制御と振動の数学/Laplace 変換/定積分の計算への応用/Fresnel積分

(2.38)

$\int _{0}^{\infty }\cos ax^{2}dx={\frac {1}{2}}{\sqrt {\frac {\pi }{2a}}}\quad (a>0)$

これは，光学や電磁気学に現れる式である．例によって，

$f(t)=\int _{0}^{\infty }\cos tx^{2}dx\quad (t>0)$

とおいて Laplace 変換すると，

${\mathcal {L}}[f(t)]=\int _{0}^{\infty }{\frac {s}{x^{4}+s^{2}}}dx$

いま $x:={\sqrt {s}}y,(s>0)$ とおけば，

${\mathcal {L}}[f(t)]={\frac {1}{\sqrt {s}}}\int _{0}^{\infty }{\frac {dy}{y^{4}+1}}={\frac {\pi }{2{\sqrt {2}}}}{\frac {1}{\sqrt {s}}}$

となる．この原像を求めれば，

$f(t)={\frac {1}{2}}{\sqrt {\frac {\pi }{2t}}}\quad (t>0)$

を得る．$t=a$ とおけばよい．

注

$\int {\frac {dx}{x^{4}+1}}={\frac {1}{2{\sqrt {2}}}}\int \left\{{\frac {x+{\sqrt {2}}}{x^{2}+{\sqrt {2}}x+1}}-{\frac {x-{\sqrt {2}}}{x^{2}-{\sqrt {2}}x+1}}\right\}dx$

$={\frac {1}{4{\sqrt {2}}}}\log |x^{2}+{\sqrt {2}}x+1|-{\frac {1}{4{\sqrt {2}}}}\log |x^{2}-{\sqrt {2}}x+1|+{\frac {1}{2{\sqrt {2}}}}\tan ^{-1}({\sqrt {2}}x+1)$$+{\frac {1}{2{\sqrt {2}}}}\tan ^{-1}({\sqrt {2}}x-1)$

であるから，

$\int _{0}^{\infty }{\frac {dx}{x^{4}+1}}={\frac {1}{2{\sqrt {2}}}}{\frac {\pi }{2}}\cdot 2={\frac {\pi }{2{\sqrt {2}}}}$

を得る．この定積分は関数論の留数計算で求めるのが簡単である．

例65$\quad$

$\int _{0}^{\infty }\sin ax^{2}dx={\frac {1}{2}}{\sqrt {\frac {\pi }{2a}}}\quad (a>0)$

を示せ．

解答例

$\int _{0}^{\infty }\cos ax^{2}dx$ の場合と同じ方針で解く．

$f(t)=\int _{0}^{\infty }\sin tx^{2}dx$

とおいて両辺を Laplace 変換すると，

${\mathcal {L}}[f]=\int _{0}^{\infty }{\frac {x^{2}}{s^{2}+x^{4}}}dx$

$x={\sqrt {s}}y,\quad s>0$ とおいて積分変数を $x$ から $y$ に変換すると $dx={\sqrt {s}}\ dy$，また，

${\mathcal {L}}[f]=\int _{0}^{\infty }{\frac {sy^{2}}{s^{2}(1+y^{4})}}{\sqrt {s}}dy$

$={\frac {1}{\sqrt {s}}}\int _{0}^{\infty }{\frac {y^{2}dy}{y^{4}+1}}$

分母 $y^{4}+1$ は

$y^{4}+1=(y^{2}+{\sqrt {2}}y+1)(y^{2}-{\sqrt {2}}y+1)$

と因数分解できるから，

${\frac {y^{2}}{y^{4}+1}}={\frac {Ay+B}{y^{2}+{\sqrt {2}}y+1}}+{\frac {Cy+D}{y^{2}-{\sqrt {2}}y+1}}$

とおいて，

$y^{2}=(Ay+B)(y^{2}-{\sqrt {2}}y+1)+(Cy+D)(y^{2}+{\sqrt {2}}y+1)$

$y^{2}=(A+C)y^{3}+{\bigg [}B+D+{\sqrt {2}}(C-A){\bigg ]}y^{2}+{\bigg [}A+C+{\sqrt {2}}(D-B){\bigg ]}y+(B+D)$

$y^{n}$ の係数を等置して，

$A+C=0$

$B+D=0$

$B+D+{\sqrt {2}}(C-A)=1$

$A+C+{\sqrt {2}}(D-B)=0$

これを解いて，

$A=-{\frac {1}{2{\sqrt {2}}}},\quad C={\frac {1}{2{\sqrt {2}}}},\quad B=D=0$

$\therefore {\frac {y^{2}}{y^{4}+1}}={\frac {1}{2{\sqrt {2}}}}\left\{{\frac {y}{y^{2}-{\sqrt {2}}y+1}}-{\frac {y}{y^{2}+{\sqrt {2}}y+1}}\right\}$

$={\frac {1}{4{\sqrt {2}}}}\left\{{\frac {2y}{y^{2}-{\sqrt {2}}y+1}}-{\frac {2y}{y^{2}+{\sqrt {2}}y+1}}\right\}$

$={\frac {1}{4{\sqrt {2}}}}\left\{{\frac {2y-{\sqrt {2}}}{y^{2}-{\sqrt {2}}y+1}}-{\frac {2y+{\sqrt {2}}}{y^{2}+{\sqrt {2}}y+1}}\right\}+{\frac {1}{4{\sqrt {2}}}}\left\{{\frac {\sqrt {2}}{y^{2}-{\sqrt {2}}y+1}}+{\frac {\sqrt {2}}{y^{2}+{\sqrt {2}}y+1}}\right\}$

$={\frac {1}{4{\sqrt {2}}}}\left\{{\frac {2y-{\sqrt {2}}}{y^{2}-{\sqrt {2}}y+1}}-{\frac {2y+{\sqrt {2}}}{y^{2}+{\sqrt {2}}y+1}}\right\}+{\frac {1}{4}}\left\{{\frac {1}{\left(y-{\frac {\sqrt {2}}{2}}\right)^{2}+\left({\sqrt {\frac {1}{2}}}\right)^{2}}}+{\frac {1}{\left(y+{\frac {\sqrt {2}}{2}}\right)^{2}+\left({\sqrt {\frac {1}{2}}}\right)^{2}}}\right\}$

$\therefore \int _{0}^{\infty }{\frac {y^{2}}{y^{4}+1}}dy={\frac {1}{4{\sqrt {2}}}}\int _{0}^{\infty }\left\{{\frac {2y-{\sqrt {2}}}{y^{2}-{\sqrt {2}}y+1}}-{\frac {2y+{\sqrt {2}}}{y^{2}+{\sqrt {2}}y+1}}\right\}dy+{\frac {1}{4}}\int _{0}^{\infty }\left\{{\frac {1}{\left(y-{\frac {\sqrt {2}}{2}}\right)^{2}+\left({\sqrt {\frac {1}{2}}}\right)^{2}}}+{\frac {1}{\left(y+{\frac {\sqrt {2}}{2}}\right)^{2}+\left({\sqrt {\frac {1}{2}}}\right)^{2}}}\right\}dy$

$={\bigg [}\log |y^{2}-{\sqrt {2}}x+1|-\log |y^{2}+{\sqrt {2}}x+1|{\bigg ]}_{0}^{\infty }+{\frac {\sqrt {2}}{4}}\left({\bigg [}\tan ^{-1}({\sqrt {2}}y){\bigg ]}_{-{\frac {\sqrt {2}}{2}}}^{\infty }+{\bigg [}\tan ^{-1}({\sqrt {2}}y){\bigg ]}_{\frac {\sqrt {2}}{2}}^{\infty }\right)$

$={\frac {1}{4{\sqrt {2}}}}{\bigg [}\log \left|{\frac {y^{2}-{\sqrt {2}}x+1}{y^{2}+{\sqrt {2}}x+1}}\right|{\bigg ]}_{0}^{\infty }+{\frac {\sqrt {2}}{4}}\left\{{\frac {\pi }{2}}-\tan ^{-1}\left(-{\sqrt {2}}\cdot {\frac {\sqrt {2}}{2}}\right)+{\frac {\pi }{2}}-\tan ^{-1}\left({\sqrt {2}}\cdot {\frac {\sqrt {2}}{2}}\right)\right\}$

$=0-0+{\frac {\sqrt {2}}{4}}\left({\frac {\pi }{2}}+{\frac {\pi }{4}}+{\frac {\pi }{2}}-{\frac {\pi }{4}}\right)$

$={\frac {\sqrt {2}}{4}}\cdot 2\cdot {\frac {\pi }{2}}={\frac {{\sqrt {2}}\pi }{4}}={\frac {\pi }{2{\sqrt {2}}}}$

$\therefore {\mathcal {L}}[f]={\frac {1}{\sqrt {s}}}\int _{0}^{\infty }{\frac {y^{2}dy}{y^{4}+1}}={\frac {\pi }{2{\sqrt {2}}}}{\frac {1}{\sqrt {s}}}$

$\therefore f(t)={\frac {1}{2}}{\sqrt {\frac {\pi }{2t}}}\quad (t>0)$

$\therefore f(a)=\int _{0}^{\infty }\sin ax^{2}dx={\frac {1}{2}}{\sqrt {\frac {\pi }{2a}}}$

$\diamondsuit$