Ratios of some definte integrals without using $\beta$-function

Question

Let $$I_1=\int_{0}^{1} (1-x^3)^{1/5} dx,~ I_2=\int_{0}^{1} (1-x^5)^{1/3} dx,~ I_3=\int_{0}^{1}x^3 (1-x^3)^{1/5} dx, ~I_4=\int_{0}^{1} x^5 (1-x^5)^{1/3} dx $$ These integral can be evaluated in terms of gamma functions using the $beta$-intrgral: $$\int_{0}^{\pi/2} \sin^{2x-1} \theta ~ \cos^{2y-1} d\theta= \frac{\Gamma(x) \Gamma(y)}{2\Gamma(x+y)}~~~~(1).$$ Here the question is: How to evaluate the ratios: $$\frac{I_1}{I_2},~~ \frac{I_2}{I_3},~~\frac{I_3}{I_4}$$ without using (1)?

Answer 1

Integrate by parts

$$I_1-I_3=\int_0^1(1-x^3)^{6/5}dx = \frac{18}5 \int_0^1x^3(1-x^3)^{1/5}dx=\frac{18}5I_3$$

which yields $\frac{I_1}{I_3}=\frac{23}5$. Similarly, $\frac{I_2}{I_4}=\frac{23}3$. Then, let $y^5+x^3=1$,

$$\begin{align} I&=I_1-I_2 =\int_0^1 [(1-x^3)^{1/5} - (1-x^5)^{1/3} ]dx =\int_0^1 y (x)dx-\int_0^1 x (y)dy\\ &= \left(y(x)x|_0^1 - \int_1^0 x(y)dy\right) -\left (x(y)y|_0^1 - \int_1^0 y(x)dx\right) =-I \end{align}$$

Thus, $I_1-I_2=0$ and

$$\frac{I_1}{I_2}=1, \>\>\>\>\> \frac{I_2}{I_3}=\frac{23}5, \>\>\>\>\> \frac{I_3}{I_4}=\frac53$$