# 菲涅耳積分

S(x)C(x)

## 定義

$S(x)=\int_0^x \sin(t^2)\,dt=\sum_{n=0}^{\infin}(-1)^n\frac{x^{4n+3}}{(2n+1)!(4n+3)},$
$C(x)=\int_0^x \cos(t^2)\,dt=\sum_{n=0}^{\infin}(-1)^n\frac{x^{4n+1}}{(2n)!(4n+1)}.$

## 羊角螺线

### 估計值

CS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

$e^{-z^2}$

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

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

$\int_{slope}^{} exp(-z^2)dz = \int_{0}^{\infty} \exp(-t^2e^{i\pi/2})e^{i\pi/4}dt = e^{i\pi/4} (\int_{0}^{\infty} \cos(-z^2)dz+i\int_{0}^{\infty} \sin(-z^2)dz)$
$\int_{0}^{\infty} \cos t^2\,\mathrm{d}t = \int_{0}^{\infty} \sin t^2\,\mathrm{d}t = \frac{\sqrt{2\pi}}{4} = \sqrt{\frac{\pi}{8}}.$

## 相关公式

• $\int_{0}^{\infty} e^{-at}sin(t^2)$$= (1/4)*\sqrt(2)*\sqrt(\pi)*(cos((1/4)*a^2)*(1-2*FresnelC((1/2)*a*\sqrt(2)/\sqrt(\pi)))+sin((1/4)*a^2)*(1-2*FresnelS((1/2)*a*\sqrt(2)/\sqrt(Pi))))$
• $\int(sin(ax^2 +2bx+c)dx=$$\frac{ \sqrt(2)*\sqrt(\pi)*(cos((b^2-a*c)/a)*FresnelS(\sqrt(2)*(a*x+b)/(\sqrt(\pi)*\sqrt(a)))-sin((b^2-a*c)/a)*FresnelC(\sqrt(2)*(a*x+b)/(\sqrt(\pi)*\sqrt(a)))) }{2\sqrt(a) }$
• $\int(FresnelC(t)dt=FresnelC(t)*t-\frac{sin((1/2)*\pi*t^2)}{\pi}$
• $\int(FesnelS(t)dt=FresnelS(t)*t+\frac{cos((1/2)*\pi*t^2)}{\pi}$
• $\frac{d FresnelC(t)}{dt}=cos((1/2)*\pi*t^2)$
• $\frac{d FresnelS(t)}{dt}=sin((1/2)*\pi*t^2)$

## 參考資料

1. ^ Beatty, Thomas. How to evaluate Fresnel Integrals. FGCU MATH - SUMMER 2013. [27 July 2013].
2. ^ Abromowitz and Stegun, Handbook of Mathematical Functions,p303-305, 1972 Natinal Bureau of Standards