Dawson function

In mathematics, the Dawson function or Dawson integral^[1] (named after H. G. Dawson ^[2] ) is the one-sided Fourier–Laplace sine transform of the Gaussian function.

The Dawson function is defined as either: $${\displaystyle D_{+}(x)=e^{-x^{2}}\int _{0}^{x}e^{t^{2}},円dt,}$${\displaystyle D_{+}(x)=e^{-x^{2}}\int _{0}^{x}e^{t^{2}},円dt,} also denoted as ${\displaystyle F(x)}${\displaystyle F(x)} or ${\displaystyle D(x),}${\displaystyle D(x),} or alternatively $${\displaystyle D_{-}(x)=e^{x^{2}}\int _{0}^{x}e^{-t^{2}},円dt.\!}$${\displaystyle D_{-}(x)=e^{x^{2}}\int _{0}^{x}e^{-t^{2}},円dt.\!}

The Dawson function is the one-sided Fourier–Laplace sine transform of the Gaussian function, $${\displaystyle D_{+}(x)={\frac {1}{2}}\int _{0}^{\infty }e^{-t^{2}/4},円\sin(xt),円dt.}$${\displaystyle D_{+}(x)={\frac {1}{2}}\int _{0}^{\infty }e^{-t^{2}/4},円\sin(xt),円dt.}

It is closely related to the error function erf, as

${\displaystyle D_{+}(x)={{\sqrt {\pi }} \over 2}e^{-x^{2}}\operatorname {erfi} (x)=-{i{\sqrt {\pi }} \over 2}e^{-x^{2}}\operatorname {erf} (ix)}${\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, $${\displaystyle D_{-}(x)={\frac {\sqrt {\pi }}{2}}e^{x^{2}}\operatorname {erf} (x)}$${\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 ${\displaystyle w(z),}${\displaystyle w(z),} the Dawson function can be extended to the entire complex plane:^[3] $${\displaystyle F(z)={{\sqrt {\pi }} \over 2}e^{-z^{2}}\operatorname {erfi} (z)={\frac {i{\sqrt {\pi }}}{2}}\left[e^{-z^{2}}-w(z)\right],}$${\displaystyle 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 $${\displaystyle D_{+}(x)=F(x)={\frac {\sqrt {\pi }}{2}}\operatorname {Im} [w(x)]}$${\displaystyle D_{+}(x)=F(x)={\frac {\sqrt {\pi }}{2}}\operatorname {Im} [w(x)]} $${\displaystyle D_{-}(x)=iF(-ix)=-{\frac {\sqrt {\pi }}{2}}\left[e^{x^{2}}-w(-ix)\right]}$${\displaystyle D_{-}(x)=iF(-ix)=-{\frac {\sqrt {\pi }}{2}}\left[e^{x^{2}}-w(-ix)\right]} for real ${\displaystyle x.}${\displaystyle x.}

For ${\displaystyle |x|}${\displaystyle |x|} near zero, F(x) ≈ x. For ${\displaystyle |x|}${\displaystyle |x|} large, F(x) ≈ 1/(2x). More specifically, near the origin it has the series expansion $${\displaystyle F(x)=\sum _{k=0}^{\infty }{\frac {(-1)^{k},2円^{k}}{(2k+1)!!}},円x^{2k+1}=x-{\frac {2}{3}}x^{3}+{\frac {4}{15}}x^{5}-\cdots ,}$${\displaystyle F(x)=\sum _{k=0}^{\infty }{\frac {(-1)^{k},2円^{k}}{(2k+1)!!}},円x^{2k+1}=x-{\frac {2}{3}}x^{3}+{\frac {4}{15}}x^{5}-\cdots ,} while for large ${\displaystyle x}${\displaystyle x} it has the asymptotic expansion $${\displaystyle F(x)={\frac {1}{2x}}+{\frac {1}{4x^{3}}}+{\frac {3}{8x^{5}}}+\cdots .}$${\displaystyle F(x)={\frac {1}{2x}}+{\frac {1}{4x^{3}}}+{\frac {3}{8x^{5}}}+\cdots .}

More precisely $${\displaystyle \left|F(x)-\sum _{k=0}^{N}{\frac {(2k-1)!!}{2^{k+1}x^{2k+1}}}\right|\leq {\frac {C_{N}}{x^{2N+3}}}.}$${\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 ${\displaystyle n!!}${\displaystyle n!!} is the double factorial.

${\displaystyle F(x)}${\displaystyle F(x)} satisfies the differential equation $${\displaystyle {\frac {dF}{dx}}+2xF=1,円\!}$${\displaystyle {\frac {dF}{dx}}+2xF=1,円\!} with the initial condition ${\displaystyle F(0)=0.}${\displaystyle F(0)=0.} Consequently, it has extrema for $${\displaystyle F(x)={\frac {1}{2x}},}$${\displaystyle F(x)={\frac {1}{2x}},} resulting in x = ±0.92413887... (OEIS: A133841 ), F(x) = ±0.54104422... (OEIS: A133842 ).

Inflection points follow for $${\displaystyle F(x)={\frac {x}{2x^{2}-1}},}$${\displaystyle F(x)={\frac {x}{2x^{2}-1}},} resulting in x = ±1.50197526... (OEIS: A133843 ), F(x) = ±0.42768661... (OEIS: A245262 ). (Apart from the trivial inflection point at ${\displaystyle x=0,}${\displaystyle x=0,} ${\displaystyle F(x)=0.}${\displaystyle F(x)=0.})

The Hilbert transform of the Gaussian is defined as $${\displaystyle H(y)=\pi ^{-1}\operatorname {P.V.} \int _{-\infty }^{\infty }{\frac {e^{-x^{2}}}{y-x}},円dx}$${\displaystyle H(y)=\pi ^{-1}\operatorname {P.V.} \int _{-\infty }^{\infty }{\frac {e^{-x^{2}}}{y-x}},円dx}

P.V. denotes the Cauchy principal value, and we restrict ourselves to real ${\displaystyle y.}${\displaystyle y.} ${\displaystyle H(y)}${\displaystyle H(y)} can be related to the Dawson function as follows. Inside a principal value integral, we can treat ${\displaystyle 1/u}${\displaystyle 1/u} as a generalized function or distribution, and use the Fourier representation $${\displaystyle {1 \over u}=\int _{0}^{\infty }dk,円\sin ku=\int _{0}^{\infty }dk,円\operatorname {Im} e^{iku}.}$${\displaystyle {1 \over u}=\int _{0}^{\infty }dk,円\sin ku=\int _{0}^{\infty }dk,円\operatorname {Im} e^{iku}.}

With ${\displaystyle 1/u=1/(y-x),}${\displaystyle 1/u=1/(y-x),} we use the exponential representation of ${\displaystyle \sin(ku)}${\displaystyle \sin(ku)} and complete the square with respect to ${\displaystyle x}${\displaystyle x} to find $${\displaystyle \pi H(y)=\operatorname {Im} \int _{0}^{\infty }dk,円\exp[-k^{2}/4+iky]\int _{-\infty }^{\infty }dx,円\exp[-(x+ik/2)^{2}].}$${\displaystyle \pi H(y)=\operatorname {Im} \int _{0}^{\infty }dk,円\exp[-k^{2}/4+iky]\int _{-\infty }^{\infty }dx,円\exp[-(x+ik/2)^{2}].}

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

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

We change variables to ${\displaystyle u=ik/2+y:}${\displaystyle u=ik/2+y:} $${\displaystyle \pi ^{1/2}H(y)=-2e^{-y^{2}}\operatorname {Im} i\int _{y}^{i\infty +y}du\ e^{u^{2}}.}$${\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 $${\displaystyle H(y)=2\pi ^{-1/2}F(y)}$${\displaystyle H(y)=2\pi ^{-1/2}F(y)} where ${\displaystyle F(y)}${\displaystyle F(y)} is the Dawson function as defined above.

The Hilbert transform of ${\displaystyle x^{2n}e^{-x^{2}}}${\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 $${\displaystyle H_{n}=\pi ^{-1}\operatorname {P.V.} \int _{-\infty }^{\infty }{\frac {x^{2n}e^{-x^{2}}}{y-x}},円dx.}$${\displaystyle H_{n}=\pi ^{-1}\operatorname {P.V.} \int _{-\infty }^{\infty }{\frac {x^{2n}e^{-x^{2}}}{y-x}},円dx.}

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

The ${\displaystyle n}${\displaystyle n}th derivative is $${\displaystyle {\partial ^{n}H_{a} \over \partial a^{n}}=(-1)^{n}\pi ^{-1}\operatorname {P.V.} \int _{-\infty }^{\infty }{\frac {x^{2n}e^{-ax^{2}}}{y-x}},円dx.}$${\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}},円dx.}

We thus find $${\displaystyle \left.H_{n}=(-1)^{n}{\frac {\partial ^{n}H_{a}}{\partial a^{n}}}\right|_{a=1}.}$${\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 ${\displaystyle a=1.}${\displaystyle a=1.} A change of variable also gives ${\displaystyle H_{a}=2\pi ^{-1/2}F(y{\sqrt {a}}).}${\displaystyle H_{a}=2\pi ^{-1/2}F(y{\sqrt {a}}).} Since ${\displaystyle F'(y)=1-2yF(y),}${\displaystyle F'(y)=1-2yF(y),} we can write ${\displaystyle H_{n}=P_{1}(y)+P_{2}(y)F(y)}${\displaystyle H_{n}=P_{1}(y)+P_{2}(y)F(y)} where ${\displaystyle P_{1}}${\displaystyle P_{1}} and ${\displaystyle P_{2}}${\displaystyle P_{2}} are polynomials. For example, ${\displaystyle H_{1}=-\pi ^{-1/2}y+2\pi ^{-1/2}y^{2}F(y).}${\displaystyle H_{1}=-\pi ^{-1/2}y+2\pi ^{-1/2}y^{2}F(y).} Alternatively, ${\displaystyle H_{n}}${\displaystyle H_{n}} can be calculated using the recurrence relation (for ${\displaystyle n\geq 0}${\displaystyle n\geq 0}) $${\displaystyle H_{n+1}(y)=y^{2}H_{n}(y)-{\frac {(2n-1)!!}{{\sqrt {\pi }}2^{n}}}y.}$${\displaystyle H_{n+1}(y)=y^{2}H_{n}(y)-{\frac {(2n-1)!!}{{\sqrt {\pi }}2^{n}}}y.}

• List of mathematical functions

• 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 Archived 2019年11月01日 at the Wayback Machine

• Gaussian function
• Special functions

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