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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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    (The last URL got butchered; you'll have to copy and paste, I guess)2010-07-29

2 Answers 2

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Note that if

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

then

$ \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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    https://en.m.wikipedia.org/wiki/Integral_of_inverse_functions2017-05-09
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$\int_0^1(1-x^m)^{(1/n)}dx=(m+n)\Gamma(1/m)\Gamma(1/n)/\Gamma(1/m+1/n)$ is symmetric in $m, n$.