tan^{2n}x[0,π/4]などの定積分

\begin{alignat}{2}
&(1) \displaystyle\int_0^{\frac{π}{2}}(\tan x)^μdx=\frac{π}{2} \sec \frac{πμ}{2}  (|μ| \lt 1)\\
&(2) \displaystyle\int_0^{\frac{π}{2}}(\tan x)^{-μ}dx=\frac{π}{2} \sec \frac{πμ}{2}  (|μ| \lt 1)\\
&(3) \displaystyle\int_0^{\frac{π}{4}}(\tan x)^μdx=\frac{1}{2}β\left(\frac{μ+1}{2}\right)  (μ \gt -1)\\
&(4) \displaystyle\int_0^{\frac{π}{4}} \tan^{2n} xdx=\frac{π(-1)^n}{4}+\displaystyle\sum_{k=0}^{n-1} \frac{(-1)^k}{2n-2k-1}\\
&(5) \displaystyle\int_0^{\frac{π}{4}} \tan^{2n+1} xdx=\frac{(-1)^n \log 2}{2}+\displaystyle\sum_{k=0}^{n-1} \frac{(-1)^k}{2n-2k}
\end{alignat}













<証明>

\((1)\) \(\tan\) を \(\sin,\cos\) に分けて、ベータ関数の公式を用います。
\begin{alignat}{2}
&\displaystyle\int_0^{\frac{π}{2}}(\tan x)^μdx=\displaystyle\int_0^{\frac{π}{2}}(\sin x)^μ (\cos x)^{-μ}dx=\frac{1}{2}B\left(\frac{1+μ}{2},\frac{1-μ}{2}\right)\\
&            =\frac{1}{2}Γ\left(\frac{μ}{2}+\frac{1}{2}\right)Γ\left(\frac{1}{2}-\frac{μ}{2}\right)\\
&            =\frac{1}{2}\cdot \frac{π}{\cos \frac{πμ}{2}}=\frac{π}{2} \sec \frac{πμ}{2}
\end{alignat}






\((2)\) \((1)\) と同様に解きます。
\begin{alignat}{2}
&\displaystyle\int_0^{\frac{π}{2}}(\tan x)^{-μ}dx=\displaystyle\int_0^{\frac{π}{2}}(\sin x)^{-μ} (\cos x)^μdx=\frac{1}{2}B\left(\frac{1-μ}{2},\frac{1+μ}{2}\right)\\
&              =\frac{1}{2}Γ\left(\frac{1}{2}-\frac{μ}{2}\right)Γ\left(\frac{μ}{2}+\frac{1}{2}\right)\\
&              =\frac{1}{2}\cdot \frac{π}{\cos \frac{πμ}{2}}=\frac{π}{2} \sec \frac{πμ}{2}
\end{alignat}







\((3)\) \(\tan x=t\) と置きます。\(\displaystyle \left(\frac{1}{\cos^2 x}dx=dt, dx=\frac{1}{1+t^2}dt\right)\)
\begin{alignat}{2}
&\displaystyle\int_0^{\frac{π}{4}}(\tan x)^μ dx=\displaystyle\int_0^1 \frac{t^μ}{1+t^2}dt=\displaystyle\int_0^1 t^μ(1-t^2+t^4-t^6+ \cdots)dt\\
&            =\displaystyle\int_0^1 t^μ \displaystyle\sum_{n=0}^{\infty} (-1)^n t^{2n}dt=\displaystyle\sum_{n=0}^{\infty} (-1)^n\displaystyle\int_0^1 t^{μ+2n}dt\\
&            =\displaystyle\sum_{n=0}^{\infty} (-1)^n\left[\frac{t^{2n+1+μ}}{2n+1+μ}\right]_0^1=\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1+μ}\\
&            =\frac{1}{2}\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{n+\frac{μ+1}{2}}=\frac{1}{2}β\left(\frac{μ+1}{2}\right)
\end{alignat}








\((4)\) \(\tan^2 x\) だけを取り出すことを繰り返し、次数を下げます。
\begin{alignat}{2}
&\displaystyle\int_0^{\frac{π}{4}} \tan^{2n}xdx=\displaystyle\int_0^{\frac{π}{4}} \tan^{2n-2}x \tan^2 xdx\\
&            =\displaystyle\int_0^{\frac{π}{4}} \tan^{2n-2}x \left(\frac{1}{\cos^2 x}-1\right)dx\\
&            =\frac{1}{2n-1}\displaystyle\int_0^{\frac{π}{4}} (\tan^{2n-1}x)’dx- \displaystyle\int_0^{\frac{π}{4}} \tan^{2n-2}x dx\\
&            =\frac{1}{2n-1}[\tan^{2n-1}x]_0^{\frac{π}{4}}- \displaystyle\int_0^{\frac{π}{4}} \tan^{2n-2}x dx\\
&            =\frac{1}{2n-1}- \displaystyle\int_0^{\frac{π}{4}} \tan^{2n-2}x dx\\
&            =\frac{1}{2n-1}-\frac{1}{2n-3} +\displaystyle\int_0^{\frac{π}{4}} \tan^{2n-4}x dx\\
&            =\frac{1}{2n-1}-\frac{1}{2n-3} +\frac{1}{2n-5}-\displaystyle\int_0^{\frac{π}{4}} \tan^{2n-6}x dx\\
&\\
&                            \cdots\\
&\\
&            =\frac{1}{2n-1}-\frac{1}{2n-3} +\frac{1}{2n-5}- \cdots +\frac{(-1)^{n-2}}{3}+(-1)^{n-1}\displaystyle\int_0^{\frac{π}{4}} \tan^2 x dx\\
\end{alignat}右の積分は$$\displaystyle\int_0^{\frac{π}{4}} \tan^2 x dx=\displaystyle\int_0^{\frac{π}{4}} \left(\frac{1}{\cos^2 x}-1\right) dx=[\tan x-x]_0^{\frac{π}{4}}=1-\frac{π}{4}$$となるので
\begin{alignat}{2}
&=\frac{1}{2n-1}-\frac{1}{2n-3} +\frac{1}{2n-5}- \cdots +\frac{(-1)^{n-2}}{3}+(-1)^{n-1}+(-1)^n \cdot \frac{π}{4}\\
&=\displaystyle\sum_{k=0}^{n-1}\frac{(-1)^k}{2n-2k-1}+\frac{π(-1)^n}{4}
\end{alignat}以上より$$\displaystyle\int_0^{\frac{π}{4}} \tan^{2n} xdx=\frac{π(-1)^n}{4}+\displaystyle\sum_{k=0}^{n-1} \frac{(-1)^k}{2n-2k-1}$$







\((5)\) \((4)\) と同様に解いていきます。
\begin{alignat}{2}
&\displaystyle\int_0^{\frac{π}{4}} \tan^{2n+1} xdx=\displaystyle\int_0^{\frac{π}{4}} \tan^{2n-1}x \tan^2 xdx\\
&             =\displaystyle\int_0^{\frac{π}{4}} \tan^{2n-1} x \left(\frac{1}{\cos^2 x}-1\right)dx\\
&             =\frac{1}{2n}\displaystyle\int_0^{\frac{π}{4}} (\tan^{2n}x)’dx- \displaystyle\int_0^{\frac{π}{4}} \tan^{2n-1} x dx\\
&             =\frac{1}{2n}[\tan^{2n}x]_0^{\frac{π}{4}}- \displaystyle\int_0^{\frac{π}{4}} \tan^{2n-1} x dx\\
&             =\frac{1}{2n}- \displaystyle\int_0^{\frac{π}{4}} \tan^{2n-1} x dx\\
&             =\frac{1}{2n}-\frac{1}{2n-2} +\displaystyle\int_0^{\frac{π}{4}} \tan^{2n-3} x dx\\
&             =\frac{1}{2n}-\frac{1}{2n-2} +\frac{1}{2n-4}-\displaystyle\int_0^{\frac{π}{4}} \tan^{2n-4} x dx\\
&\\
&                            \cdots\\
&\\
&             =\frac{1}{2n}-\frac{1}{2n-2} +\frac{1}{2n-4}- \cdots +\frac{(-1)^{n-1}}{2}+(-1)^n\displaystyle\int_0^{\frac{π}{4}} \tan x dx\\
\end{alignat}右の積分は$$\displaystyle\int_0^{\frac{π}{4}} \tan x dx=[-\log (\cos x)]_0^{\frac{π}{4}}=-\log \frac{1}{\sqrt{2}}=\frac{1}{2}\log 2$$となるので
\begin{alignat}{2}
&=\frac{1}{2n}-\frac{1}{2n-2} +\frac{1}{2n-4}- \cdots +\frac{(-1)^{n-1}}{2}+\frac{(-1)^n \log 2}{2}\\
&=\displaystyle\sum_{k=0}^{n-1} \frac{(-1)^k}{2n-2k}+\frac{(-1)^n \log 2}{2}\\
\end{alignat}以上より$$\displaystyle\int_0^{\frac{π}{4}} \tan^{2n+1} xdx=\frac{(-1)^n \log 2}{2}+\displaystyle\sum_{k=0}^{n-1} \frac{(-1)^k}{2n-2k}$$

“tan^{2n}x[0,π/4]などの定積分” への2件の返信

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