Taking an integral of an integrand consisting of different power radical functions

So I have this integral, $$\int_0^1 (1-x^7)^{1/3}-(1-x^3)^{1/7} dx$$

and I don't know where to start with this. I tried doing some Algebra, but I'm not recognizing any patterns. (Maybe it is my tired brain, but I'm completely lost.)

If any of you can point me in the right direction, give me some useful hints, or explain how to solve it, I would be forever grateful.

Thank you!

Solution 1:

Note that enforcing the substitution $x\to x^{1/7}$ yields

$$\begin{align} \int_0^1 (1-x^7)^{1/3}\,dx&=\int_0^1(1-x)^{1/3}\frac{1}{7}x^{-6/7}\,dx\\\\ &=\int_0^1 x^{1/3}\frac17 (1-x)^{-6/7}\,dx \tag 1\\\\ \end{align}$$

Integrating by parts the integral on the right-hand side of $(1)$ with $u=x^{1/3}$ and $v=-(1-x)^{1/7}$ reveals

$$\begin{align} \int_0^1 (1-x^7)^{1/3}\,dx&=\int_0^1 (1-x)^{1/7}\frac13x^{-2/3}\,dx\tag2\\\\ &=\int_0^1(1-x^3)^{1/7}\,dx\tag 3 \end{align}$$

where we enforced the substitution $x\to x^3$ to go from $(2)$ to $(3)$.

And we are done!

NOTE: There is nothing special about the numbers $3$ and $7$ in the previous development. Let $I(r,s)=\int_0^1 (1-x^r)^s\,dx$, with $r>0$ and $s>-1$.

Then, enforcing the substitution $x\to x^{1/r}$ yields

$$\begin{align} I(r,s)&=\int_0^1 (1-x)^s \frac1r x^{1/r-1}\,dx\\\\ &=\int_0^1x^s\frac1r (1-x)^{1/r-1}\,dx \end{align}$$

Integrating by parts with $u=x^s$ and $v=-(1-x)^{1/r}$ reveals

$$I(r,s)=\int_0^1 sx^{s-1}(1-x)^{1/r}$$

Finally, enforcing the substitution $x\to x^{1/s}$, we obtain

$$I(r,s)=\int_0^1(1-x^{1/s})^{1/r}\,dx=I(1/s,1/r)$$

as was to be shown.

Solution 2:

HINT:

If $(1-x^7)^{1/3}=y,y^3+x^7=1$ and if $x=1,y=0;$ if $x=1,y=0$

and if $(1-x^3)^{1/7}=y,y^7+x^3=1$ and if $x=1,y=0;$ if $x=1,y=0$

So, both represent the same area of the curve between $[0,1]$

Solution 3:

Note that $$ \begin{align} \int_0^1\left(1-x^7\right)^{1/3}\,\mathrm{d}x &=\color{#C00}{\int_0^1(1-x)^{1/3}\,\mathrm{d}x^{1/7}}\\ &=\frac17\int_0^1(1-x)^{1/3}x^{-6/7}\,\mathrm{d}x\\ &=\frac17\frac{\Gamma(4/3)\,\Gamma(1/7)}{\Gamma(31/21)}\\ &=\frac{\Gamma(4/3)\,\Gamma(8/7)}{\Gamma(31/21)} \end{align} $$ and $$ \begin{align} \int_0^1\left(1-x^3\right)^{1/7}\,\mathrm{d}x &=\color{#C00}{\int_0^1(1-x)^{1/7}\,\mathrm{d}x^{1/3}}\\ &=\frac13\int_0^1(1-x)^{1/7}x^{-2/3}\,\mathrm{d}x\\ &=\frac13\frac{\Gamma(8/7)\,\Gamma(1/3)}{\Gamma(31/21)}\\ &=\frac{\Gamma(8/7)\,\Gamma(4/3)}{\Gamma(31/21)} \end{align} $$ Therefore, $$ \int_0^1\left[\left(1-x^7\right)^{1/3}-\left(1-x^3\right)^{1/7}\right]\,\mathrm{d}x=0 $$ This result can be arrived at without the Beta integrals by noting that the red integrals are equal by the change of variables $x\mapsto1-x$ and integration by parts.

Generalization

As Claude Leibovici points out in a comment, we can generalize the preceding as $$ \begin{align} \int_0^1\left(1-x^m\right)^{\frac1n}\,\mathrm{d}x &=\int_0^1\left(1-x\right)^{\frac1n}\,\mathrm{d}x^{\frac1m}\\ &=\frac1m\int_0^1\left(1-x\right)^{\frac1n}x^{\frac1m-1}\,\mathrm{d}x\\ &=\frac1m\frac{\Gamma\left(1+\frac1n\right)\Gamma\left(\frac1m\right)}{\Gamma\left(1+\frac1n+\frac1m\right)}\\ &=\frac{\Gamma\left(1+\frac1n\right)\Gamma\left(1+\frac1m\right)}{\Gamma\left(1+\frac1n+\frac1m\right)} \end{align} $$ and use the symmetry in the preceding formula to show that $$ \int_0^1\left(1-x^m\right)^{\frac1n}\,\mathrm{d}x =\int_0^1\left(1-x^n\right)^{\frac1m}\,\mathrm{d}x $$