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@sounansya_
さんがツイートした
こちらの問題
の解説です.
以下の式を求めてください.
$\displaystyle\int_0^\frac{\pi}{2}\sin^{\frac{1}{3}}\theta\cos^{\frac{5}{3}}\theta\ln\tan\theta\mathrm{d}\theta$
解答
解説
\begin{align*}
&\int_0^\frac{\pi}{2}\sin^{\frac{1}{3}}\theta\cos^{\frac{5}{3}}\theta\ln\tan\theta\mathrm{d}\theta\\
=&\frac{1}{2}\int_0^\frac{\pi}{2}\left(2\ln\sin\theta\right)\sin^{2\cdot\frac{2}{3}-1}\theta\cos^{2\cdot\frac{4}{3}-1}\theta\mathrm{d}\theta-\frac{1}{2}\int_0^\frac{\pi}{2}\sin^{2\cdot\frac{2}{3}-1}\theta\left(2\ln\cos\theta\right)\cos^{2\cdot\frac{4}{3}-1}\theta\mathrm{d}\theta\\
=&\frac{1}{4}\lim_{x\rightarrow\frac{2}{3}}2\int_0^\frac{\pi}{2}\left(\frac{\mathrm{d}}{\mathrm{d}x}\sin^{2x-1}\theta\right)\cos^{2\cdot\frac{4}{3}-1}\theta\mathrm{d}\theta-\frac{1}{4}\lim_{y\rightarrow\frac{4}{3}}2\int_0^\frac{\pi}{2}\sin^{2\cdot\frac{2}{3}-1}\theta\left(\frac{\mathrm{d}}{\mathrm{d}y}\cos^{2y-1}\theta\right)\mathrm{d}\theta\\
=&\frac{1}{4}\lim_{x\rightarrow\frac{2}{3}}\frac{\mathrm{d}}{\mathrm{d}x}B\left(x,\frac{4}{3}\right)-\frac{1}{4}\lim_{y\rightarrow\frac{4}{3}}\frac{\mathrm{d}}{\mathrm{d}y}B\left(\frac{2}{3},y\right)\\
=&\frac{1}{4}\lim_{x\rightarrow\frac{2}{3}}\frac{\mathrm{d}}{\mathrm{d}x}\frac{\Gamma\left(x\right)\Gamma\left(\frac{4}{3}\right)}{\Gamma\left(x+\frac{4}{3}\right)}-\frac{1}{4}\lim_{y\rightarrow\frac{4}{3}}\frac{\mathrm{d}}{\mathrm{d}y}\frac{\Gamma\left(\frac{2}{3}\right)\Gamma\left(y\right)}{\Gamma\left(\frac{2}{3}+y\right)}\\
=&\frac{1}{4}\left(\psi\left(\frac{2}{3}\right)-\psi\left(\frac{2}{3}+\frac{4}{3}\right)\right)\frac{\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{4}{3}\right)}{\Gamma\left(\frac{2}{3}+\frac{4}{3}\right)}-\frac{1}{4}\left(\psi\left(\frac{4}{3}\right)-\psi\left(\frac{2}{3}+\frac{4}{3}\right)\right)\frac{\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{4}{3}\right)}{\Gamma\left(\frac{2}{3}+\frac{4}{3}\right)}\\
=&\frac{1}{4}\left(\psi\left(\frac{2}{3}\right)-\psi\left(\frac{4}{3}\right)\right)\frac{\Gamma\left(\frac{2}{3}\right)\Gamma\left(\frac{4}{3}\right)}{\Gamma\left(\frac{2}{3}+\frac{4}{3}\right)}\\
=&\frac{1}{4}\left(\psi\left(1-\frac{1}{3}\right)-\psi\left(\frac{1}{3}\right)-\frac{1}{\frac{1}{3}}\right)\frac{\Gamma\left(1-\frac{1}{3}\right)\Gamma\left(\frac{1}{3}\right)\frac{1}{3}}{\Gamma\left(2\right)}\\
=&\frac{1}{12}\left(\pi\cot\frac{\pi}{3}-3\right)\frac{\pi}{\sin\frac{\pi}{3}}\\
=&\frac{1}{12}\left(\frac{\pi}{\sqrt{3}}-3\right)\frac{\pi}{\frac{\sqrt{3}}{2}}\\
=&\frac{\pi^2}{18}-\frac{\pi}{2\sqrt{3}}
\end{align*}
なので,$\displaystyle\int_0^\frac{\pi}{2}\sin^{\frac{1}{3}}\theta\cos^{\frac{5}{3}}\theta\ln\tan\theta\mathrm{d}\theta=\frac{\pi^2}{18}-\frac{\pi}{2\sqrt{3}}$です.