2008年10月23日木曜日

1/(1-(sinh(x))^2)の積分

$=\int\frac{1}{(1+\sinh x)(1-\sinh x)}dx$

$\sinh x=\frac{e^x-e^{-x}}{2}$ より,
$=\int\frac{1}{(1+\frac{e^x-e^{-x}}{2})(1-\frac{e^x-e^{-x}}{2})}dx$

$e^x=t$ とすれば, $e^x\,dx=dt$ より $dx=\frac{1}{e^x}dt=\frac{1}{t}dt$
$=\int\frac{1}{(1+\frac{t-\frac{1}{t}}{2})(1-\frac{t-\frac{1}{t}}{2})}\frac{1}{t}dt\\=\int\frac{-4t}{(t^2+2t-1)(t^2-2t-1)}dt\\=\int\frac{-4t}{(t+1+\sqrt{2})(t+1-\sqrt{2})(t-1+\sqrt{2})(t-1-\sqrt{2})}dt$

部分分数分解
$=\frac{a}{t+1+\sqrt{2}}+\frac{b}{t+1-\sqrt{2}}+\frac{c}{t-1+\sqrt{2}}+\frac{d}{t-1-\sqrt{2}}$
両辺を$(t+1+\sqrt{2})(t+1-\sqrt{2})(t-1+\sqrt{2})(t-1-\sqrt{2})$倍
$4t=a(t+1-\sqrt{2})(t-1+\sqrt{2})(t-1-\sqrt{2})\\\hspace{10mm}+b(t+1+\sqrt{2})(t-1+\sqrt{2})(t-1-\sqrt{2})\\\hspace{10mm}+c(t+1+\sqrt{2})(t+1-\sqrt{2})(t-1-\sqrt{2}\\\hspace{10mm}+d(t+1+\sqrt{2})(t+1-\sqrt{2})(t-1+\sqrt{2})\\=(a+b+c+d)t^3\\\hspace{10mm}+((1+\sqrt{2})a+(1-\sqrt{2})b+(-1+\sqrt{2})c+(-1-\sqrt{2})d)t^2\\\hspace{10mm}+((-3+2\sqrt{2})a+(-3-2\sqrt{2})b+(-3-2\sqrt{2})c+(-3+2\sqrt{2})d)t\\\hspace{10mm}+(-1+\sqrt{2})a+(-1-\sqrt{2})b+(1+\sqrt{2})c+(1-\sqrt{2})d$
係数比較で,
$a+b+c+d=0$
$(1+\sqrt{2})a+(1-\sqrt{2})b+(-1+\sqrt{2})c+(-1-\sqrt{2})d=0$
$(-3+2\sqrt{2})a+(-3-2\sqrt{2})b+(-3-2\sqrt{2})c+(-3+2\sqrt{2})d=-4$
$(-1+\sqrt{2})a+(-1-\sqrt{2})b+(1+\sqrt{2})c+(1-\sqrt{2})d=0$
をといて,
$a=\frac{-1}{2\sqrt{2}}$,$b=\frac{1}{2\sqrt{2}}$,$c=\frac{1}{2\sqrt{2}}$,$d=\frac{-1}{2\sqrt{2}}$

$=\frac{1}{2\sqrt{2}}\int\left(\frac{-1}{t+1+\sqrt{2}}+\frac{1}{t+1-\sqrt{2}}+\frac{1}{t-1+\sqrt{2}}+\frac{-1}{t-1-\sqrt{2}}\right)dt\\=\frac{1}{2\sqrt{2}}\left(-\log(t+1+\sqrt{2})+\log(t+1-\sqrt{2})+\log(t-1+\sqrt{2})-\log(t-1-\sqrt{2})\right)\\=\frac{1}{2\sqrt{2}}\left(\log(t+1-\sqrt{2})(t-1+\sqrt{2})-\log(t+1+\sqrt{2})(t-1-\sqrt{2})\right)\\=\frac{1}{2\sqrt{2}}\left(\log(t^2-3+2\sqrt{2})-\log(t^2-3-2\sqrt{2})\right)\\=\frac{1}{2\sqrt{2}}\left(\log(e^{2x}-3+2\sqrt{2})-\log(e^{2x}-3-2\sqrt{2})\right)$

追記>双曲三角関数の性質をちゃんと使えば,もっと簡単.

>>積分の記事

3 件のコメント:

  1. えーと、ハイパーサイン...
    じゃなくて、ハイパボリックサイン・・・あれ?
    ううむ・・・おかしいな.

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  2. 実は,計算が違っているはずなのです.
    グラフの形がぜんぜん違うので,間違っているのは確実なのですが,忙しくてゆっくり確かめる時間がないのです.

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  3. なんとか合わせました.
    双曲三角関数だけの式変形を使った,(つまり指数対数を使わない)本来の計算結果とつじつまがあわせられなかったのです.
    近いうちに本来の計算を記事にします.

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