(sin^{μ}2x-csc^{μ}2x)tan{(π/4)+x}[0,π/4]などの定積分

\begin{alignat}{2}
&(1) \displaystyle\int_0^{\frac{π}{4}}(\sin^μ 2x-\csc^μ 2x)\tan \left(\frac{π}{4}+x\right)dx=-\frac{1}{2μ}+\frac{π}{2}\cot μπ\\
&(2) \displaystyle\int_0^{\frac{π}{4}}(\cos^μ 2x-\sec^μ 2x)\cot xdx=-\frac{1}{2μ}+\frac{π}{2}\cot μπ\\
&(3) \displaystyle\int_0^{\frac{π}{4}}(\sin^{μ-1} 2x+\csc^μ 2x)\cot \left(\frac{π}{4}+x\right)dx=\frac{π}{2}\csc μπ\\
&(4) \displaystyle\int_0^{\frac{π}{4}}(\cos^{μ-1} 2x+\sec^μ 2x)\tan xdx=\frac{π}{2}\csc μπ\\
&(5) \displaystyle\int_0^{\frac{π}{4}}(\sin^{μ-1} 2x-\csc^μ 2x)\tan \left(\frac{π}{4}+x\right)dx=\frac{π}{2}\cot μπ\\
&(6) \displaystyle\int_0^{\frac{π}{4}}(\cos^{μ-1} 2x-\sec^μ 2x)\cot xdx=\frac{π}{2}\cot μπ\\
\end{alignat}ただし、全て \(0 \lt μ \lt 1\)







<証明>

\((1)(3)(5)\) は \(\sin 2x=t\) と置きます。\((2 \cos 2xdx=dt)\)

\((2)(4)(6)\) は \(\displaystyle x=\frac{π}{4}-t\) と置きます。\((dx=-dt)\)


\begin{alignat}{2}
&(1) \displaystyle\int_0^{\frac{π}{4}}(\sin^μ 2x-\csc^μ 2x)\tan \left(\frac{π}{4}+x\right)dx\\
&=\displaystyle\int_0^1 (t^μ-t^{-μ})\sqrt{\frac{1+t}{1-t}}\cdot \frac{1}{2\sqrt{1-t^2}}dt\\
&=\frac{1}{2}\displaystyle\int_0^1 \frac{t^μ-t^{-μ}}{1-t}dt=\frac{1}{2}\{ψ(1-μ)-ψ(1+μ)\}\\
&=\frac{1}{2}\left\{ψ(1-μ)-ψ(μ)-\frac{1}{μ}\right\}=-\frac{1}{2μ}+\frac{π}{2}\cot μπ\\
\end{alignat}以上より$$\displaystyle\int_0^{\frac{π}{4}}(\sin^μ 2x-\csc^μ 2x)\tan \left(\frac{π}{4}+x\right)dx=-\frac{1}{2μ}+\frac{π}{2}\cot μπ$$







\begin{alignat}{2}
&(2) \displaystyle\int_0^{\frac{π}{4}}(\cos^μ 2x-\sec^μ 2x)\cot xdx\\
&=\displaystyle\int_{\frac{π}{4}}^0 (\sin^μ 2t-\csc^μ 2t)\cot \left(\frac{π}{4}-t\right)(-dt)\\
&=\displaystyle\int_0^{\frac{π}{4}} (\sin^μ 2t-\csc^μ 2t)\tan \left(\frac{π}{4}+t\right)dt=-\frac{1}{2μ}+\frac{π}{2}\cot μπ\\
\end{alignat}以上より$$\displaystyle\int_0^{\frac{π}{4}}(\cos^μ 2x-\sec^μ 2x)\cot xdx=-\frac{1}{2μ}+\frac{π}{2}\cot μπ$$







\begin{alignat}{2}
&(3) \displaystyle\int_0^{\frac{π}{4}}(\sin^{μ-1} 2x+\csc^μ 2x)\cot \left(\frac{π}{4}+x\right)dx\\
&=\displaystyle\int_0^1 (t^{μ-1}+t^{-μ})\sqrt{\frac{1-t}{1+t}}\cdot \frac{1}{2\sqrt{1-t^2}}dt\\
&=\frac{1}{2}\displaystyle\int_0^1 \frac{t^{μ-1}+t^{-μ}}{1+t}dt=\frac{1}{2}B(μ,1-μ)=\frac{π}{2}\csc μπ\\
\end{alignat}以上より$$\displaystyle\int_0^{\frac{π}{4}}(\sin^{μ-1} 2x+\csc^μ 2x)\cot \left(\frac{π}{4}+x\right)dx=\frac{π}{2}\csc μπ$$







\begin{alignat}{2}
&(4) \displaystyle\int_0^{\frac{π}{4}}(\cos^{μ-1} 2x+\sec^μ 2x)\tan xdx\\
&=\displaystyle\int_{\frac{π}{4}}^0 (\sin^{μ-1} 2t+\csc^μ 2t)\tan \left(\frac{π}{4}-t\right)(-dt)\\
&=\displaystyle\int_0^{\frac{π}{4}} (\sin^{μ-1} 2t+\csc^μ 2t)\cot \left(\frac{π}{4}+t\right)dt=\frac{π}{2}\csc μπ\\
\end{alignat}以上より$$\displaystyle\int_0^{\frac{π}{4}}(\cos^{μ-1} 2x+\sec^μ 2x)\tan xdx=\frac{π}{2}\csc μπ$$








\begin{alignat}{2}
&(5) \displaystyle\int_0^{\frac{π}{4}}(\sin^{μ-1} 2x-\csc^μ 2x)\tan \left(\frac{π}{4}+x\right)dx\\
&=\displaystyle\int_0^1 (t^{μ-1}-t^{-μ})\sqrt{\frac{1+t}{1-t}}\cdot \frac{1}{2\sqrt{1-t^2}}dt\\
&=\frac{1}{2}\displaystyle\int_0^1 \frac{t^{μ-1}-t^{-μ}}{1-t}dt=\frac{1}{2}\{ψ(1-μ)-ψ(μ)\}=\frac{π}{2}\cot μπ\\
\end{alignat}以上より$$\displaystyle\int_0^{\frac{π}{4}}(\sin^{μ-1} 2x-\csc^μ 2x)\tan \left(\frac{π}{4}+x\right)dx=\frac{π}{2}\cot μπ$$








\begin{alignat}{2}
&(6) \displaystyle\int_0^{\frac{π}{4}}(\cos^{μ-1} 2x-\sec^μ 2x)\cot xdx\\
&=\displaystyle\int_{\frac{π}{4}}^0 (\sin^{μ-1} 2t-\csc^μ 2t)\cot \left(\frac{π}{4}-t\right)(-dt)\\
&=\displaystyle\int_0^{\frac{π}{4}} (\sin^{μ-1} 2t-\csc^μ 2t)\tan \left(\frac{π}{4}+t\right)dt=\frac{π}{2}\cot μπ\\
\end{alignat}以上より$$\displaystyle\int_0^{\frac{π}{4}}(\cos^{μ-1} 2x-\sec^μ 2x)\cot xdx=\frac{π}{2}\cot μπ$$

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