菲涅耳积分

菲涅耳积分可由下面两个级数求得，对所有x均收敛。

${\displaystyle S(x)=\int _{0}^{x}\sin(t^{2})\,dt=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{4n+3}}{(2n+1)!(4n+3)}},}$ 

${\displaystyle C(x)=\int _{0}^{x}\cos(t^{2})\,dt=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{4n+1}}{(2n)!(4n+1)}}.}$ 

估计值

C和S的值当变数趋近于无穷大时，可用复变分析的方法求得。用以下这个函数的路径积分：

${\displaystyle e^{-z^{2}}}$ 

在复数平面上的一个扇型的边界，其中下边绕着正x轴，上半边是沿着y = x, x ≥ 0的路径，外圈则是一个半径为R，中心在原点的弧形。

当R趋近于无穷大时，路径积分沿弧形的部分将趋近于零^[1]，而实数轴部分的积分将可由高斯积分

${\displaystyle \int _{y-axis}^{}e^{-z^{2}}dz=\int _{0}^{\infty }e^{-t^{2}}dt={\frac {\sqrt {\pi }}{2}},}$ 

并且经过简单的计算后，第一象限平分线的那条积分便可以变成菲涅耳积分。

${\displaystyle \int _{slope}^{}\exp(-z^{2})dz=\int _{0}^{\infty }\exp(-t^{2}e^{i\pi /2})e^{i\pi /4}dt=e^{i\pi /4}(\int _{0}^{\infty }\cos(-z^{2})dz+i\int _{0}^{\infty }\sin(-z^{2})dz)}$ 

${\displaystyle \int _{0}^{\infty }\cos t^{2}\,\mathrm {d} t=\int _{0}^{\infty }\sin t^{2}\,\mathrm {d} t={\frac {\sqrt {2\pi }}{4}}={\sqrt {\frac {\pi }{8}}}.}$ 

下列一些包含菲涅耳积分的关系式^[2]

• ${\displaystyle \int _{0}^{\infty }e^{-at}\sin(t^{2})\mathrm {d} t={\frac {1}{4}}*{\sqrt {2\pi }}*(\cos {\frac {a^{2}}{4}}*(1-2*{\rm {FresnelC}}((1/2)*a*{\sqrt {2}}/{\sqrt {\pi }}))+\sin {\frac {a^{2}}{4}}*(1-2*\mathrm {FresnelS} ((1/2)*a*{\sqrt {2}}/{\sqrt {\pi }})))}$ 
• ${\displaystyle \int \sin(ax^{2}+2bx+c)\mathrm {d} x={\frac {{\sqrt {2\pi }}*(\cos((b^{2}-a*c)/a)*{\rm {FresnelS}}({\sqrt {2}}(ax+b)/({\sqrt {\pi a}}))-\sin((b^{2}-a*c)/a)*{\rm {FresnelC}}({\sqrt {2}}(ax+b)/({\sqrt {\pi a}})))}{2{\sqrt {a}}}}}$ 
• ${\displaystyle \int \mathrm {FresnelC} (t)\mathrm {d} t=\mathrm {FresnelC} (t)*t-{\frac {\sin {\frac {\pi t^{2}}{2}}}{\pi }}}$ 
• ${\displaystyle \int \mathrm {FresnelS} (t)\mathrm {d} t=\mathrm {FresnelS} (t)*t+{\frac {\cos {\frac {\pi t^{2}}{2}}}{\pi }}}$ 
• ${\displaystyle {\frac {\mathrm {d} ~\mathrm {FresnelC} (t)}{\mathrm {d} t}}=\cos {\frac {\pi t^{2}}{2}}}$ 
• ${\displaystyle {\frac {\mathrm {d} ~\mathrm {FresnelS} (t)}{\mathrm {d} t}}=\sin {\frac {\pi t^{2}}{2}}}$ 

• 奥古斯丁·菲涅耳
• 羊角螺线