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Evaluate the integral: $$ \int_0^1 \left(\sqrt[3]{1-x^7} - \sqrt[7]{1-x^3}\right)\;dx$$

The answer is $0,$ but I am unable to get it. There is some symmetry I can not see.

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Generally speaking, $$\int_0^1 \sqrt[m]{1-x^n}dx = \int_0^1 \sqrt[n]{1-x^m}dx = \frac{\frac{1}{m} ! \cdot \frac{1}{n} !}{\left(\frac{1}{m} + \frac{1}{n}\right) !}$$ where $$\frac{1}{n} ! = \int_0^\infty e^{-x^n}dx$$ See here and here for more details. – Lucian Oct 14 '13 at 23:16
up vote 55 down vote accepted

Let $m, n > 0$. Then observe that $$ \int_{0}^{1} \sqrt[n]{1-x^m} \; dx$$ is the area of the region given by inequalities $$ 0 \leq x \leq 1 \quad \text{and} \quad 0 \leq y \leq \sqrt[n]{1-x^m}.$$ But the last inequality is equivalent to $0 \leq x^m + y^n \leq 1$. Thus $$ \int_{0}^{1} \sqrt[n]{1-x^m} \; dx = [\text{Area given by} \ 0 \leq x^m + y^n \leq 1, \ 0 \leq x, y \leq 1 ]$$ Thus by interchanging the role of $x$ and $y$, we have $$ \int_{0}^{1} \sqrt[n]{1-x^m} \; dx = \int_{0}^{1} \sqrt[m]{1-x^n} \; dx.$$

Of course, we can give a purely analytic approach. Let $y = \sqrt[3]{1 - x^7}$. Then $x = \sqrt[7]{1 - y^3}$ and hence by integration by substitution, $$\begin{align*} \int_{0}^{1} \sqrt[3]{1 - x^7} \; dx &= \int_{0}^{1} y(x) \; dx \\ &= \int_{1}^{0} y \; dx(y) \\ &= [y x(y)]_{1}^{0} - \int_{1}^{0} x(y) \; dy \\ &= \int_{0}^{1} \sqrt[7]{1 - y^3} \; dy. \end{align*}$$

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Formally (and not only formally), we know that $d(xy)=x\, dy + y \, dx$ Hence $xy = \int x \, dy + y \, dx$. This is an abstract way to compute the integral of a function provided we know the integral of its inverse. – Siminore May 1 '12 at 15:37
@Siminore: The way you demonstrated is very clear and systematic. All the scattered understandings on this kind of integral are now integrated. Thanks. – Sangchul Lee May 1 '12 at 15:58

Another way to go is to use $\beta$-function

$$ \mathrm{\beta}(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}\,dt,\quad {Re}(x),{Re(y)}>0. $$

$$ \int_0^1 \sqrt[3]{1-x^7} \mathrm dx = \frac{1}{7}\int_0^1 u^{-6/7}(1-u)^{1/3} \mathrm dx=\dots\,. $$

Now, you can finish the problem.

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