A calculus problem by Refaat M. Sayed

The substitution $u = x^3$ makes this integral $I \; = \; \int_0^1 \frac{dx}{\sqrt{1-x^3}} \; = \; \tfrac13 \int_0^1 u^{-\frac23}(1-u)^{-\frac12}\,du \; = \; \tfrac13B(\tfrac13,\tfrac12) \; = \; \frac{\Gamma(\tfrac13)\Gamma(\tfrac12)}{3\Gamma(\tfrac56)} \; = \; \frac{\sqrt{\pi}\Gamma(\tfrac13)}{3\Gamma(\tfrac56)}$ But $\begin{array}{rcl} \Gamma(\tfrac23) & = & \displaystyle \frac{1}{\sqrt{2\pi}} 2^{\frac16} \Gamma(\tfrac13)\Gamma(\tfrac56) \\ \pi \mathrm{cosec}\tfrac13\pi \; = \; \Gamma(\tfrac13)\Gamma(\tfrac23) & = & \displaystyle \frac{1}{\sqrt{2\pi}} 2^{\frac16} \Gamma(\tfrac13)^2 \Gamma(\tfrac56) \\ \displaystyle \frac{1}{\Gamma(\tfrac56)} & = & \displaystyle \frac{\Gamma(\tfrac13)^2 2^{\frac16} 3^{\frac12}}{(2\pi)^{\frac32}} \\ I & = & \displaystyle \frac{\sqrt{\pi}}{3} \times \frac{\Gamma(\tfrac13)^3 2^{\frac16} 3^{\frac12}}{(2\pi)^{\frac32}} \; = \; \frac{1}{2\pi \sqrt{3} \sqrt[3]{2}} \Gamma(\tfrac13)^3 \end{array}$ making the answer $2+3+3+2+3+3 = \boxed{16}$ .

Put $x^3=sin^2t \\ 3x^2. dx=2 .sin t. cos t. dt$

and $x^2=sin^{\frac{4}{3}}t$

Substituting this in the integral it becomes:

$\displaystyle I=\int_{0}^{\frac{\pi}{2}} \dfrac{2\,sint\,cost}{3\,sin^{4/3}t\,cost} dt \\ =\displaystyle \frac{2}{3} \int_{0}^{\frac{\pi}{2}} sin^{-\frac{1}{3}}t dt \\ =\displaystyle \frac{1}{3} \times 2 \int_{0}^{\frac{\pi}{2}} sin^{-\frac{1}{3}}t \, cos^{0} t dt \\ =\displaystyle \frac{1}{3}B(\frac{1}{3},\frac{1}{2}) \\ =\displaystyle \frac{\Gamma(\frac{1}{3})\,\Gamma(\frac{1}{2})}{3\, \Gamma(\frac{5}{6})}.$

We know that :

(1) $\Gamma(z)\, \Gamma(1-z)=\dfrac{\pi}{sin(\pi z)}$

(2) $\Gamma(z)\, \Gamma(z+1/2)= 2^{1-2z}\, \sqrt{\pi} \, \Gamma(2z)$

Putting $z=1/3$ in the first equation : $\Gamma(2/3)\, \Gamma(1/3)=\pi \, cosec(\pi/3)=\frac{2\pi}{\sqrt{3}})$

Again putting $z=1/3$ in the second equation:

$\Gamma(1/3) \, \Gamma(5/6)=2^{1/3}\sqrt{\pi} \Gamma(2/3) \\ \Gamma(1/3) \, \Gamma(5/6)=2^{1/3}\sqrt{\pi} \times \dfrac{2\pi}{\sqrt{3} \, \Gamma(1/3)} \\ \Gamma(5/6)=\dfrac{2^{1/3}\,\sqrt{\pi}\,2\pi}{\sqrt{3}\, \Gamma^{2}(1/3)}$

Putting this value in the first equation:

$I=\dfrac{1}{2\,\pi\,\sqrt{3}\,\sqrt[3]{2}}. \Gamma^{3}(\frac{1}{3})$