18
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ 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 $\endgroup$ – Mariano Suárez-Álvarez Jul 29 '10 at 15:06
  • 1
    $\begingroup$ (The last URL got butchered; you'll have to copy and paste, I guess) $\endgroup$ – Mariano Suárez-Álvarez Jul 29 '10 at 15:07
45
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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 '10 at 9:07
  • $\begingroup$ en.m.wikipedia.org/wiki/Integral_of_inverse_functions $\endgroup$ – Peter Szilas May 9 '17 at 8:55
7
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.