【微分積分】4-5-8 置換積分の広義積分計算｜要点まとめ
このページでは、大学数学の微分積分で扱う「置換積分の広義積分計算」について整理します。置換積分を用いた広義積分の計算手順や、収束条件の確認方法を例題を通してわかりやすく解説します。変数変換のコツや極限の扱い方など、計算を進める上で重要なポイントを押さえ、広義積分の理解と計算力をしっかり身につけましょう。
広義積分における置換積分の基本
【例題】次の広義積分を求めなさい。
(1)\(\displaystyle \int_0^\infty\frac{1}{e^x+1}dx\)
\(\displaystyle =\lim_{t\to\infty}\int_0^t\frac{1}{e^x+1}dx\)
\(u=e^x\)とおくと、\(\displaystyle dx=\frac{1}{u}du\)
\(\displaystyle =\lim_{t\to\infty}\int_1^t\left(\frac{1}{u}-\frac{1}{u+1}\right)du\)
\(\displaystyle =\lim_{t\to\infty}\left[\log\left|\frac{u}{u+1}\right|\right]_1^t\)
\(\displaystyle =\lim_{t\to\infty}\log\frac{2t}{t+1}\)
\(\displaystyle =\log2\)
\(u=e^x\)とおくと、\(\displaystyle dx=\frac{1}{u}du\)
\(\displaystyle =\lim_{t\to\infty}\int_1^t\left(\frac{1}{u}-\frac{1}{u+1}\right)du\)
\(\displaystyle =\lim_{t\to\infty}\left[\log\left|\frac{u}{u+1}\right|\right]_1^t\)
\(\displaystyle =\lim_{t\to\infty}\log\frac{2t}{t+1}\)
\(\displaystyle =\log2\)
(2)\(\displaystyle \int_{-\infty}^\infty\frac{1}{\cosh x}dx\)
\(\displaystyle =\int_{-\infty}^\infty\frac{2e^x}{e^{2x}+1}dx\)
\(\displaystyle =\int_{-\infty}^0\frac{2e^x}{e^{2x}+1}dx+\int_0^\infty\frac{2e^x}{e^{2x}+1}dx\)
\(\displaystyle =\lim_{t\to-\infty}\int_t^0\frac{2e^x}{e^{2x}+1}dx+\lim_{t\to\infty}\int_0^t\frac{2e^x}{e^{2x}+1}dx\)
\(u=e^x\)とおくと、\(\displaystyle dx=\frac{1}{u}du\)
\(\displaystyle =\lim_{t\to-\infty}\int_t^0\frac{2}{u^2+1}du+\lim_{t\to\infty}\int_0^t\frac{2}{u^2+1}du\)
\(\displaystyle =\lim_{t\to-\infty}[2\tan^{-1}e^x]_t^0+\lim_{t\to\infty}[2\tan^{-1}e^x]_0^t\)
\(\displaystyle =0+2・\frac{\pi}{2}\)
\(\displaystyle =\pi\)
\(\displaystyle =\int_{-\infty}^0\frac{2e^x}{e^{2x}+1}dx+\int_0^\infty\frac{2e^x}{e^{2x}+1}dx\)
\(\displaystyle =\lim_{t\to-\infty}\int_t^0\frac{2e^x}{e^{2x}+1}dx+\lim_{t\to\infty}\int_0^t\frac{2e^x}{e^{2x}+1}dx\)
\(u=e^x\)とおくと、\(\displaystyle dx=\frac{1}{u}du\)
\(\displaystyle =\lim_{t\to-\infty}\int_t^0\frac{2}{u^2+1}du+\lim_{t\to\infty}\int_0^t\frac{2}{u^2+1}du\)
\(\displaystyle =\lim_{t\to-\infty}[2\tan^{-1}e^x]_t^0+\lim_{t\to\infty}[2\tan^{-1}e^x]_0^t\)
\(\displaystyle =0+2・\frac{\pi}{2}\)
\(\displaystyle =\pi\)
(3)\(\displaystyle \int_0^1\frac{x\sin^{-1}x}{\sqrt{1-x^2}}dx\)
\(u=\sin^{-1}x\)とおくと、\(\displaystyle dx=\sqrt{1-x^2}du\)
\(\displaystyle =\int_0^{\frac{\pi}{2}}u\sin udu\)
\(\displaystyle =[-u\cos u+\sin u]_0^{\frac{\pi}{2}}\)
\(\displaystyle =1\)
\(\displaystyle =\int_0^{\frac{\pi}{2}}u\sin udu\)
\(\displaystyle =[-u\cos u+\sin u]_0^{\frac{\pi}{2}}\)
\(\displaystyle =1\)
(4)\(\displaystyle \int_{-\infty}^\infty\frac{1}{(x^2+1)^{n+1}}dx\)
\(\displaystyle \frac{1}{(x^2+1)^{n+1}}\)は\(x\to\infty\)のとき、\(\displaystyle \frac{1}{x^{2n+2}}\)と同位の無限小である。
\(x\to-\infty\)も同様なので、\(\displaystyle \int_{-\infty}^\infty\frac{1}{(x^2+1)^{n+1}}dx\)は収束する。
\(x=\tan\theta\)とおくと、\(\displaystyle dx=(x^2+1)d\theta\)
よって、
\(\displaystyle \int_{-\infty}^\infty\frac{1}{(x^2+1)^{n+1}}dx\)
\(\displaystyle =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^{2n}\theta d\theta\)
\(\displaystyle =2\int_0^{\frac{\pi}{2}}\cos^{2n}\theta d\theta\)
\(\displaystyle =2\frac{(2n-1)!!}{(2n)!!}\frac{\pi}{2}\)
\(\displaystyle =\frac{(2n-1)!!}{(2n)!!}\pi\)
\(x\to-\infty\)も同様なので、\(\displaystyle \int_{-\infty}^\infty\frac{1}{(x^2+1)^{n+1}}dx\)は収束する。
\(x=\tan\theta\)とおくと、\(\displaystyle dx=(x^2+1)d\theta\)
よって、
\(\displaystyle \int_{-\infty}^\infty\frac{1}{(x^2+1)^{n+1}}dx\)
\(\displaystyle =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^{2n}\theta d\theta\)
\(\displaystyle =2\int_0^{\frac{\pi}{2}}\cos^{2n}\theta d\theta\)
\(\displaystyle =2\frac{(2n-1)!!}{(2n)!!}\frac{\pi}{2}\)
\(\displaystyle =\frac{(2n-1)!!}{(2n)!!}\pi\)
(5)\(\displaystyle \int_{-1}^1\frac{1}{(1+x^2)\sqrt{1-x^2}}dx\)
\(\displaystyle \lim_{x\to1-0}\frac{\frac{1}{(1+x^2)\sqrt{1-x^2}}}{\frac{1}{\sqrt{1-x}}}\)
\(\displaystyle \ \ \ =\lim_{x\to1-0}\frac{1}{(1+x^2)\sqrt{1+x}}=\frac{1}{2\sqrt{2}}\)
\(\displaystyle \frac{1}{(1+x^2)\sqrt{1-x^2}}\)は\(x\to1-0\)のとき、\(\displaystyle \frac{1}{\sqrt{1-x}}\)と同位の無限大である。
\(\displaystyle \int_0^1\frac{1}{\sqrt{1-x}}dx\)は収束するので、\(\displaystyle \int_{-1}^1\frac{1}{(1+x^2)\sqrt{1-x^2}}dx\)も収束する。
\(\displaystyle t=\sqrt{\frac{1-x}{1+x}}\)とおくと、\(\displaystyle dx=\frac{-4t}{(t^2+1)^2}dt\)
よって、
\(\displaystyle \int_{-1}^1\frac{1}{(1+x^2)\sqrt{1-x^2}}dx\)
\(\displaystyle =\int_{\infty}^0\frac{(t^2+1)^3}{4t(t^4+1)}・\frac{-4t}{(t^2+1)^2}dt\)
\(\displaystyle =\int_0^\infty\frac{t^2+1}{t^4+1}dt\)
\(\displaystyle =\frac{1}{2}\int_0^\infty\left(\frac{1}{t^2+\sqrt{2}t+1}+\frac{1}{t^2-\sqrt{2}t+1}\right)dt\)
\(\displaystyle =\frac{\sqrt{2}}{2}\left[\tan^{-1}(\sqrt{2}t+1)+\tan^{-1}(\sqrt{2}t-1)\right]_0^\infty\)
\(\displaystyle =\frac{1}{\sqrt{2}}\left(\frac{\pi}{2}+\frac{\pi}{2}\right)\)
\(\displaystyle =\frac{\pi}{\sqrt{2}}\)
\(\displaystyle \ \ \ =\lim_{x\to1-0}\frac{1}{(1+x^2)\sqrt{1+x}}=\frac{1}{2\sqrt{2}}\)
\(\displaystyle \frac{1}{(1+x^2)\sqrt{1-x^2}}\)は\(x\to1-0\)のとき、\(\displaystyle \frac{1}{\sqrt{1-x}}\)と同位の無限大である。
\(\displaystyle \int_0^1\frac{1}{\sqrt{1-x}}dx\)は収束するので、\(\displaystyle \int_{-1}^1\frac{1}{(1+x^2)\sqrt{1-x^2}}dx\)も収束する。
\(\displaystyle t=\sqrt{\frac{1-x}{1+x}}\)とおくと、\(\displaystyle dx=\frac{-4t}{(t^2+1)^2}dt\)
よって、
\(\displaystyle \int_{-1}^1\frac{1}{(1+x^2)\sqrt{1-x^2}}dx\)
\(\displaystyle =\int_{\infty}^0\frac{(t^2+1)^3}{4t(t^4+1)}・\frac{-4t}{(t^2+1)^2}dt\)
\(\displaystyle =\int_0^\infty\frac{t^2+1}{t^4+1}dt\)
\(\displaystyle =\frac{1}{2}\int_0^\infty\left(\frac{1}{t^2+\sqrt{2}t+1}+\frac{1}{t^2-\sqrt{2}t+1}\right)dt\)
\(\displaystyle =\frac{\sqrt{2}}{2}\left[\tan^{-1}(\sqrt{2}t+1)+\tan^{-1}(\sqrt{2}t-1)\right]_0^\infty\)
\(\displaystyle =\frac{1}{\sqrt{2}}\left(\frac{\pi}{2}+\frac{\pi}{2}\right)\)
\(\displaystyle =\frac{\pi}{\sqrt{2}}\)
次の学習に進もう！