$$$$
ツイート
した自作問題(積分)の解説をします
$\displaystyle\int_0^\frac{\pi}{3}\frac{dx}{\sqrt[3]{\cos^2x}}= ?$
(解説)
\begin{align}
\int_0^\frac{\pi}{3}\frac{dx}{\sqrt[3]{\cos^2x}}
&=\int_0^\frac{\pi}{3}(1+\tan^2x)^\frac{1}{3}dx\\
&=\int_0^\sqrt{3}\frac{dx}{(1+x^2)^{\frac{2}{3}}}\\
&=\sqrt{3}\int_0^1\frac{dx}{(1+3x^2)^{\frac{2}{3}}}\\
&=\frac{\sqrt{3}}{2}\int_{-1}^1\frac{dx}{(1+3x^2)^\frac{2}{3}}\\
&=\frac{\sqrt{3}}{\sqrt[3]{2}}\int_{-1}^1\frac{dx}{((1-x)^3+(1+x)^3)^{\frac{2}{3}}}\\
&=\frac{\sqrt{3}}{2^\frac{4}{3}}\int_{-1}^1\left(1+\left(\frac{1-x}{1+x}\right)^3\right)^{-\frac{2}{3}}\frac{2dx}{(1+x)^2}\\
&=\frac{\sqrt{3}}{2^\frac{4}{3}}\int_0^\infty\frac{dx}{(1+x^3)^\frac{2}{3}}\\
&=\frac{1}{2^\frac{4}{3}\sqrt{3}}\int_0^\infty\frac{x^{\frac{1}{3}-1}}{(1+x)^\frac{2}{3}}dx\\
&=\frac{1}{2^\frac{4}{3}\sqrt{3}}\frac{\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{1}{3}\right)}{\Gamma\left(\frac{2}{3}\right)}
\end{align}
これを簡単な形にします。
次の式を使います。
(補題)
$$\frac{\Gamma\left(\frac{1}{3}\right)^2}{\Gamma\left(\frac{1}{6}\right)}=\frac{2^\frac{1}{3}\sqrt{\pi}}{\sqrt{3}}$$
(証明)
\begin{align}
\frac{\Gamma\left(\frac{1}{3}\right)^2}{\Gamma\left(\frac{1}{6}\right)}
&=\frac{\Gamma\left(\frac{1}{3}\right)^2\Gamma\left(\frac{2}{3}\right)}{\Gamma\left(\frac{1}{6}\right)\Gamma\left(\frac{2}{3}\right)}\\
&=\frac{\Gamma\left(\frac{1}{3}\right)\frac{2\pi}{\sqrt{3}}}{2^\frac{2}{3}\sqrt{\pi}\Gamma\left(\frac{1}{3}\right)}\\
&=\frac{2^\frac{1}{3}\sqrt{\pi}}{\sqrt{3}}
\end{align}
これより、
\begin{align}
\int_0^\frac{\pi}{3}\frac{dx}{\sqrt[3]{\cos^2x}}
&=\frac{1}{2^\frac{4}{3}\sqrt{3}}\frac{2^\frac{1}{3}\sqrt{\pi}}{\sqrt{3}}\frac{\Gamma\left(\frac{1}{6}\right)}{\Gamma\left(\frac{2}{3}\right)}\\
&=\frac{\sqrt{\pi}}{6}\frac{\Gamma\left(\frac{1}{6}\right)}{\Gamma\left(\frac{2}{3}\right)}\\
&=\frac{\sqrt{\pi}\Gamma\left(\frac{7}{6}\right)}{\Gamma\left(\frac{2}{3}\right)}
\end{align}