√tan の積分から作った分数式の積分

$\int_0^{\frac{\pi}{4}}\sqrt{\tan x}\,dx$ を$\sqrt{\tan x}=t$とおいて分数式に置換した積分．

$\int_0^{\frac{\pi}{4}}\sqrt{\tan x}\,dx=\int_0^{1}\,\frac{2t^2}{t^4+1}\,dt$

$t^4+1=t^4+2t^2+1-2t^2=(t^2+1)^2-(\sqrt{2}t)^2=(t^2-\sqrt{2}t+1)(t^2+\sqrt{2}t+1)$

より，$\frac{2t^2}{t^4+1}$ を部分分数に分解する．

恒等式，$\frac{2t^2}{t^4+1}=\frac{at+b}{t^2-\sqrt{2}t+1}+\frac{ct+d}{t^2+\sqrt{2}t+1}$ の分母を払って，

$2t^2=(at+b)(t^2+\sqrt{2}t+1)+(ct+d)(t^2-\sqrt{2}t+1)$

$=(a+c)t^3+(\sqrt{2}a+b-\sqrt{2}c+d)t^2+(a+\sqrt{2}b+c-\sqrt{2}d)t+b+d$

係数比較で，

$a+c=0$ より，$c=-a$

$b+d=0$ より，$d=-b$

$\sqrt{2}a+b-\sqrt{2}c+d=2$ より，$\sqrt{2}a+b-\sqrt{2}(-a)+(-b)=2$, $2\sqrt{2}a=2$, $a=\frac{1}{\sqrt{2}}$, $c=-\frac{1}{\sqrt{2}}$

$a+\sqrt{2}b+c-\sqrt{2}d=0$ より，$a+\sqrt{2}b+(-a)-\sqrt{2}(-b)=0$, $2\sqrt{2}b=0$, $b=0$, $d=0$

$\frac{2t^2}{t^4+1}=\frac{\frac{1}{\sqrt{2}}t}{t^2-\sqrt{2}t+1}+\frac{-\frac{1}{\sqrt{2}}t}{t^2+\sqrt{2}t+1}$

$\int_0^{\frac{\pi}{4}}\sqrt{\tan x}\,dx=\int_0^{1}\,\frac{2t^2}{t^4+1}\,dt=\int_0^{1} \frac{\frac{1}{\sqrt{2}}t}{t^2-\sqrt{2}t+1} \,dt+\int_0^{1}\frac{-\frac{1}{\sqrt{2}}t}{t^2+\sqrt{2}t+1} \,dt$

１つ目の分数式の積分

分母 $(t^2-\sqrt{2}t+1)'=2t-\sqrt{2}$ で，分子$\frac{1}{\sqrt{2}}t$を割ると

$\frac{1}{\sqrt{2}}t=\frac{1}{2\sqrt{2}}(2t-\sqrt{2})+\frac{1}{2}$

$\int_0^{1} \frac{\frac{1}{\sqrt{2}}t}{t^2-\sqrt{2}t+1} \,dt = \int_0^{1} \frac{\frac{1}{2\sqrt{2}}(2t-\sqrt{2})+\frac{1}{2}}{t^2-\sqrt{2}t+1} \,dt$

$=\frac{1}{2\sqrt{2}}\int_0^{1} \frac{(t^2-\sqrt{2}t+1)'}{t^2-\sqrt{2}t+1} \,dt +\frac{1}{2}\int_0^{1} \frac{1}{t^2-\sqrt{2}t+1} \,dt$

と分ける．

（１つ目前半）

$\frac{1}{2\sqrt{2}}\int_0^{1} \frac{(t^2-\sqrt{2}t+1)'}{t^2-\sqrt{2}t+1} \,dt =\frac{1}{2\sqrt{2}}[\log(t^2-\sqrt{2}t+1)]_0^{1}$

$=\frac{1}{2\sqrt{2}}\log(1^2-\sqrt{2}+1)-\frac{1}{2\sqrt{2}}\log(0^2-\sqrt{2}\times 0+1)=\frac{\log(2-\sqrt{2})}{2\sqrt{2}}$

（１つ目後半）

分母 $t^2-\sqrt{2}t+1=(t-\frac{\sqrt{2}}{2})^2-\frac{1}{2}+1=(t-\frac{1}{\sqrt{2}})^2+\frac{1}{2}$

より，$t-\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\tan\theta$ とすると，

$(t-\frac{1}{\sqrt{2}})^2+\frac{1}{2}=(\frac{1}{\sqrt{2}}\tan\theta)^2+\frac{1}{2}=\frac{1}{2}(\tan^2\theta+1)=\frac{1}{2}\times\frac{1}{\cos^2\theta}$

積分区間は $t=0$ のとき，$0-\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\tan\theta$, $\tan\theta=-1$, $\theta=-\frac{\pi}{4}$

$t=1$ のとき，$1-\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\tan\theta$, $\tan\theta=\sqrt{2}-1$, $\theta=\arctan(\sqrt{2}-1)$ 

$dt=\frac{1}{\sqrt{2}}\frac{1}{\cos^2\theta}\,d\theta$

 で置換すると，

$\frac{1}{2}\int_0^{1} \frac{1}{t^2-\sqrt{2}t+1} \,dt=\frac{1}{2}\int_{-\frac{\pi}{4}}^{\arctan(\sqrt{2}-1)} \frac{1}{\frac{1}{2}\times\frac{1}{\cos^2\theta}} \frac{1}{\sqrt{2}}\frac{1}{\cos^2\theta}\,d\theta$

$=\frac{1}{\sqrt{2}}\int_{-\frac{\pi}{4}}^{\arctan(\sqrt{2}-1)} 1\,d\theta=\frac{1}{\sqrt{2}}[\theta]_{-\frac{\pi}{4}}^{\arctan(\sqrt{2}-1)}=\frac{1}{\sqrt{2}}(\arctan(\sqrt{2}-1)-(-\frac{\pi}{4}))$

１つ目の分数式の積分は

$\int_0^{1} \frac{\frac{1}{\sqrt{2}}t}{t^2-\sqrt{2}t+1} \,dt=\frac{\log(2-\sqrt{2})}{2\sqrt{2}}+\frac{1}{\sqrt{2}}(\arctan(\sqrt{2}-1)+\frac{\pi}{4})$

2つ目の分数式の積分も同様にすると，

$\int_0^{1} \frac{-\frac{1}{\sqrt{2}}t}{t^2+\sqrt{2}t+1} \,dt=\frac{-\log(2+\sqrt{2})}{2\sqrt{2}}-\frac{1}{\sqrt{2}}(\frac{\pi}{4}-\arctan(\sqrt{2}+1))$

したがって，

$\int_0^{\frac{\pi}{4}}\sqrt{\tan x}\,dx=\int_0^{1}\,\frac{2t^2}{t^4+1}\,dt$

$=\frac{\log(2-\sqrt{2})}{2\sqrt{2}}+\frac{1}{\sqrt{2}}(\arctan(\sqrt{2}-1)+\frac{\pi}{4})+\frac{-\log(2+\sqrt{2})}{2\sqrt{2}}+\frac{1}{\sqrt{2}}(\frac{\pi}{4}-\arctan(\sqrt{2}+1))$

$=\frac{\log(2-\sqrt{2})}{2\sqrt{2}}+\frac{-\log(2+\sqrt{2})}{2\sqrt{2}}+\frac{1}{\sqrt{2}}(\arctan(\sqrt{2}-1)+\frac{\pi}{4}-\frac{\pi}{4}-\arctan(\sqrt{2}+1))$

$=\frac{1}{2\sqrt{2}}(\log(2-\sqrt{2})-\log(2+\sqrt{2})) + \frac{1}{\sqrt{2}}(\arctan(\sqrt{2}-1)+\arctan(\sqrt{2}+1))$

$=\frac{1}{2\sqrt{2}}(\log(2-\sqrt{2})-\log(2+\sqrt{2})) + \frac{1}{\sqrt{2}}(\frac{\pi}{2})$ ・・・①

$=\frac{1}{2\sqrt{2}}(\pi +\log(2-\sqrt{2})-\log(2+\sqrt{2}))$

$=\frac{1}{2\sqrt{2}}(\pi +\log\frac{2-\sqrt{2}}{2+\sqrt{2}})$

$=\frac{1}{2\sqrt{2}}(\pi +\log\frac{\sqrt{2}-1}{\sqrt{2}+1})$

>検算

$\frac{\sqrt{2}-1}{\sqrt{2}+1}=\frac{(\sqrt{2}-1)(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{3-2\sqrt{2}}{2-1}=3-2\sqrt{2}$

だから合ってる．

分数式の $\int_0^{1}\,\frac{2t^2}{t^4+1}\,dt$ を検算したら，＞wolfram

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