積分備忘録

問題

広義積分を使う解法

広義積分の場合は雑魚問題です。
$$I={\int }_0^{\pi }\frac{dx}{a+{\cos }^{2}x}$$

$y=x-{\displaystyle \frac{\pi }{2}}$とおく。

$$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{dy}{a+{\sin }^{2}y}$$
$$=2{\int }_0^{\frac{\pi }{2}}\frac{dy}{a+{\sin }^{2}y}$$

$t=\tan y$とおく。

$${\displaystyle \frac{I}{2}}={\int }_0^{\mathrm{\infty }}\frac{1}{a+\frac{y^2}{1+y^2}}\cdot \frac{1}{1+y^2}dy$$
$$={\int }_0^{\mathrm{\infty }}\frac{dy}{(a+1)y^2+a}$$
$$=\frac{1}{a}{\int }_0^{\mathrm{\infty }}{\displaystyle \frac{dy}{(\frac{\sqrt{a^2+a}}{a}y{)}^2+1}}$$
$$=\frac{1}{a}\cdot \frac{a}{\sqrt{a^2+a}}[{\tan }^{-1}\frac{\sqrt{a^2+a}}{a}y{]}_0^{\mathrm{\infty }}$$

$(i)0<a$のとき

$${\displaystyle \frac{I}{2}}=\frac{1}{a}\cdot \frac{a}{\sqrt{a^2+a}}\cdot {\displaystyle \frac{\pi }{2}}=\frac{\pi }{2\sqrt{a^2+a}}$$

$(ii)a<-1$のとき

$${\displaystyle \frac{I}{2}}=\frac{1}{a}\cdot \frac{a}{\sqrt{a^2+a}}\cdot (-{\displaystyle \frac{\pi }{2}})=-\frac{\pi }{2\sqrt{a^2+a}}$$

$(i),(ii)$より

$$\therefore I=\frac{\left|a\right|\pi }{a\sqrt{a^2+a}}=\frac{\pi \mathrm{s}\mathrm{g}\mathrm{n}\mathit{\text{a}}}{\sqrt{a^2+a}}$$

広義積分を使わない解法

$$I={\int }_0^{\pi }\frac{dx}{a+{\cos }^{2}x}$$

対称性より

$${\displaystyle \frac{I}{2}}={\int }_0^{\frac{\pi }{2}}\frac{dx}{a+{\cos }^{2}x}$$

$(i)0<a$のとき

$${\displaystyle \frac{I}{2}}={\int }_0^{\frac{\pi }{2}}\frac{dx}{a+1-{\sin }^{2}x}$$
$$I\sqrt{a+1}={\int }_0^{\frac{\pi }{2}}\frac{1}{\sqrt{a+1}-\sin x}+\frac{1}{\sqrt{a+1}+\sin x}dx$$

$t=\tan {\displaystyle \frac{x}{2}}$とおく。
$$I\sqrt{a+1}={\int }_0^1(\frac{1}{\sqrt{a+1}-\frac{2t}{1+t^2}}+\frac{1}{\sqrt{a+1}+\frac{2t}{1+t^2}})\cdot {\displaystyle \frac{2}{1+t^2}}dt$$
$${\displaystyle \frac{\sqrt{a+1}}{2}}I={\int }_0^1\frac{1}{(1+t^2)\sqrt{a+1}-2t}+\frac{1}{(1+t^2)\sqrt{a+1}+2t}dt$$
$${\displaystyle \frac{a+1}{2}}I={\int }_0^1\frac{1}{(t-\frac{1}{\sqrt{a+1}}{)}^2+\frac{a}{1+a}}+\frac{1}{(t+\frac{1}{\sqrt{a+1}}{)}^2+\frac{a}{1+a}}dt$$
$$={\displaystyle \frac{a+1}{a}}{\int }_0^1\frac{1}{(\frac{\sqrt{a^2+a}}{a}t-\frac{1}{\sqrt{a}}{)}^2+1}+\frac{1}{(\frac{\sqrt{a^2+a}}{a}t+\frac{1}{\sqrt{a}}{)}^2+1}dt$$
$${\displaystyle \frac{a}{2}}I=\frac{a}{\sqrt{a^2+a}}[{\tan }^{-1}(\frac{\sqrt{a^2+a}}{a}t-{\displaystyle \frac{1}{\sqrt{a}}})+{\tan }^{-1}(\frac{\sqrt{a^2+a}}{a}t+{\displaystyle \frac{1}{\sqrt{a}}}){]}_0^1$$
$${\displaystyle \frac{\sqrt{a^2+a}}{2}}I=\{{\tan }^{-1}(\frac{\sqrt{a^2+a}-\sqrt{a}}{a})+{\tan }^{-1}(\frac{\sqrt{a^2+a}+\sqrt{a}}{a})\}-\{{\tan }^{-1}(-{\displaystyle \frac{1}{\sqrt{a}}})+{\tan }^{-1}({\displaystyle \frac{1}{\sqrt{a}}})\}$$

$({\displaystyle \frac{\sqrt{a^2+a}-\sqrt{a}}{a}})({\displaystyle \frac{\sqrt{a^2+a}+\sqrt{a}}{a}})=1$より
${\tan }^{-1}({\displaystyle \frac{\sqrt{a^2+a}-\sqrt{a}}{a}})+{\tan }^{-1}({\displaystyle \frac{\sqrt{a^2+a}+\sqrt{a}}{a}})={\displaystyle \frac{\pi }{2}}$
また
${\tan }^{-1}(-{\displaystyle \frac{1}{\sqrt{a}}})+{\tan }^{-1}({\displaystyle \frac{1}{\sqrt{a}}})=0$より

$${\displaystyle \frac{\sqrt{a^2+a}}{2}}I={\displaystyle \frac{\pi }{2}}$$
$$\therefore I=\frac{\pi }{\sqrt{a^2+a}}$$

$(ii)a<-1$のとき

$${\displaystyle \frac{I}{2}}={\int }_0^{\frac{\pi }{2}}\frac{dx}{a+{\cos }^{2}x}$$
$$=-{\displaystyle \frac{1}{2\sqrt{-a}}}{\int }_0^{\frac{\pi }{2}}\frac{1}{\sqrt{-a}-\cos x}+\frac{1}{\sqrt{-a}+\cos x}dx$$

$t=\tan {\displaystyle \frac{x}{2}}$とおく。

$$-I\sqrt{-a}={\int }_0^1(\frac{1}{\sqrt{-a}-\frac{1-t^2}{1+t^2}}+\frac{1}{\sqrt{-a}+\frac{1-t^2}{1+t^2}})\cdot {\displaystyle \frac{2}{1+t^2}}dt$$
$$-{\displaystyle \frac{\sqrt{-a}}{2}}I={\int }_0^1\frac{1}{(\sqrt{-a}+1)t^2+\sqrt{-a}-1}+\frac{1}{(\sqrt{-a}-1)t^2+\sqrt{-a}+1}dt$$
$$={\int }_0^1{\displaystyle \frac{1}{\sqrt{-a}-1}}\cdot \frac{1}{\frac{\sqrt{-a}+1}{\sqrt{-a}-1}{t}^2+1}+{\displaystyle \frac{1}{\sqrt{-a}+1}}\cdot \frac{1}{\frac{\sqrt{-a}-1}{\sqrt{-a}+1}{t}^2+1}dt$$
$$=[{\displaystyle \frac{1}{\sqrt{-a}-1}}\cdot \sqrt{{\displaystyle \frac{\sqrt{-a}-1}{\sqrt{-a}+1}}}\cdot {\tan }^{-1}(t\sqrt{{\displaystyle \frac{\sqrt{-a}+1}{\sqrt{-a}-1}}})+{\displaystyle \frac{1}{\sqrt{-a}+1}}\cdot \sqrt{{\displaystyle \frac{\sqrt{-a}+1}{\sqrt{-a}-1}}}\cdot {\tan }^{-1}(t\sqrt{{\displaystyle \frac{\sqrt{-a}-1}{\sqrt{-a}+1}}}){]}_0^1$$
$$=[{\displaystyle \frac{1}{\sqrt{-a-1}}}({\tan }^{-1}(t\sqrt{{\displaystyle \frac{\sqrt{-a}+1}{\sqrt{-a}-1}}})+{\tan }^{-1}(t\sqrt{{\displaystyle \frac{\sqrt{-a}-1}{\sqrt{-a}+1}}})){]}_0^1$$
$$={\displaystyle \frac{1}{\sqrt{-a-1}}}({\tan }^{-1}\sqrt{{\displaystyle \frac{\sqrt{-a}+1}{\sqrt{-a}-1}}}+{\tan }^{-1}\sqrt{{\displaystyle \frac{\sqrt{-a}-1}{\sqrt{-a}+1}}})$$

$(\sqrt{{\displaystyle \frac{\sqrt{-a}+1}{\sqrt{-a}-1}}})(\sqrt{{\displaystyle \frac{\sqrt{-a}-1}{\sqrt{-a}+1}}})=1$より

$$-{\displaystyle \frac{\sqrt{-a}}{2}}I={\displaystyle \frac{\pi }{2\sqrt{-a-1}}}$$
$$\therefore I=-{\displaystyle \frac{\pi }{\sqrt{a^2+a}}}$$

$(i),(ii)$より

$$\therefore I=\frac{\left|a\right|\pi }{a\sqrt{a^2+a}}=\frac{\pi \mathrm{s}\mathrm{g}\mathrm{n}\mathit{\text{a}}}{\sqrt{a^2+a}}$$

積分ガチャ超級に応募しなかった例の問題をそのまま出したところ、友人に想定してない方法で解かれまして、これならズルできまいと作った問題です。解き方は例の問題と一緒ですので、途中は省略して解答は以下の通りです。

$$J={\displaystyle \frac{\pi }{6}}-2+\sqrt{3}+\log (1+\sqrt{3})-{\displaystyle \frac{1}{2}}\log 2-\sqrt{2}\log (\sqrt{2}-1)(\sqrt{3}+\sqrt{2})$$