Dawson integral 道森積分 （Dawson integral)由下式定義
F
(
x
)
=
e
−
x
2
∫
0
x
e
t
2
d
t
{\displaystyle F(x)=e^{-x^{2}}\int _{0}^{x}e^{t^{2}}\,dt}
道森積分的拐點為
-.92413887300459176701
+.92413887300459176701
Anti symmetric Dawson Integral 道森積分是反對稱函數
F
(
−
x
)
=
−
F
(
x
)
{\displaystyle F(-x)=-F(x)}
Derivative of Dawson Integral 道森積分的微商是
d
F
(
x
)
d
x
=
1
−
2
∗
x
∗
F
(
x
)
{\displaystyle {\frac {dF(x)}{dx}}=1-2*x*F(x)}
F
′
(
y
)
=
∫
0
∞
e
−
x
2
2
x
cos
(
2
x
y
)
d
x
{\displaystyle F'(y)=\int _{0}^{\infty }e^{-x^{2}}2x{\cos }(2xy)dx}
F
′
(
y
)
+
2
y
F
(
y
)
=
∫
0
∞
2
e
−
x
2
[
x
cos
(
2
x
y
)
+
y
sin
(
2
x
y
)
]
d
x
=
−
e
−
x
2
cos
(
2
x
y
)
|
0
∞
=
1
{\displaystyle F'(y)+2yF(y)=\int _{0}^{\infty }2e^{-x^{2}}[x{\cos }(2xy)+y{\sin }(2xy)]dx=-e^{-x^{2}}{\cos }(2xy)|_{0}^{\infty }=1}
而且顯然
F
(
0
)
=
0
{\displaystyle F(0)=0}
F
(
y
)
{\displaystyle F(y)}
f
′
(
y
)
+
2
y
f
(
y
)
=
1
{\displaystyle f'(y)+2yf(y)=1}
在初始條件
f
(
0
)
=
0
{\displaystyle f(0)=0}
柯西-利普希茨定理 解是唯一的.
F
(
y
)
=
∫
0
y
e
t
2
−
y
2
d
t
{\displaystyle F\left(y\right)=\int _{0}^{y}{{e^{{t^{2}}-{y^{2}}}}{\text{d}}t}}
證明:
只要證明也滿足它的微分方程即可
對
y
{\displaystyle y}
積分符號內取微分 有
F
′
(
y
)
=
d
d
y
e
−
y
2
∫
0
y
e
t
2
d
t
=
−
2
y
e
−
y
2
∫
0
y
e
t
2
d
t
+
1
{\displaystyle F'(y)={\frac {d}{dy}}e^{-y^{2}}\int _{0}^{y}e^{t^{2}}dt=-2ye^{-y^{2}}\int _{0}^{y}e^{t^{2}}dt+1}
Press, WH; Teukolsky, SA; Vetterling, WT; Flannery, BP, Section 6.9. Dawson's Integral , Numerical Recipes: The Art of Scientific Computing 3rd, New York: Cambridge University Press, 2007 [2015-01-27 ] , ISBN 978-0-521-88068-8原始內容 存檔於2011-08-11） Temme, N. M., Error Functions, Dawson’s and Fresnel Integrals , Olver, Frank W. J. ; Lozier, Daniel M.; Boisvert, Ronald F.; Clark, Charles W. (編), NIST Handbook of Mathematical Functions , Cambridge University Press, 2010, ISBN 978-0521192255MR 2723248 
![{\displaystyle F'(y)+2yF(y)=\int _{0}^{\infty }2e^{-x^{2}}[x{\cos }(2xy)+y{\sin }(2xy)]dx=-e^{-x^{2}}{\cos }(2xy)|_{0}^{\infty }=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f8198a6b7134aa7950af3ca5612a006214509f3)
