S(x)與C(x)。
菲涅耳積分,常被寫作 S(x)和C(x)。以奧古斯丁·菲涅耳為名。
菲涅耳積分可由下面兩個級數求得,對所有x均收斂。
![{\displaystyle S(x)=\int _{0}^{x}\sin(t^{2})\,dt=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{4n+3}}{(2n+1)!(4n+3)}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5376792e6de305beffeb0f04725fc70ad6ac43a0)
![{\displaystyle C(x)=\int _{0}^{x}\cos(t^{2})\,dt=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{4n+1}}{(2n)!(4n+1)}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/09c6c3332ffea33fdcf84abd87741b075f3bdaab)
用來計算Fresnel integrals的扇形路徑
C和S的值當變數趨近於無窮大時,可用複變分析的方法求得。用以下這個函數的路徑積分:
![{\displaystyle e^{-z^{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/294fe8388f1e37e66d26dab964051307bd12156d)
在複數平面上的一個扇型的邊界,其中下邊繞着正x軸,上半邊是沿着y = x, x ≥ 0的路徑,外圈則是一個半徑為R,中心在原點的弧形。
當R趨近於無窮大時,路徑積分沿弧形的部分將趨近於零[1],而實數軸部分的積分將可由高斯積分
![{\displaystyle \int _{y-axis}^{}e^{-z^{2}}dz=\int _{0}^{\infty }e^{-t^{2}}dt={\frac {\sqrt {\pi }}{2}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd84442399e56077eb42c6fe2d1427de1a495ba2)
並且經過簡單的計算後,第一象限平分線的那條積分便可以變成菲涅耳積分。
![{\displaystyle \int _{slope}^{}\exp(-z^{2})dz=\int _{0}^{\infty }\exp(-t^{2}e^{i\pi /2})e^{i\pi /4}dt=e^{i\pi /4}(\int _{0}^{\infty }\cos(-z^{2})dz+i\int _{0}^{\infty }\sin(-z^{2})dz)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef143b6ef0ada74c2e3c922080fbb9ce25b25971)
![{\displaystyle \int _{0}^{\infty }\cos t^{2}\,\mathrm {d} t=\int _{0}^{\infty }\sin t^{2}\,\mathrm {d} t={\frac {\sqrt {2\pi }}{4}}={\sqrt {\frac {\pi }{8}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d2f4b9ef355a64629211a614e2fa06e551786ed)
下列一些包含菲涅耳積分的關係式[2]
![{\displaystyle \int _{0}^{\infty }e^{-at}\sin(t^{2})\mathrm {d} t={\frac {1}{4}}*{\sqrt {2\pi }}*(\cos {\frac {a^{2}}{4}}*(1-2*{\rm {FresnelC}}((1/2)*a*{\sqrt {2}}/{\sqrt {\pi }}))+\sin {\frac {a^{2}}{4}}*(1-2*\mathrm {FresnelS} ((1/2)*a*{\sqrt {2}}/{\sqrt {\pi }})))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8e3b54c62e81bd6896b31dc119e807813a6fd56)
![{\displaystyle \int \sin(ax^{2}+2bx+c)\mathrm {d} x={\frac {{\sqrt {2\pi }}*(\cos((b^{2}-a*c)/a)*{\rm {FresnelS}}({\sqrt {2}}(ax+b)/({\sqrt {\pi a}}))-\sin((b^{2}-a*c)/a)*{\rm {FresnelC}}({\sqrt {2}}(ax+b)/({\sqrt {\pi a}})))}{2{\sqrt {a}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a21c4e7d211cf709458ecd7ac39578069cd32af)
![{\displaystyle \int \mathrm {FresnelC} (t)\mathrm {d} t=\mathrm {FresnelC} (t)*t-{\frac {\sin {\frac {\pi t^{2}}{2}}}{\pi }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0501ac7d2203982fccc6c28169bf6d1a4aa4e007)
![{\displaystyle \int \mathrm {FresnelS} (t)\mathrm {d} t=\mathrm {FresnelS} (t)*t+{\frac {\cos {\frac {\pi t^{2}}{2}}}{\pi }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/821cbfce50594af7ba1c36c269d405ba7398fb01)
![{\displaystyle {\frac {\mathrm {d} ~\mathrm {FresnelC} (t)}{\mathrm {d} t}}=\cos {\frac {\pi t^{2}}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7dd4cd95eb659d04e537b4b86999eca524442dda)
![{\displaystyle {\frac {\mathrm {d} ~\mathrm {FresnelS} (t)}{\mathrm {d} t}}=\sin {\frac {\pi t^{2}}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d4699655dc8ad44a3b4f0643c9d00718bae8be2)
- ^ Beatty, Thomas. How to evaluate Fresnel Integrals (PDF). FGCU MATH - SUMMER 2013. [27 July 2013]. (原始內容存檔 (PDF)於2015-02-07).
- ^ Abromowitz and Stegun, Handbook of Mathematical Functions,p303-305, 1972 Natinal Bureau of Standards