1. Fresnel積分★

フレネル積分　　$I=\displaystyle \int_{0}^{ \infty }e^{ix^{2 } }dx=\displaystyle \lim_{R \to \infty }\displaystyle \int_{0}^{R}e^{ix^{2 } }dx$を求めなさい． 
$S(x)=\displaystyle \int_{0}^{x}\textrm{sin}t^{2}dt$, $C(x)=\displaystyle \int_{0}^{x}\textrm{cos}t^{2}dt$もFresnel積分と呼ばれる．
ただし，$S( \infty )=C( \infty )=\displaystyle \int_{0}^{ \infty }\textrm{sin}x^{2}dx=\displaystyle \int_{0}^{ \infty }\textrm{cos}x^{2}dx=\displaystyle \frac{1}{2}\sqrt{\displaystyle \frac{\pi }{2 } }$を既知とすると，
$I=\displaystyle \int_{0}^{ \infty }e^{ix^{2 } }dx=\displaystyle \int_{0}^{ \infty }\textrm{cos}x^{2}dx+i\displaystyle \int_{0}^{ \infty }\textrm{sin}x^{2}dx=\displaystyle \frac{1}{2}\sqrt{\displaystyle \frac{\pi }{2 } }\left( 1+i \right)$が得られます．
（参考）Gauss積分　$\displaystyle \int_{0}^{ \infty }\textrm{exp}[-t^{2}]dt=\sqrt{\pi }/2$

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これを複素積分法で計算してみます： 
$I=\displaystyle \int_{0}^{ \infty }\textrm{exp}[ix^{2}]dx=\displaystyle \int_{0}^{ \infty }\textrm{exp}[-(e^{-i\pi /2})x^{2}]dx$
$z=xe^{-i\pi /4}$とおくと，$dz=dxe^{-i\pi /4}$だから，　$I=e^{i\pi /4}\displaystyle \lim_{R \to \infty }\displaystyle \int_{0}^{Re^{-i\pi /4 } }\textrm{exp}[-z^{2}]dz$

被積分関数$\textrm{exp}[-z^{2}]$は，周回積分の領域$C$で正則だから，コーシー・リーマンの定理により周回積分は０になる：
$0=\displaystyle \int_{(C)}^{}\textrm{exp}[-z^{2}]dz=\displaystyle \int_{(C_{1})}^{}\textrm{exp}[-z^{2}]dz+\displaystyle \int_{(C_{2})}^{}\textrm{exp}[-z^{2}]dz+\displaystyle \int_{(C_{3})}^{}\textrm{exp}[-z^{2}]dz$
$z=te^{i\theta }$, $dz=dte^{-i\pi /4}$, $dz=d \theta iRe^{i\theta }$
$C_{1}$
$\displaystyle \lim_{R \to \infty }\displaystyle \int_{(C_{1})}^{}\textrm{exp}[-z^{2}]dz = e^{-i\pi /4}\displaystyle \lim_{R \to \infty }\displaystyle \int_{0}^{R}\textrm{exp}[-t^{2}e^{-i\pi /2}]dt=e^{-i\pi /4}\displaystyle \lim_{R \to \infty }\displaystyle \int_{0}^{R}e^{it^{2 } }dt$
$C_{2}$
$\displaystyle \lim_{R \to \infty }\displaystyle \int_{(C_{2})}^{}\textrm{exp}[-z^{2}]dz = \displaystyle \lim_{R \to \infty }iR\displaystyle \int_{-\pi /4}^{0}\textrm{exp}[-R^{2}e^{i2\theta }]e^{i\theta }d\theta =$
$=\displaystyle \lim_{R \to \infty }iR\displaystyle \int_{-\pi /4}^{0}e^{-R^{2}\textrm{2cos}2\theta }e^{-iR^{2}\textrm{sin}2\theta }e^{i\theta }d\theta \leqq \displaystyle \lim_{R \to \infty }iR\displaystyle \int_{-\pi /4}^{0}|e^{-R^{2}\textrm{cos}2\theta }|d\theta \leqq$
$\leqq \displaystyle \lim_{R \to \infty }iR\displaystyle \int_{-\pi /4}^{0}|e^{-R^{2}(4\theta /\pi +1)}|d\theta =\displaystyle \lim_{R \to \infty }iRe^{-R^{2 } }\displaystyle \int_{-\pi /4}^{0}e^{-(4R^{2}/\pi )\theta }d\theta =\displaystyle \lim_{R \to \infty }iRe^{-R^{2 } }\left[ \displaystyle \frac{e^{-(4R^{2}/\pi )\theta } }{-4R^{2}/\pi } \right] _{-\pi /4}^{0}$
$=\displaystyle \lim_{R \to \infty }iRe^{-R^{2 } }\left( 1-e^{R^{2 } } \right) /\left( -4R^{2}/\pi \right) =\displaystyle \lim_{R \to \infty }i\pi (1-e^{-R^{2 } })/4R=0$

$\displaystyle \lim_{R \to \infty }\displaystyle \int_{(C_{3})}^{}\textrm{exp}[-z^{2}]dz = \displaystyle \lim_{R \to \infty }-\displaystyle \int_{0}^{R}\textrm{exp}[-t^{2}]dt=-\sqrt{\pi }/2$
従って，$\displaystyle \lim_{R \to \infty }\displaystyle \int_{(C_{1})}^{}=-\displaystyle \lim_{R \to \infty }\displaystyle \int_{(C_{2})}^{}$ より，
$\displaystyle \frac{\left( 1-i \right) }{\sqrt{2 } }\displaystyle \int_{0}^{ \infty }e^{it^{2 } }dt=\sqrt{\pi }/2$
$\displaystyle \int_{0}^{ \infty }e^{it^{2 } }dt=\displaystyle \frac{1}{2}\sqrt{\displaystyle \frac{\pi }{2 } }(1+i)$