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How to prove that $$\int_{0}^{1}(1+x^n)^{-1-\frac{1}{n}}dx=2^{-\frac{1}{n}}$$ I have tried letting $t=x^n$,and then convert it into a beta function, but I failed. Is there any hints or solutions? Thanks for attention.

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2 Answers 2

up vote 8 down vote accepted

The change of variable $t=x^{-n}$ yields $\mathrm dt=-nx^{-n-1}\mathrm dx$, that is, $\mathrm dx=-\frac1nt^{-1-1/n}\mathrm dt$. Thus, the integral is $$ \int_1^{+\infty}\frac1n\frac{\mathrm dt}{(1+t)^{1+1/n}}=\left[\frac{-1}{(1+t)^{1/n}}\right]_1^{+\infty}=\frac1{2^{1/n}}. $$

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Let $x=(\tan{t})^{2/n}$. Then the integral becomes

$$\begin{align} \int_0^1 dx \: (1+x^n)^{-\left ( 1 + \frac{1}{n} \right )} &= \frac{2}{n} \int_0^{\pi/4} dt \: \cot{t} (\tan{t})^{2/n} (\sec{t})^{-2/n} \\ &= \frac{2}{n} \int_0^{\pi/4} dt \: \cot{t} (\sin{t})^{2/n} \\ &= \frac{2}{n} \int_0^{\pi/4} d(\sin{t}) \: (\sin{t})^{2/n - 1} \\ &= [(\sin{t})^{2/n}]_0^{\pi/4} \\ &= 2^{-1/n} \end{align}$$

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