Weird Integral

Let us define the integral $\displaystyle I = \int_0^1 {\left( 1 - {\left( 1 - x^3 \right)}^p \right)}^q x^2 \, dx$

Now, we try to solve the integral in terms of $p$ and $q$ .

First substitute in $x^3 = t \implies x^2 \, dx = \dfrac{dt}{3}$ with limits $\displaystyle \int_0^1 \rightarrow \int_0^1$ to get

$\displaystyle I = \dfrac{1}{3} \int_0^1 {\left( 1 - {\left( 1 - t \right)}^p \right)}^q \, dt$

Now substituting $(1-t) = u \implies \,dt = -\,du$ with limits $\displaystyle \int_0^1 \rightarrow \int_1^0$ to get

$\begin{aligned} \displaystyle I &= - \dfrac{1}{3} \int_1^0 {\left( 1 - u^p \right)}^q \, du \\ &= \dfrac{1}{3} \int_0^1 {\left( 1 - u^p \right)}^q \, du \end{aligned}$

Let $u^p = y \implies p u^{p-1} \,du = \,dy \implies \,du = \dfrac{dy}{p u^{p-1}} \implies \,du = \dfrac{y^{\frac{1}{p} -1} \,dy}{p}$ with limits $\displaystyle \int_0^1 \rightarrow \int_0^1$ to get

$I = \dfrac{1}{3p} \int_0^1 y^{\frac{1}{p} -1} {\left( 1 - y \right)}^q \,dy$

This corresponds to the Beta function given by

$B(m,n) = \int_0^1 y^{m-1} (1-y)^{n-1} \,dy$

and hence

$I = \dfrac{1}{3p} B \left( \dfrac{1}{p} , q+1 \right)$

On comparison with the integrals given in the problem, we have

$I_1 = \dfrac{1}{3\sqrt{2}} B \left( \dfrac{1}{\sqrt{2}} , \sqrt{3}+1 \right) \\ I_2 = \dfrac{1}{3\sqrt{2}} B \left( \dfrac{1}{\sqrt{2}} , \sqrt{3}+2 \right)$

We shall use the following identities to get the required result

$(1) \ B(m,n) = \dfrac{\Gamma(m) \Gamma(n)}{\Gamma(m+n)} \\ (2) \ \Gamma(s+1) = s \Gamma(s)$

where $\Gamma(\cdot)$ denotes the Gamma function.

Therefore, we have

$\begin{aligned} \dfrac{I_1}{I_2} &= \dfrac{B \left( \frac{1}{\sqrt{2}} , \sqrt{3}+1 \right)}{B \left( \frac{1}{\sqrt{2}} , \sqrt{3}+2 \right)} \\ &= \dfrac{ \frac{\Gamma({1}/{\sqrt{2}}) \Gamma(\sqrt{3}+1)}{\Gamma(({1}/{\sqrt{2}}) + \sqrt{3} +1)}} { \frac{\Gamma({1}/{\sqrt{2}}) \Gamma(\sqrt{3}+2)}{\Gamma(({1}/{\sqrt{2}}) + \sqrt{3} +2)}} \\ \\ &= \dfrac{ \Gamma(\sqrt{3}+1) }{\Gamma(({1}/{\sqrt{2}}) + \sqrt{3} +1)} \cdot \dfrac{\Gamma(({1}/{\sqrt{2}}) + \sqrt{3} +2)}{ \Gamma(\sqrt{3}+2)} \\ &= \dfrac{ \Gamma(\sqrt{3}+1) }{\Gamma(({1}/{\sqrt{2}}) + \sqrt{3} +1)} \cdot \dfrac{(({1}/{\sqrt{2}}) + \sqrt{3} +1) \ \Gamma(({1}/{\sqrt{2}}) + \sqrt{3} +1)}{(\sqrt{3}+1) \ \Gamma(\sqrt{3}+1)} \\ &= \dfrac{\left( \frac{1}{\sqrt{2}} + \sqrt{3} +1 \right)}{ \sqrt{3} +1} \\ &= 1 + \dfrac{1}{\sqrt{2} \left( \sqrt{3} +1 \right)} \\ &= 1 + \dfrac{\sqrt{3} -1}{2 \sqrt{2}} \end{aligned}$

The answer evaluates to be

$\dfrac{1}{10} \left( 1 + \dfrac{\sqrt{3} -1}{2 \sqrt{2}} - \dfrac{\sqrt{3} -1}{2 \sqrt{2}} + \dfrac{1}{5} \right) = \dfrac{1}{10} \cdot \dfrac{6}{5} = \dfrac{3}{25} = \boxed{0.12}$