log(sinx)sin^{p-1}x/cos^{p+1}x[0,π/2]などの定積分

\begin{alignat}{2}
&(1)  \displaystyle\int_0^{\frac{π}{2}}\frac{\log (\sin x) \sin^{p-1} x}{\cos^{p+1} x}dx=-\frac{π}{2p}\csc \frac{pπ}{2}  (0 \lt p \lt 2)\\
&(2)  \displaystyle\int_0^{\frac{π}{2}}\frac{\log (\sin x)}{\tan^{p-1} x \sin 2x}dx=\frac{π}{4(p-1)}\sec \frac{pπ}{2}  (|p| \lt 1)\\
\end{alignat}







<証明>

次の定積分の結果を用います。(詳細はこちらです)
\begin{alignat}{2}
&(A) \displaystyle\int_0^{\frac{π}{2}}(\tan x)^μdx=\frac{π}{2} \sec \frac{πμ}{2}  (|μ| \lt 1)\\
&(B) \displaystyle\int_0^{\frac{π}{2}}(\tan x)^{-μ}dx=\frac{π}{2} \sec \frac{πμ}{2}  (|μ| \lt 1)\\
\end{alignat}







\begin{alignat}{2}
&(1)  \displaystyle\int_0^{\frac{π}{2}}\frac{\log (\sin x) \sin^{p-1} x}{\cos^{p+1} x}dx\\
&=\displaystyle\int_0^{\frac{π}{2}} \log (\sin x) \tan^{p-1} x \cdot \frac{1}{\cos^2 x}dx\\
&=\frac{1}{p}\displaystyle\int_0^{\frac{π}{2}}\log (\sin x)(\tan^p x)’dx\\
&=\frac{1}{p}\left\{\left[\log (\sin x)\tan^p x\right]_0^{\frac{π}{2}}-\displaystyle\int_0^{\frac{π}{2}} \frac{\cos x}{\sin x} \cdot \tan^p xdx\right\}\\
&=-\frac{1}{p}\displaystyle\int_0^{\frac{π}{2}}\tan^{p-1} xdx=-\frac{1}{p}\cdot \frac{π}{2}\sec \frac{π(p-1)}{2}=-\frac{π}{2p}\csc \frac{pπ}{2}\\
\end{alignat}以上より$$\displaystyle\int_0^{\frac{π}{2}}\frac{\log (\sin x) \sin^{p-1} x}{\cos^{p+1} x}dx=-\frac{π}{2p}\csc \frac{pπ}{2}$$







\begin{alignat}{2}
&(2)  \displaystyle\int_0^{\frac{π}{2}}\frac{\log (\sin x)}{\tan^{p-1} x \sin 2x}dx\\
&=\displaystyle\int_0^{\frac{π}{2}} \log (\sin x) \cdot \frac{\tan^{1-p} x}{2 \sin x \cos x}dx\\
&=\frac{1}{2}\displaystyle\int_0^{\frac{π}{2}} \log (\sin x) \cdot \frac{\tan^{1-p} x}{\tan x} \cdot \frac{1}{\cos^2 x}dx\\
&=\frac{1}{2}\displaystyle\int_0^{\frac{π}{2}} \log (\sin x) (\tan x)^{-p} \cdot \frac{1}{\cos^2 x}dx\\
&=\frac{1}{2(1-p)}\displaystyle\int_0^{\frac{π}{2}} \log (\sin x)\{(\tan x)^{1-p}\}’dx\\
&=\frac{1}{2(1-p)}\left\{[\log (\sin x)(\tan x)^{1-p}]_0^{\frac{π}{2}}-\displaystyle\int_0^{\frac{π}{2}} \frac{\cos x}{\sin x} \cdot (\tan x)^{1-p})\right\}dx\\
&=\frac{1}{2(p-1)}\displaystyle\int_0^{\frac{π}{2}}(\tan x)^{-p}dx=\frac{1}{2(p-1)} \cdot \frac{π}{2}\sec \frac{pπ}{2}=\frac{π}{4(p-1)}\sec \frac{pπ}{2}\\
\end{alignat}以上より$$\displaystyle\int_0^{\frac{π}{2}}\frac{\log (\sin x)}{\tan^{p-1} x \sin 2x}dx=\frac{π}{4(p-1)}\sec \frac{pπ}{2}$$

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