y 2 Im 0 d k exp [ ( k / 2 i y ) 2 ] . {\displaystyle \pi ^{1/2}H(y)=e^{-y^{2}}\operatorname {Im} \int _{0}^{\infty }dk\,\exp.}

We change variables to u = i k / 2 + y : {\displaystyle u=ik/2+y:} π 1 / 2 H ( y ) = 2 e y 2 Im i y i + y d u   e u 2 . {\displaystyle \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 π 1 / 2 F ( y ) {\displaystyle H(y)=2\pi ^{-1/2}F(y)} 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 H n = π 1 P . V . x 2 n e x 2 y x d x . {\displaystyle H_{n}=\pi ^{-1}\operatorname {P.V.} \int _{-\infty }^{\infty }{\frac {x^{2n}e^{-x^{2}}}{y-x}}\,dx.}

Introduce H a = π 1 P . V . e a x 2 y x d x . {\displaystyle H_{a}=\pi ^{-1}\operatorname {P.V.} \int _{-\infty }^{\infty }{e^{-ax^{2}} \over y-x}\,dx.}

The n {\displaystyle n} th derivative is n H a a n = ( 1 ) n π 1 P . V . x 2 n e a x 2 y x d x . {\displaystyle {\partial ^{n}H_{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 H n = ( 1 ) n n H a a n | a = 1 . {\displaystyle \left.H_{n}=(-1)^{n}{\frac {\partial ^{n}H_{a}}{\partial a^{n}}}\right|_{a=1}.}

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 H n = P 1 ( y ) + P 2 ( y ) F ( y ) {\displaystyle H_{n}=P_{1}(y)+P_{2}(y)F(y)} where P 1 {\displaystyle P_{1}} and P 2 {\displaystyle P_{2}} are polynomials. For example, H 1 = π 1 / 2 y + 2 π 1 / 2 y 2 F ( y ) . {\displaystyle H_{1}=-\pi ^{-1/2}y+2\pi ^{-1/2}y^{2}F(y).} Alternatively, H n {\displaystyle H_{n}} can be calculated using the recurrence relation (for n 0 {\displaystyle n\geq 0} ) H n + 1 ( y ) = y 2 H n ( y ) ( 2 n 1 ) ! ! π 2 n y . {\displaystyle H_{n+1}(y)=y^{2}H_{n}(y)-{\frac {(2n-1)!!}{{\sqrt {\pi }}2^{n}}}y.}

See also

References

  1. Temme, N. M. (2010), "Error Functions, Dawson's and Fresnel Integrals", in Olver, Frank W. J.; Lozier, Daniel M.; Boisvert, Ronald F.; Clark, Charles W. (eds.), NIST Handbook of Mathematical Functions, Cambridge University Press, ISBN 978-0-521-19225-5, MR 2723248.
  2. Dawson, H. G. (1897). "On the Numerical Value of 0 h exp ( x 2 ) d x {\displaystyle \textstyle \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.
  3. 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.

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