【微分積分】4-5-1 広義積分の計算|要点まとめ
このページでは、大学数学の微分積分で扱う「広義積分」について整理します。積分区間が無限に伸びる場合や、被積分関数が特異点をもつ場合の積分を、定義・収束判定の考え方・典型例題を通してわかりやすく解説します。広義積分の収束・発散を判断する力を身につけ、計算の基礎力を確実に強化しましょう。
広義積分の定義
【広義積分】
\(a,b\)は実数とする。
(1)関数\(f(x)\)は\([a,b)\)において、不定積分をもつとき
\(\displaystyle \int_a^bf(x)dx=\lim_{t\to b-0}\int_a^tf(x)dx\)
(2)関数\(f(x)\)は\((a,b]\)において、不定積分をもつとき
\(\displaystyle \int_a^bf(x)dx=\lim_{t\to a+0}\int_t^bf(x)dx\)
(3)関数\(f(x)\)は\([a,\infty)\)において、不定積分をもつとき
\(\displaystyle \int_a^\infty f(x)dx=\lim_{t\to\infty}\int_a^tf(x)dx\)
(4)関数\(f(x)\)は\((-\infty,b]\)において、不定積分をもつとき
\(\displaystyle \int_{-\infty}^bf(x)dx=\lim_{t\to-\infty}\int_t^bf(x)dx\)
これらを広義積分という。右辺の極限値が存在する場合に広義積分は収束するといい、そうでない場合は発散するという。
【例題】次の広義積分を求めなさい。
\(\displaystyle =\lim_{t\to+0}(2-2\sqrt{t})\)
\(\displaystyle =2\)
\(\displaystyle =\lim_{t\to+0}(-\log{t})\)
\(\displaystyle =\infty\)
\(\displaystyle =\lim_{t\to\infty}\left(-\frac{1}{t}+1\right)\)
\(\displaystyle =1\)
\(\displaystyle =\lim_{t\to\infty}\log t\)
\(\displaystyle =\infty\)
\(\displaystyle =\lim_{t\to\infty}\left([x・(-e^{-x})]_0^t+\int_0^t e^{-x}dx\right)\)
\(\displaystyle =\lim_{t\to\infty}(-te^{-t}+[-e^{-x}]_0^t)\)
\(\displaystyle =\lim_{t\to\infty}(-te^{-t}-e^{-t}+1)\)
\(\displaystyle =1\)
\(\displaystyle =\lim_{t\to+0}[x\log x-x]_t^1\)
\(\displaystyle =\lim_{t\to+0}(-1-t\log t+t)\)
\(\displaystyle =-1\)
\(\displaystyle =\lim_{t\to1-0}[\sin^{-1}x]_0^t\)
\(\displaystyle =\lim_{t\to1-0}(\sin^{-1}t)\)
\(\displaystyle =\frac{\pi}{2}\)
\(\displaystyle =\lim_{t\to\infty}\left[\frac{1}{2}\tan^{-1}\frac{x}{2}\right]_0^t\)
\(\displaystyle =\lim_{t\to\infty}\frac{1}{2}\tan^{-1}\frac{t}{2}\)
\(\displaystyle =\frac{1}{2}・\frac{\pi}{2}\)
\(\displaystyle =\frac{\pi}{4}\)
\(\displaystyle =\lim_{t\to\infty}\left[-\frac{1}{2}e^{-x^2}\right]_0^t\)
\(\displaystyle =\lim_{t\to\infty}\frac{1}{2}(1-e^{-t^2})\)
\(\displaystyle =\frac{1}{2}\)
\(\displaystyle =\lim_{t\to\infty}\left(\left[-\frac{1}{x}\log x\right]_e^t+\int_e^t\frac{1}{x^2}dx\right)\)
\(\displaystyle =\lim_{t\to\infty}\left(-\frac{1}{t}\log t+\frac{1}{e}+\left[-\frac{1}{x}\right]_e^t\right)\)
\(\displaystyle =\lim_{t\to\infty}\left(-\frac{\log t}{t}-\frac{1}{t}+\frac{2}{e}\right)\)
\(\displaystyle =\frac{2}{e}\)
\(\displaystyle =\lim_{t\to1-0}\left[-\sqrt{1-x^2}\sin^{-1}x\right]_0^t\)
\(\displaystyle \ \ \ +\lim_{t\to1-0}\int_0^t\sqrt{1-x^2}\frac{1}{\sqrt{1-x^2}}dx\)
\(\displaystyle =\lim_{t\to1-0}(-\sqrt{1-t^2}\sin^{-1}t+t)\)
\(\displaystyle =1\)
\(\displaystyle =\lim_{t\to\infty}\int_1^t\frac{1}{2}\left(\frac{1}{x}-\frac{1}{x+2}\right)dx\)
\(\displaystyle =\lim_{t\to\infty}\left[\frac{1}{2}\log\left|\frac{x}{x+2}\right|\right]_1^t\)
\(\displaystyle =\lim_{t\to\infty}\left(\frac{1}{2}\log\frac{1}{1+\frac{2}{t}}+\frac{1}{2}\log3\right)\)
\(\displaystyle =\frac{1}{2}\log3\)
\(\displaystyle =\lim_{t\to\infty}\frac{1}{2}\left[\log|x+1|-\frac{1}{2}\log(x^2+1)+\tan^{-1}x\right]_0^t\)
\(\displaystyle =\lim_{t\to\infty}\left(\frac{1}{4}\log\frac{(t+1)^2}{t^2+1}+\frac{1}{2}\tan^{-1}t\right)\)
\(\displaystyle =\frac{1}{4}\log1+\frac{1}{2}・\frac{\pi}{2}\)
\(\displaystyle =\frac{\pi}{4}\)
\(\displaystyle \ \ \ +\frac{9}{2(x+1)^3}-\frac{3}{4(x^2+3)}-\frac{3}{x^2+3})dx\)
\(\displaystyle =\lim_{t\to\infty}[\frac{3}{4}\log(x+1)-\frac{15}{4(x+1)}-\frac{9}{4(x+1)^2}\)
\(\displaystyle \ \ \ -\frac{3}{8}\log(x^2+3)-\frac{3}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}}]_3^t\)
\(\displaystyle =\lim_{t\to\infty}(\frac{3}{8}\log\frac{(t+1)^2}{t^2+3}-\frac{15}{4(t+1)}-\frac{9}{4(t+1)^2}\)
\(\displaystyle \ \ \ -\sqrt{3}\tan^{-1}\frac{t}{\sqrt{3}}-\frac{3}{8}\log\frac{4}{3}+\frac{69}{64}+\frac{\pi}{\sqrt{3}})\)
\(\displaystyle =\frac{3}{8}\log1-\sqrt{3}・\frac{\pi}{2}-\frac{3}{8}\log\frac{4}{3}+\frac{69}{64}+\frac{\pi}{\sqrt{3}}\)
\(\displaystyle =-\frac{3}{8}\log\frac{4}{3}+\frac{69}{64}-\frac{\sqrt{3}}{6}\pi\)