When I tried to approximate $$\int_{0}^{1} (1-x^7)^{1/5}-(1-x^5)^{1/7}\ dx$$ I kept getting answers that were really close to $0$, so I think it might be true. But why? When I ask Mathematica, I get a bunch of symbols I don't understand!

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    $\begingroup$ The Wolfram Integrator computes only indefinite integrals, as far as I can see, so you were getting the primitive for your integrand, which is understandably obscure (Those ${}_2F_1$ functions that show up there are hypergeometric functions (see functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1) which are a simply amazing family of functions) To compute definite integrals like the one you want, you can use Alpha: for example, http://www.wolframalpha.com/input/?i=integrate+%281-x^7%29^{1%2F5}+-+%281-x^5%29^{1%2F7}+from+0+to+1 $\endgroup$ Jul 29, 2010 at 15:06
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    $\begingroup$ (The last URL got butchered; you'll have to copy and paste, I guess) $\endgroup$ Jul 29, 2010 at 15:07

2 Answers 2


Note that if

$$ y = \left(1 - x^7\right)^{1/5} $$


$$ \left(1 - y^5\right)^{1/7} = x $$

This means $(1-x^7)^{1/5}$ is the inverse function of $(1-x^5)^{1/7}$. In the graph, one will be the same as the other when reflected along the diagonal line y = x.

Also, both functions

  1. share the same range [0, 1] and domain [0, 1] and
  2. monotonically decreasing,

Therefore, the area under the graph in [0, 1] will be the same for both functions:

$$ \int_0^1 \left(1-x^7\right)^{1/5} dx = \int_0^1 \left(1-y^5\right)^{1/7} dy $$

Grouping the two integrals yield the equation in the title.

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    $\begingroup$ The point that the ranges are (0,1] and the domains are [0,1) tripped me up at first - I thought x^2 and x^(1/2) would form a counterexample. I would have stated it as: "Both these functions have value 1 at x=0 and value 0 at x=1", that way the geometry becomes clearer, IMHO. $\endgroup$
    – yatima2975
    Jul 29, 2010 at 9:07
  • $\begingroup$ en.m.wikipedia.org/wiki/Integral_of_inverse_functions $\endgroup$ May 9, 2017 at 8:55

$\int_0^1(1-x^m)^{(1/n)}dx=(m+n)\Gamma(1/m)\Gamma(1/n)/\Gamma(1/m+1/n)$ is symmetric in $m, n$.


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