prove $ \int_0^1 (1-x^p)^\frac{1}{q} \,dx = \int_0^1 (1-x^q)^\frac{1}{p} \,dx$

Question

prove that for every $ p,q \gt 0$

$$ \int_0^1 (1-x^p)^\frac{1}{q} \,dx = \int_0^1 (1-x^q)^\frac{1}{p} \,dx$$

I tried to start from one side and change variables to get something similiar to the right side, but it got my no where.

Answer 1

We have $\int_0^1\,\left(1-x^p\right)^{\frac1q}\,\text{d}x=\int_0^1\,(1-t)^{\frac1q}\,\frac{1}{p}t^{\frac{1}{p}-1}\,\text{d}t$, where $t:=x^{\frac1p}$. Now, with $u:=1-t$, we get $\int_0^1\,\left(1-x^p\right)^{\frac1q}\,\text{d}x=\frac1p\,\int_0^1\,u^{\frac1q}(1-u)^{\frac1p-1}\,\text{d}u$. Using the technique of Integration by Parts, we obtain $$\int_0^1\,\left(1-x^p\right)^{\frac1q}\,\text{d}x=\left(\left.-u^{\frac1q}(1-u)^{\frac1p}\right|_{u=0}^{u=1}\right)+\int_0^1\,(1-u)^{\frac1p}\,\frac{1}{q}\,u^{\frac{1}{q}-1}\,\text{d}u\,.$$ Finally, with $y:=u^{\frac1q}$, we have $$\int_0^1\,\left(1-x^p\right)^{\frac1q}\,\text{d}x=(0-0)+\int_0^1\,(1-y^q)^{\frac1p}\,\text{d}y=\int_0^1\,(1-y^q)^{\frac1p}\,\text{d}y\,.$$

Answer 2

Hint: Notice that both integrals measure the area described by the same geometric shape, namely $x^p+y^q=1$.