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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Dawson's integral ： ウィキペディア英語版 Dawson function In mathematics, the Dawson function or Dawson integral (named for H. G. Dawson) is either :$F(x)\; =\; D\_+(x)\; =\; e^\; \backslash int\_0^x\; e^\backslash ,dt$, also denoted as ''F''(''x'') or ''D''(''x''), or alternatively :$D\_-(x)\; =\; e^\; \backslash int\_0^x\; e^\backslash ,dt\backslash !$. The Dawson function is the one-sided Fourier-Laplace sine transform of the Gaussian function, :$D\_+(x)\; =\; \backslash frac12\; \backslash int\_0^\backslash infty\; e^\backslash ,\backslash sin\backslash ,dt.$ It is closely related to the error function erf, as :$D\_+(x)\; =\; e^\; \backslash mathrm\; (x)\; =\; -\; e^\; \backslash mathrm\; (ix)$ where erfi is the imaginary error function, Similarly, :$D\_-(x)\; =\; \backslash frac\; e^\; \backslash mathrm(x)$ 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:〔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 ).〕 :$F(z)\; =\; e^\; \backslash mathrm\; (z)\; =\; \backslash frac\; \backslash left(e^\; -\; w(z)\; \backslash right)$, which simplifies to :$D\_+(x)\; =\; F(x)\; =\; \backslash frac\; \backslash operatorname(w(x))$ :$D\_-(x)\; =\; i\; F(-ix)\; =\; -\backslash frac\; \backslash left(e^\; -\; w(-ix)\; \backslash right)$ for real ''x''. For |''x''| near zero, and for |''x''| large, More specifically, near the origin it has the series expansion :$F(x)\; =\; \backslash sum\_^\; \backslash frac\; \backslash ,\; x^$ = x - \frac x^3 + \frac x^5 - \cdots, while for large ''x'' it has the asymptotic expansion :$F(x)\; =\; \backslash sum\_^\; \backslash frac\}$ = \frac + \frac + \frac + \cdots, where ''n''!! is the double factorial. ''F''(''x'') satisfies the differential equation :$\backslash frac\; +\; 2xF=1\backslash ,\backslash !$ with the initial condition ''F''(0) = 0. Consequently, it has extrema for :$F(x)\; =\; \backslash frac$, resulting in ''x'' = ±''0.92413887''… (), ''F''(''x'') = ±''0.54104422''… (). Inflection points follow for :$F(x)\; =\; \backslash frac$, 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 :$H(y)\; =\; \backslash pi^\; P.V.\; \backslash int\_^\backslash infty\; dx$ 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 :$=\; \backslash int\_0^\backslash infty\; dk\; \backslash sin\; ku\; =\; \backslash int\_0^\backslash infty\; dk\; \backslash Im\; e^$ With $u=1/(y-x)$, we use the exponential representation of $\backslash sin(ku)$ and complete the square with respect to $x$ to find :$\backslash pi\; H(y)\; =\; \backslash Im\; \backslash int\_0^\backslash infty\; dk\; \backslash exp()\; \backslash int\_^\backslash infty\; dx\; \backslash exp()$ We can shift the integral over $x$ to the real axis, and it gives $\backslash pi^$. Thus :$\backslash pi^\; H(y)\; =\; \backslash Im\; \backslash int\_0^\backslash infty\; dk\; \backslash exp()$ We complete the square with respect to $k$ and obtain :$\backslash pi^H(y)\; =\; e^\; \backslash Im\; \backslash int\_0^\backslash infty\; dk\; \backslash exp()$ We change variables to $u=ik/2+y$: :$\backslash pi^H(y)\; =\; -2e^\; \backslash Im\; \backslash int\_y^\; du\; e^$ The integral can be performed as a contour integral around a rectangle in the complex plane. Taking the imaginary part of the result gives :$H(y)\; =\; 2\backslash pi^\; F(y)$ where $F(y)$ is the Dawson function as defined above. The Hilbert transform of $x^e^$ is also related to the Dawson function. We see this with the technique of differentiating inside the integral sign. Let :$H\_n\; =\; \backslash pi^\; P.V.\; \backslash int\_^\backslash infty\; \backslash over\; y-x\}\; dx$ Introduce :$H\_a\; =\; \backslash pi^\; P.V.\; \backslash int\_^\backslash infty\; dx$ The nth derivative is :$=\; (-1)^n\backslash pi^\; P.V.\; \backslash int\_^\backslash infty\; \backslash over\; y-x\}\; dx$ We thus find :$H\_n=(-1)^n\; |\_$ The derivatives are performed first, then the result evaluated at $a=1$. A change of variable also gives $H\_a=2\backslash pi^F(y\backslash sqrt\; a)$. Since $F\text{'}(y)=1-2yF(y)$, we can write $H\_n\; =\; P\_1(y)+P\_2(y)F(y)$ where $P\_1$ and $P\_2$ are polynomials. For example, $H\_1=-\backslash pi^y+2\backslash pi^y^2F(y)$. Alternatively, $H\_n$ can be calculated using the recurrence relation (for $n\; \backslash geq\; 0$) :$H\_(y)\; =\; y^2\; H\_n(y)\; -\; \backslash frac\; y$. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Dawson function」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. 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