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show that $$I=\int_0^1 (\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}) \, dx=0$$

I find this is Nice equalition!

My try: let $$\sqrt[3]{1-x^7}=t\Longrightarrow x=\sqrt[7]{1-t^3}$$ so $$dx=-\dfrac{3}{7}t^2(1-t^3)^{-\dfrac{6}{7}} \, dt$$ so $$I=\frac{3}{7}\int_0^1 \frac{t^3}{\sqrt[7]{(1-t^3)^6}} \, dt-\int_0^1 \sqrt[7]{1-x^3} \, dx$$

By parts,we have $$\int_0^1 \sqrt[7]{1-x^3} dx=\dfrac{3}{7}\int_0^1 \frac{x^3}{\sqrt[7]{(1-x^3)^6}} \, dt$$ so $$I=0$$

this problem maybe have more other nice methods!Thank you

math110
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    Why not try to split the integral into two parts, i.e. $\displaystyle \int_0^1 \sqrt[3]{1-x^7} dx$ and $\displaystyle \int_0^1 \sqrt[7]{1-x^3} dx$, and then setting them equal? – 2012ssohn Dec 27 '13 at 05:00
  • Related http://math.stackexchange.com/questions/139393/how-to-show-that-int-01-left-sqrt31-x7-sqrt71-x3-right-dx/139398#139398 – clark Dec 27 '13 at 05:32

3 Answers3

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Both parts of the integral express the area under the curve given by $x^7+y^3=1$.

Ryan Reich
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  • Cool! I was going to ask if the OP knew about $beta$ functions. I need sleep – user44197 Dec 27 '13 at 05:13
  • I learned this trick in my first analysis class in the proof of the Holder inequality. It's magical. – Ryan Reich Dec 27 '13 at 05:19
  • the second integral is of the curve: y^7 + x^3 = 1. So are these curves the same? or can be use interchangeably through some tranformation? – DeepSea Dec 27 '13 at 05:21
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    More generally: if $f$ is a strictly decreasing continuous function with $f(0)=1$ and $f(1)=0$, then $\int_0^1 f(x),dx = \int_0^1 f^{-1}(y),dy$ - because both quantities equal the area between the axes and the graph of $y=f(x)$ - and hence $\int_0^1 (f(x)-f^{-1}(x)),dx = 0$. – Greg Martin Dec 27 '13 at 05:22
  • It truly is magical! I am going to add it to my list of dreams (to with my other favorite: http://en.wikipedia.org/wiki/Sophomore%27s_dream ) – user44197 Dec 27 '13 at 05:23
  • @GregMartin,It's nice! – math110 Dec 27 '13 at 05:26
  • @Matik Since the equations are the same but for exchanging $x$ and $y$, the area under the second is the same as the area to the left of the first. The curve being monotonic, as Greg said, this is the same as the area under the first curve. – Ryan Reich Dec 27 '13 at 05:38
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You can write the two parts of the integrals as follows: $$ \int_0^1 \sqrt[3]{1-x^7} \, dx = \int_0^1 \int_0^\sqrt[3]{1-x^7} \, dy\,dx $$ and $$ \int_0^1 \sqrt[7]{1-y^3}) \, dy = \int_0^1 \int_0^\sqrt[7]{1-y^3} \, dx\,dy. $$ Now you can show that the regions of both these integral are the same. Namely the sets $\{(x, y): 0 \leq x \leq 1, 0 \leq \sqrt[3]{1-x^7}\}$ and $\{(x, y): 0 \leq y \leq 1, 0 \leq x \leq \sqrt[7]{1-y^3}\}$ are the same.

Pratyush Sarkar
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The antiderivative of $(1-x^a)^{1/b}$, at least for integer values of $a$ and $b$, is given by $x{\rm Hypergeometric2F1}[1/a, -(1/b), 1 + 1/a, x^a]$.
The integral of $(1-x^a)^{1/b}$ between 0 and 1 is then given by
$(\Gamma[1 + 1/a] \Gamma[1 + 1/b]) / \Gamma[1 + 1/a + 1/b]$
Then, the two integrals are equal and your identity is proved.

Claude Leibovici
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