Integrate $\int_{0}^{\pi/2}\cos^{-\frac{2}{3}}(x) dx$

Question

Calculate $$\int_{0}^{\pi/2}\cos^{-\frac{2}{3}}(x) dx$$ Attempt $$\int_0^{\pi/2}\sin^m(x) \cos^n(x) dx =\frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})} $$ Can we apply this formula to $m=0 $ and $n=-2/3$ which is a negative rational number? Please provide any other method if the above does not hold.

Answer 1

The alternative is to use the binomial expansion theorem, the duplication formula of the $\Gamma$-function and known formulas for the hypergeometric series: $$ I\equiv \int \frac{1}{\cos^{2/3}(x)} dx = \int \frac{1}{(1-\sin^2 x)^{1/3}} dx = \sum_{l\ge 0} \binom{-1/3}{l}\int (-\sin^2 x)^l dx $$ $$ = \sum_{l\ge 0} (-)^l \frac{(-1/3)(-4/3)(-7/3)\cdots (-1/3-l+1)}{l!}\int \sin^{2l} x dx $$ $$ = \sum_{l\ge 0} \frac{(1/3)(4/3)(7/3)\cdots (1/3+l-1)}{l!}\int \sin^{2l} x dx $$ $$ = \sum_{l\ge 0} \frac{(1/3)_l}{l!}\int \sin^{2l} x dx $$ where $(a)_n=\Gamma(a+n)/\Gamma(a)$ is Pochhammer's symbol. In general $$ \int \sin^{2n}x dx = \frac{1}{2^{2n}}\binom{2n}{n}x+\frac{(-)^n}{2^{2n-1}}\sum_{k=0}^{n-1}(-)\binom{2n}{k}\frac{\sin(2n-2k)x}{2n-2k}, $$ so $$\int_0^{\pi/2} \sin^{2m}x dx =\frac{\pi}{2}\frac{(2m)!}{(2^m m!)^2} . $$ $$ I= \frac{\pi}{2}\sum_{l\ge 0} \frac{(1/3)_l}{l!}\frac{(2l)!}{2^{2l}(l!)^2} = \frac{\pi}{2}\sum_{l\ge 0} \frac{(1/3)_l}{l!}\frac{\Gamma(2l+1)}{2^{2l}\Gamma^2(l+1)} $$ $$ = \frac{\pi}{2\sqrt{2\pi}}\sum_{l\ge 0} \frac{(1/3)_l}{l!}\frac{\Gamma(l+1/2)\Gamma(l+1)2^{2l+1/2}}{2^{2l}\Gamma^2(l+1)} $$ $$ = \frac{\pi}{2\sqrt{\pi}}\sum_{l\ge 0} \frac{(1/3)_l}{l!}\frac{\Gamma(l+1/2)}{\Gamma(l+1)} $$ $$ = \frac{\surd \pi}{2}\Gamma(1/2)\sum_{l\ge 0} \frac{(1/3)_l}{l!}\frac{(1/2)_l}{(1)_l} $$ $$ = \frac{\pi}{2}{}_2F_1(1/3,1/2 ; 1; 1) =\frac{\pi^{3/2}}{\Gamma(5/6)\Gamma(2/3)}. $$ This is the same result as with the original formula.