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}}}

^[1]

となる．この原像を求めれば，

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

^[2]

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)

^[3]

+{\frac {1}{2{\sqrt {2}}}}\tan ^{-1}({\sqrt {2}}x-1)

であるから^[4]，

\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．

^ $\int _{0}^{\infty }{\frac {dx}{x^{4}+1}}={\frac {\pi }{2{\sqrt {2}}}}$ は後に説明される．
^ 公式5の $\lambda ={\frac {1}{2}}$ の場合より，
${\frac {1}{\sqrt {s}}}\sqsubset {\frac {1}{\sqrt {\pi t}}}$
$\therefore {\frac {\pi }{2{\sqrt {2}}}}{\frac {1}{\sqrt {s}}}\sqsubset {\frac {1}{2{\sqrt {2}}}}{\frac {\pi }{\sqrt {\pi t}}}={\frac {1}{2}}{\sqrt {\frac {\pi }{2t}}}$
^ ${\sqrt {2}}\left(x\pm {\frac {\sqrt {2}}{2}}\right)={\sqrt {2}}x\pm 1$
^ まず $x^{4}+1$ を因数分解する．
$x^{4}+1=(x^{2}+1)^{2}-2x^{2}=(x^{2}+{\sqrt {2}}x+1)(x^{2}-{\sqrt {2}}x+1)$
次に，
${\frac {1}{x^{4}+1}}={\frac {Ax+B}{x^{2}+{\sqrt {2}}x+1}}+{\frac {Cx+D}{x^{2}-{\sqrt {2}}x+1}}$
とおいて，
$={\frac {(Ax+B)(x^{2}-{\sqrt {2}}x+1)+(Cx+D)(x^{2}+{\sqrt {2}}x+1)}{(x^{2}+{\sqrt {2}}x+1)(x^{2}-{\sqrt {2}}x+1)}}$

$(Ax+B)(x^{2}-{\sqrt {2}}x+1)=Ax^{3}+(B-{\sqrt {2}}A)x^{2}+(A-{\sqrt {2}}B)x+B$
$-{\sqrt {2}}$ を $+{\sqrt {2}}$ に換えて
$(Cx+D)(x^{2}+{\sqrt {2}}x+1)=Cx^{3}+(D+{\sqrt {2}}C)x^{2}+(C+{\sqrt {2}}D)x+D$
よって
$1=(A+C)x^{3}+{\bigg [}B+D+{\sqrt {2}}(C-A){\bigg ]}x^{2}+{\bigg [}A+C+{\sqrt {2}}(D-B){\bigg ]}x+(B+D)$
両辺の $x^{n}$ の係数を比較し等置すると，
$B+D=1$ より $D=1-B$ ，

$A+C=0$ より $C=-A$ ，

$B+D+{\sqrt {2}}(C-A)=1+{\sqrt {2}}(-2A)=0\therefore A={\frac {1}{2{\sqrt {2}}}}$

$A+C+{\sqrt {2}}(D-B)={\sqrt {2}}(1-2B)=0\therefore B={\frac {1}{2}}$ .

$\therefore C=-{\frac {1}{2{\sqrt {2}}}},\quad D={\frac {1}{2}}$

$\therefore {\frac {1}{x^{4}+1}}={\frac {{\frac {1}{2{\sqrt {2}}}}x+{\frac {1}{2}}}{x^{2}+{\sqrt {2}}x+1}}+{\frac {{\frac {-1}{2{\sqrt {2}}}}x+{\frac {1}{2}}}{x^{2}-{\sqrt {2}}x+1}}$

$={\frac {1}{2{\sqrt {2}}}}\left({\frac {x+{\sqrt {2}}}{x^{2}+{\sqrt {2}}x+1}}-{\frac {x-{\sqrt {2}}}{x^{2}-{\sqrt {2}}x+1}}\right)$
$\int {\frac {f'(x)}{f(x)}}dx=\log |f(x)|+C$ を視野にいれてさらに変形する．
${\frac {1}{x^{4}+1}}={\frac {1}{4{\sqrt {2}}}}\left({\frac {2x+2{\sqrt {2}}}{x^{2}+{\sqrt {2}}x+1}}-{\frac {2x-2{\sqrt {2}}}{x^{2}-{\sqrt {2}}x+1}}\right)$

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

$={\frac {1}{4{\sqrt {2}}}}\left({\frac {2x+{\sqrt {2}}}{x^{2}+{\sqrt {2}}x+1}}-{\frac {2x-{\sqrt {2}}}{x^{2}-{\sqrt {2}}x+1}}\right)+{\frac {1}{4}}\left({\frac {1}{x^{2}+{\sqrt {2}}x+1}}+{\frac {1}{x^{2}-{\sqrt {2}}x+1}}\right)$
後半部分を平方完全形とし $\int {\frac {dx}{x^{2}+\alpha ^{2}}}={\frac {1}{\alpha }}\tan ^{-1}{\frac {x}{\alpha }}+C$ が適用できるようにする。
${\frac {1}{x^{4}+1}}={\frac {1}{4{\sqrt {2}}}}\left({\frac {2x+{\sqrt {2}}}{x^{2}+{\sqrt {2}}x+1}}-{\frac {2x-{\sqrt {2}}}{x^{2}-{\sqrt {2}}x+1}}\right)+{\frac {1}{4}}\left\{{\frac {1}{\left(x+{\frac {\sqrt {2}}{2}}\right)^{2}+\left({\frac {1}{\sqrt {2}}}\right)^{2}}}+{\frac {1}{\left(x-{\frac {\sqrt {2}}{2}}\right)^{2}+\left({\frac {1}{\sqrt {2}}}\right)^{2}}}\right\}$

$\therefore I_{1}=\int _{0}^{\infty }{\frac {dx}{x^{4}+1}}={\frac {1}{4{\sqrt {2}}}}\int _{0}^{\infty }\left({\frac {2x+{\sqrt {2}}}{x^{2}+{\sqrt {2}}x+1}}-{\frac {2x-{\sqrt {2}}}{x^{2}-{\sqrt {2}}x+1}}\right)dx+{\frac {1}{4}}\int _{0}^{\infty }\left\{{\frac {1}{\left(x+{\frac {\sqrt {2}}{2}}\right)^{2}+\left({\frac {1}{\sqrt {2}}}\right)^{2}}}+{\frac {1}{\left(x-{\frac {\sqrt {2}}{2}}\right)^{2}+\left({\frac {1}{\sqrt {2}}}\right)^{2}}}\right\}dx$
先に $I_{2}=\int _{0}^{\infty }{\frac {dx}{(x+\alpha )^{2}+\beta ^{2}}}$ を求めておく． $x+\alpha =u$ と置いて， $x$ から $u$ に積分変数を変換すると、 $dx=du$ ， $x:0\to \infty$ のとき $u:\alpha \to \infty$ ．
$\therefore I_{2}=\int _{\alpha }^{\infty }{\frac {du}{u^{2}+\beta ^{2}}}={\frac {1}{\beta }}{\bigg [}\tan ^{-1}{\frac {u}{\beta }}{\bigg ]}_{\alpha }^{\infty }$
よって，
$I_{1}={\frac {1}{4{\sqrt {2}}}}{\bigg [}\log |x^{2}+{\sqrt {2}}x+1|-\log |x^{2}-{\sqrt {2}}x+1|{\bigg ]}_{0}^{\infty }+{\frac {\sqrt {2}}{4}}\left({\bigg [}\tan ^{-1}({\sqrt {2}}x){\bigg ]}_{\frac {\sqrt {2}}{2}}^{\infty }+{\bigg [}\tan ^{-1}({\sqrt {2}}x){\bigg ]}_{-{\frac {\sqrt {2}}{2}}}^{\infty }\right)$

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

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

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