積分備忘録

問1

広義積分を使う解法

広義積分の場合は雑魚問題です。
$$ I=\int_0^\pi \frac{dx}{a+\cos^2{x}} $$

$y=x-\dfrac{\pi}{2}$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。

$$ \dfrac{I}{2}=\int_0^{\infty} \frac{1}{a+\frac{y^2}{1+y^2}}\cdot\frac{1}{1+y^2} dy $$
$$ =\int_0^{\infty} \frac{dy}{(a+1)y^2+a} $$
$$ =\frac{1}{a} \int_0^{\infty} \dfrac{dy}{\big( \frac{\sqrt{a^2+a}}{a}y\big)^2+1} $$
$$ =\frac{1}{a}\cdot\frac{a}{\sqrt{a^2+a}}\bigg[\tan^{-1}{\frac{\sqrt{a^2+a}}{a}y} \bigg]_0^{\infty} $$

$(i)\ 0< a$のとき

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

$(ii)\ a<-1$のとき

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

$(i),\ (ii)$より

$$ \therefore I=\frac{\abs{a}\pi}{a\sqrt{a^2+a}} =\frac{\pi\ \rm{sgn}{\textit{a}}}{\sqrt{a^2+a}} $$

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

$$ \dfrac{I}{2}=\int_0^\frac{\pi}{2} \frac{dx}{a+\cos^2{x}} $$

$(i)\ 0< a$のとき

$$ \dfrac{I}{2}=\int_0^\frac{\pi}{2} \frac{dx}{a+1-\sin^2{x}} $$
$$ I\sqrt{a+1}=\int_0^\frac{\pi}{2} \frac{1}{\sqrt{a+1}-\sin{x}}+\frac{1}{\sqrt{a+1}+\sin{x}}dx $$

$t=\tan{\dfrac{x}{2}}$とおく。
$$ I\sqrt{a+1}=\int_0^1 \bigg(\frac{1}{\sqrt{a+1}-\frac{2t}{1+t^2}}+\frac{1}{\sqrt{a+1}+\frac{2t}{1+t^2}}\bigg)\cdot\dfrac{2}{1+t^2}dt $$
$$ \dfrac{\sqrt{a+1}}{2}I=\int_0^1 \frac{1}{(1+t^2)\sqrt{a+1}-2t}+\frac{1}{(1+t^2)\sqrt{a+1}+2t}dt $$
$$ \dfrac{a+1}{2}I=\int_0^1 \frac{1}{\big(t-\frac{1}{\sqrt{a+1}}\big)^2+\frac{a}{1+a}}+\frac{1}{\big(t+\frac{1}{\sqrt{a+1}}\big)^2+\frac{a}{1+a}}dt $$
$$ =\dfrac{a+1}{a}\int_0^1 \frac{1}{\big(\frac{\sqrt{a^2+a}}{a} t-\frac{1}{\sqrt{a}}\big)^2+1}+\frac{1}{\big(\frac{\sqrt{a^2+a}}{a}t+\frac{1}{\sqrt{a}}\big)^2+1}dt $$
$$ \dfrac{a}{2}I=\cfrac{a}{\sqrt{a^2+a}}\bigg[\tan^{-1}\bigg( \frac{\sqrt{a^2+a}}{a} t-\dfrac{1}{\sqrt{a}} \bigg)+\tan^{-1}\bigg( \frac{\sqrt{a^2+a}}{a} t+\dfrac{1}{\sqrt{a}} \bigg) \bigg]_0^1 $$
$$ \dfrac{\sqrt{a^2+a}}{2} I=\bigg\{\tan^{-1}\bigg( \frac{\sqrt{a^2+a}-\sqrt{a}}{a} \bigg)+\tan^{-1}\bigg( \frac{\sqrt{a^2+a}+\sqrt{a}}{a} \bigg) \bigg\} -\bigg\{\tan^{-1}\bigg( -\dfrac{1}{\sqrt{a}} \bigg)+\tan^{-1}\bigg( \dfrac{1}{\sqrt{a}} \bigg) \bigg\} $$

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

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

$(ii)\ a<-1$のとき

$$ \dfrac{I}{2}=\int_0^\frac{\pi}{2} \frac{dx}{a+\cos^2{x}} $$
$$ =-\dfrac{1}{2\sqrt{-a}}\int_0^\frac{\pi}{2} \frac{1}{\sqrt{-a}-\cos{x}}+\frac{1}{\sqrt{-a}+\cos{x}} dx $$

$t=\tan{\dfrac{x}{2}}$とおく。

$$ -I\sqrt{-a}=\int_0^1 \bigg(\frac{1}{\sqrt{-a}-\frac{1-t^2}{1+t^2}}+\frac{1}{\sqrt{-a}+\frac{1-t^2}{1+t^2}}\bigg)\cdot\dfrac{2}{1+t^2} dt $$
$$ -\dfrac{\sqrt{-a}}{2}I=\int_0^1 \frac{1}{(\sqrt{-a}+1)t^2+\sqrt{-a}-1}+\frac{1}{(\sqrt{-a}-1)t^2+\sqrt{-a}+1}dt $$
$$ =\int_0^1 \dfrac{1}{\sqrt{-a}-1}\cdot\frac{1}{\frac{\sqrt{-a}+1}{\sqrt{-a}-1}t^2+1}+\dfrac{1}{\sqrt{-a}+1}\cdot\frac{1}{\frac{\sqrt{-a}-1}{\sqrt{-a}+1}t^2+1}dt $$
$$ =\bigg[\dfrac{1}{\sqrt{-a}-1}\cdot\sqrt{\dfrac{\sqrt{-a}-1}{\sqrt{-a}+1}}\cdot\tan^{-1}\bigg( t\ \sqrt{\dfrac{\sqrt{-a}+1}{\sqrt{-a}-1}} \bigg) +\dfrac{1}{\sqrt{-a}+1}\cdot\sqrt{\dfrac{\sqrt{-a}+1}{\sqrt{-a}-1}}\cdot\tan^{-1}\bigg( t\ \sqrt{\dfrac{\sqrt{-a}-1}{\sqrt{-a}+1}}\bigg)\bigg]_0^1 $$
$$ =\bigg[\dfrac{1}{\sqrt{-a-1}}\bigg(\tan^{-1}\bigg( t\ \sqrt{\dfrac{\sqrt{-a}+1}{\sqrt{-a}-1}} \bigg) +\tan^{-1}\bigg( t\ \sqrt{\dfrac{\sqrt{-a}-1}{\sqrt{-a}+1}}\bigg)\bigg)\bigg]_0^1 $$
$$ =\dfrac{1}{\sqrt{-a-1}}\bigg(\tan^{-1}\sqrt{\dfrac{\sqrt{-a}+1}{\sqrt{-a}-1}} +\tan^{-1}\sqrt{\dfrac{\sqrt{-a}-1}{\sqrt{-a}+1}}\bigg) $$

$\bigg(\ \sqrt{\dfrac{\sqrt{-a}+1}{\sqrt{-a}-1}} \bigg)\bigg(\ \sqrt{\dfrac{\sqrt{-a}-1}{\sqrt{-a}+1}} \bigg)=1$より

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

$(i),\ (ii)$より

$$ \therefore I=\frac{\abs{a}\pi}{a\sqrt{a^2+a}} =\frac{\pi\ \rm{sgn}{\textit{a}}}{\sqrt{a^2+a}} $$

問2

前に出した記事 の係数を一般に対応させたやつです。解法は同じ流れなので、コピペします。

解法

$$ \int_0^\pi \dfrac{\sin{x}}{a+\sqrt{b+\sin{2x}}} \ dx $$
積分区間を分けて、それぞれ$I_1,\ I_2$とおく。
$$ I_1=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin{x}}{a+\sqrt{b+\sin{2x}}} \ dx,\ I_2=\int_{\dfrac{\pi}{2}}^{\pi} \dfrac{\sin{x}}{a+\sqrt{b+\sin{2x}}} \ dx $$
King Propertyより
$$ I_1=\int_0^{\dfrac{\pi}{2}} \dfrac{\cos{x}}{a+\sqrt{b+\sin{2x}}} \ dx,\ I_2=\int_{\dfrac{\pi}{2}}^{\pi} \dfrac{-\cos{x}}{a+\sqrt{b+\sin{2x}}} \ dx $$
等しい２式をそれぞれ合わせて
$$ 2I_1=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin{x}+\cos{x}}{a+\sqrt{b+\sin{2x}}} \ dx,\ 2I_2=\int_{\dfrac{\pi}{2}}^{\pi} \dfrac{\sin{x}-\cos{x}}{a+\sqrt{b+\sin{2x}}} \ dx $$

$(i) \ I_1$を計算する。

$y=\sin{x}-\cos{x}$とおく。
$x:0\to\dfrac{\pi}{2} \Leftrightarrow y:-1\to1,$
$dy=\sin{x}+\cos{x} \ dx,\ \sin{2x}=2\sin{x}\cos{x}=1-y^2$より

$$ 2I_1=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin{x}+\cos{x}}{a+\sqrt{b+\sin{2x}}} \ dx $$

$$ =\int_{-1}^1 \dfrac{\sin{x}+\cos{x}}{a+\sqrt{b+(1-y^2)}} \cdot \dfrac{1}{\sin{x}+\cos{x}} \ dy $$
$$ =\int_{-1}^1 \dfrac{dy}{a+\sqrt{b+1-y^2}} $$
偶関数だから
$$ I_1=\int_0^1 \dfrac{dy}{a+\sqrt{b+1-y^2}} $$

$y=\sqrt{b+1}\sin{\theta} \ \bigg( |\theta|\leq\dfrac{\pi}{2} \bigg)$とおく。$y:0\to1 \Leftrightarrow \theta:0\to\dfrac{\pi}{4}, \ dy=\sqrt{b+1}\cos{\theta} \ d\theta$より
$$ =\int_0^\dfrac{\pi}{4} \dfrac{\sqrt{b+1}\cos{\theta}}{a+\sqrt{b+1} \cos{\theta}} \ d\theta=\dfrac{\pi}{4}-a \int_0^\dfrac{\pi}{4} \dfrac{d\theta}{a+\sqrt{b+1} \cos{\theta}} $$

$z=\tan{\dfrac{\theta}{2}} \ ( |z|\leq 1 )$とおく。

$\theta:0\to\dfrac{\pi}{4} \Leftrightarrow z:0\to\tan{\dfrac{\pi}{8}}, \ d\theta=\dfrac{2}{1+z^2} \ dz$より
$$ \dfrac{\pi}{4}-I_1=a\int_0^\dfrac{\pi}{4} \dfrac{d\theta}{a+\sqrt{b+1} \cos{\theta}} $$
$$ =a\int_0^{\tan{\dfrac{\pi}{8}}} \dfrac{1}{a+\sqrt{b+1} \cdot \dfrac{1-z^2}{1+z^2}} \cdot \dfrac{2}{1+z^2} \ dz $$
$$ ＝2a \int_0^{\tan{\dfrac{\pi}{8}}} \dfrac{dz}{a(1+z^2)+\sqrt{b+1}\ (1-z^2)} $$
$$ ＝2a \int_0^{\tan{\dfrac{\pi}{8}}} \dfrac{dz}{a+\sqrt{b+1}+(a-\sqrt{b+1})z^2} $$

$(i1)$ $a-\sqrt{b+1}=0$のとき

$$ \dfrac{\pi}{4}-I_1 ＝2a \int_0^{\tan{\dfrac{\pi}{8}}} \dfrac{dz}{a+\sqrt{b+1}} =\dfrac{2a(\sqrt{2}-1)}{a+\sqrt{b+1}} $$
$$ \therefore I_1 =\dfrac{\pi}{4}-\dfrac{2a(\sqrt{2}-1)}{a+\sqrt{b+1}} $$

$(i2)$ $0<\dfrac{a-\sqrt{b+1}}{a+\sqrt{b+1}}$のとき

範囲は$a<-\sqrt{b+1},\ \sqrt{b+1}< a$

$$ \dfrac{\pi}{4}-I_1 ＝2a \int_0^{\tan{\dfrac{\pi}{8}}} \dfrac{dz}{a+\sqrt{b+1}+(a-\sqrt{b+1})z^2} $$
$$ ＝\dfrac{2a}{a+\sqrt{b+1}} \int_0^{\tan{\dfrac{\pi}{8}}} \dfrac{dz}{1+\frac{a-\sqrt{b+1}}{a+\sqrt{b+1}}z^2} $$
$$ =\dfrac{2a}{a+\sqrt{b+1}}\cdot\sqrt{\dfrac{a+\sqrt{b+1}}{a-\sqrt{b+1}}}\bigg[ \arctan(\dfrac{a-\sqrt{b+1}}{\sqrt{a^2-b-1}}\ z) \bigg]_0^{\tan{\dfrac{\pi}{8}}} $$
$\tan{\dfrac{\pi}{8}}=\sqrt{2}-1$より
$$ =\dfrac{2a}{\sqrt{a^2-b-1}}\ \arctan(\dfrac{(a-\sqrt{b+1})(\sqrt{2}-1)}{\sqrt{a^2-b-1}}) $$
$$ \therefore I_1=\dfrac{\pi}{4}-\dfrac{2a}{\sqrt{a^2-b-1}}\ \arctan(\dfrac{(a-\sqrt{b+1})(\sqrt{2}-1)}{\sqrt{a^2-b-1}}) $$

$(i3)$ $\dfrac{a-\sqrt{b+1}}{a+\sqrt{b+1}}<0$のとき

範囲は$-\sqrt{b-1}< a<\sqrt{b+1}$

$$ \dfrac{\pi}{4}-I_1 ＝2a \int_0^{\tan{\dfrac{\pi}{8}}} \dfrac{dz}{a+\sqrt{b+1}+(a-\sqrt{b+1})z^2} $$
$$ ＝\dfrac{2a}{\sqrt{b+1}-a} \int_0^{\tan{\dfrac{\pi}{8}}} \dfrac{dz}{\frac{\sqrt{b+1}+a}{\sqrt{b+1}-a}-z^2} $$
$$ =\dfrac{2a}{\sqrt{b+1}-a}\cdot\dfrac{1}{2}\sqrt{\frac{\sqrt{b+1}-a}{\sqrt{b+1}+a}} \int_0^{\tan{\dfrac{\pi}{8}}} \dfrac{1}{z+\sqrt{\frac{\sqrt{b+1}+a}{\sqrt{b+1}-a}}}-\dfrac{1}{z-\sqrt{\frac{\sqrt{b+1}+a}{\sqrt{b+1}-a}}} dz $$
$$ =\dfrac{a}{\sqrt{b+1-a^2}}\bigg[\log\abs{z+\sqrt{\frac{\sqrt{b+1}+a}{\sqrt{b+1}-a}}}-\log\abs{z-\sqrt{\frac{\sqrt{b+1}+a}{\sqrt{b+1}-a}}}\bigg]_0^{\tan{\frac{\pi}{8}}} $$

$$ =\dfrac{a}{\sqrt{b+1-a^2}}\ \log\abs{\dfrac{(\sqrt2-1)\sqrt{b+1-a^2}+a+\sqrt{b+1}}{(\sqrt2-1)\sqrt{b+1-a^2}-a-\sqrt{b+1}}} $$
$$ \therefore I_1=\dfrac{\pi}{4}-\dfrac{a}{\sqrt{b+1-a^2}}\ \log\abs{\dfrac{(\sqrt2-1)\sqrt{b+1-a^2}+a+\sqrt{b+1}}{(\sqrt2-1)\sqrt{b+1-a^2}-a-\sqrt{b+1}}} $$

$(ii) \ I_2$を計算する。

$s=\sin{x}+\cos{x}$とおく。
$x:\dfrac{\pi}{2}\to\pi \Leftrightarrow s:1\to−1,$
$ds=\cos{x}-\sin{x} \ dx,\ \sin{2x}=2\sin{x}\cos{x}=s^2−1$より

$$ 2I_2=\int_{\dfrac{\pi}{2}}^{\pi} \dfrac{\sin{x}−\cos{x}}{1+\sqrt{1+\sin{2x}}} \ dx $$
$$ =\int_1^{-1}　\dfrac{\sin{x}-\cos{x}}{a+\sqrt{b+(s^2-1)}} \cdot \dfrac{1}{\cos{x}-\sin{x}} \ ds $$
$$ =\int_{-1}^1 \dfrac{ds}{a+\sqrt{b-1+s^2}} $$
偶関数だから
$$ I_2=\int_0^1 \dfrac{ds}{a+\sqrt{b-1+s^2}} $$

ここで、$I_2$の積分は$I_1$の積分の$b$を$b-2$にしたものである。$a,\ b$で表される数の大小に気を付けると
$(ii1)\ a=\sqrt{b-1}$のとき
$$ \therefore I_2 =\dfrac{\pi}{4}-\dfrac{2a(\sqrt{2}-1)}{a+\sqrt{b-1}} $$

$(ii2)\ 0<\dfrac{a-\sqrt{b-1}} {a+\sqrt{b-1}}$のとき
範囲は$a<-\sqrt{b+1},\ \sqrt{b-1}< a$

$$ I_2=\dfrac{\pi}{4}-\dfrac{2a}{\sqrt{a^2-b+1}}\ \arctan(\dfrac{(a-\sqrt{b-1})(\sqrt{2}-1)}{\sqrt{a^2-b+1}}) $$

$(ii3)\ \dfrac{a-\sqrt{b-1}}{a+\sqrt{b-1}}<0$のとき $-\sqrt{b-1}< a<\sqrt{b-1}$
$$ I_2=\dfrac{\pi}{4}-\dfrac{a}{\sqrt{b-1-a^2}}\ \log\abs{\dfrac{(\sqrt2-1)\sqrt{b-1-a^2}+a+\sqrt{b-1}}{(\sqrt2-1)\sqrt{b-1-a^2}-a-\sqrt{b-1}}} $$

以上より