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I have found that the value of the integral below is always $1/3$ for all positive $n$. $$\int_0^1 (\sqrt[n]{1-x^n}-x)^2 dx$$ Can anyone prove this for me? Thank you.

jimjim
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Zjjorsia
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  • Idea: if the integral is $I(n)$, take a derivative wrt $n$ and show that $I'(n) = 0$ (which it is). Then you can just evaluate for one easy value of $n$, say $n=1$. – kipf Mar 22 '24 at 10:18
  • What is the source of the question? What have you tried ? This is a very low quality way of asking questions heree – jimjim Mar 22 '24 at 12:23
  • @jimjim I am sorry, I haven't used this cite very much yet. Thank you for your advice – Zjjorsia Mar 23 '24 at 05:29

2 Answers2

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put $x^n=y$ then $dx=\frac{1}{n} y^{\frac{1}{n}-1} dy$ So $$ I=\int_0^1 \left((1-x^n)^{\frac{1}{n}}-x\right)^2dx=\int_0^1 \left((1-y)^{\frac{1}{n}}-y^{\frac{1}{n}}\right)^2\frac{1}{n} y^{\frac{1}{n}-1} dy $$ So $$ I=\frac{1}{n}\int_0^1 \left((1-y)^{\frac{2}{n}}y^{\frac{1}{n}-1} -2(1-y)^{\frac{1}{n}}y^{\frac{2}{n}-1}+y^{\frac{3}{n}-1} \right) dy$$ then $$ I=\frac{1}{n} \left(\beta\left(\frac{2}{n}+1,\frac{1}{n}\right)-2\beta\left(\frac{1}{n}+1,\frac{2}{n}\right)+\frac{n}{3} \right) $$ now we have from beta function $$ \beta\left(\frac{2}{n}+1,\frac{1}{n}\right)-2\beta\left(\frac{1}{n}+1,\frac{2}{n}\right)=0$$ So $$I=\frac{1}{3}$$ also we can get this result without using beta function we have $$ \int_0^1 (1-y)^{\frac{2}{n}}y^{\frac{1}{n}-1} dy =\int_0^1 y^{\frac{2}{n}}(1-y)^{\frac{1}{n}-1} dy $$ now using integral by part $$\int_0^1 (1-y)^{\frac{2}{n}}y^{\frac{1}{n}-1} dy=2\int_0^1 (1-y)^{\frac{1}{n}}y^{\frac{2}{n}-1}dy$$ therefore $$I=\frac{1}{n} \int_0^1 y^{\frac{3}{n}-1} dy=\frac{1}{3}$$

Faoler
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Explore the symmetry to integrate the general case

\begin{align} &\int_0^1 (\sqrt[n]{1-x^n}-x)^{2k} \overset{1-x^n \to x^n}{dx}\\ = &\int_0^1 (x-\sqrt[n]{1-x^n})^{2k} d(-\sqrt[n]{1-x^n})\\ =&\ \frac12 \int_0^1 (x-\sqrt[n]{1-x^n})^{2k} d(x-\sqrt[n]{1-x^n}) =\frac1{2k+1}\\ \end{align}

Quanto
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  • Thank you for your answer. Can you please explain the second equality? And if there is $1/2$ in front of the integral, isn't the answer $1\over{2(2k+1)}$? – Zjjorsia Mar 23 '24 at 05:33
  • @Zjjorsia Second equality \begin{align} I =&\frac12 \int_0^1 (x-\sqrt[n]{1-x^n})^{2k} dx+\frac12\int_0^1 (x-\sqrt[n]{1-x^n})^{2k} d(-\sqrt[n]{1-x^n})\ =&\ \frac12 \int_0^1 (x-\sqrt[n]{1-x^n})^{2k} d(x-\sqrt[n]{1-x^n}) \end{align} Also $$(x-\sqrt[n]{1-x^n})^{2k+1}|_0^1 =1-(-1)=2$$ which cancels $\frac12$ – Quanto Mar 23 '24 at 09:10
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    Wow I didn't know I can do $\int y da+\int y db=\int y d(a+b)$. I've never seen it before, can you please me tell the name of the theorem or somewhere to look for? – Zjjorsia Mar 23 '24 at 09:33
  • @Zjjorsia it is the derivative addition rule i.e. $(f(x)+g(x))’=f’(x)+g’(x)$ – Quanto Mar 23 '24 at 11:12