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How to compute the integral,

$$\int_0^\infty \frac{1}{(1+x^{\varphi})^{\varphi}}\,dx$$

where, $\varphi = \dfrac{\sqrt{5}+1}{2}$ is the Golden Ratio?

Martin Sleziak
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sciona
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    Is there a reason to believe it can be done in some closed form? – Thomas Andrews Jun 17 '15 at 05:05
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    It can apparently be shown [using the gamma function](http://www.wolframalpha.com/input/?i=integral+0+to+infinity+1%2F%281+%2B+x%5E%28golden+ratio%29%29%5E%28golden+ratio%29+dx) that this integral comes out to $1$. – Ben Grossmann Jun 17 '15 at 05:11
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    What a nice question! – tired Jun 17 '15 at 19:14
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    By the way, the antiderivative is elementary: ${\large\int}\frac{1}{(1+x^{\varphi})^{\varphi}}\,dx=\frac{x}{\left(1+x^{\varphi}\right)^{1/{\varphi}}}+C$ that can be verified by direct differentiation and simplification using the identity $1/\varphi=\varphi-1$. – Vladimir Reshetnikov Jun 24 '15 at 17:03

5 Answers5

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Since $\frac1\varphi=\varphi-1$, $$ \begin{align} \int_0^\infty\frac1{(1+x^\varphi)^\varphi}\,\mathrm{d}x &=(\varphi-1)\int_0^\infty\frac{x^{\varphi-2}}{(1+x)^\varphi}\,\mathrm{d}x\tag{1}\\[6pt] &=(\varphi-1)\mathrm{B}(\varphi-1,1)\tag{2}\\[6pt] &=(\varphi-1)\frac{\Gamma(\varphi-1)}{\Gamma(\varphi)}\tag{3}\\ &=\frac{\Gamma(\varphi)}{\Gamma(\varphi)}\tag{4}\\[6pt] &=1 \end{align} $$ Explanation:
$(1)$: substitute $x\mapsto x^{\varphi-1}$ noting that $\varphi(\varphi-1)=1$
$(2)$: $\int_0^\infty\frac{x^{\alpha-1}}{(1+x)^\beta}\,\mathrm{d}x=\mathrm{B}(\alpha,\beta-\alpha)$
$(3)$: $\mathrm{B}(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$
$(4)$: $\alpha\,\Gamma(\alpha)=\Gamma(\alpha+1)$

robjohn
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Hint: Make $x \mapsto \dfrac{1}{x}$ and use $\phi^2 = \phi + 1$ to further simplify. The final result should be $1$.

$$\int_0^{\infty} \frac{1}{(1+x^{\phi})^{\phi}}\,dx = \int_0^{\infty} \frac{x^{\phi^2}}{(1+x^{\phi})^{\phi}}\frac{dx}{x^2} = \int_0^{\infty} \frac{x^{\phi - 1}}{(1+x^{\phi})^{\phi}}\,dx = \, \cdots$$

r9m
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    @Winther when you put it that way it's obvious too :-) after all $\displaystyle \Gamma (\phi) = \int_0^{\infty} x^{\phi -1}e^{-x}\,dx = \int_0^{\infty} xe^{-x^{\phi}}\,dx^{\phi} = \Gamma \left(1+\frac{1}{\phi}\right)$ .. !! :-) – r9m Jun 17 '15 at 05:44
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    Well, there you go. Now you have two proofs for the statement. – Winther Jun 17 '15 at 05:46
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The antiderivative invokes an hypergeometric function $$\int \frac{dx}{(1+x^{a})^{a}}=x \, _2F_1\left(\frac{1}{a},a;1+\frac{1}{a};-x^a\right) $$ For the definite integral, as Omnomnomnom commented, the result expresses using the gamma function $$I(a)=\int_0^{\infty} \frac{dx}{(1+x^{a})^{a}} =\frac{\Gamma \left(1+\frac{1}{a}\right) \Gamma \left(a-\frac{1}{a}\right)}{\Gamma (a)}$$ and $I(\phi)=1$ since $\phi-\frac{1}{\phi}=1$ and $1+\frac{1}{\phi}=\phi$.

Amazing are $I(2)=\frac{\pi} 4$, $I(6)=\frac{124729 }{559872}\pi$ and $I(\infty)=1$.

Claude Leibovici
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    Filling in the missing integers: $I(3)=\frac{10\pi\sqrt3}{81}$ and $I(4)=\frac{77\pi\sqrt2}{512}$ and $I(5)=\frac{399\pi\sqrt{50+10\sqrt5}}{15625}$ – robjohn Apr 02 '16 at 08:34
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Hint: let $x^{\varphi}=u$ then $$\dfrac{1}{\varphi}\int_{0}^{+\infty}\dfrac{u^{1/\varphi-1}}{(1+u)^{\varphi}}du=\dfrac{1}{\varphi}B(\dfrac{1}{\varphi},\varphi-\dfrac{1}{\varphi})=\dfrac{1}{\varphi}B(\dfrac{1}{\varphi},1)$$

math110
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  • Here, B denotes the [Beta function](https://en.wikipedia.org/wiki/Beta_function) – Ben Grossmann Jun 17 '15 at 05:13
  • This is the path I would also present. What can be added is: \begin{align} \frac{1}{\alpha} \, B\left(\frac{1}{\alpha}, 2 \right) &= \frac{1}{\alpha} \end{align} where $2 \alpha = 1 + \sqrt{5}$. – Leucippus Jun 17 '15 at 05:31
  • I believe the last two terms should be $$\frac1\varphi\mathrm{B}\left(\frac1\varphi,\varphi-\frac1\varphi\right) =\frac1\varphi\mathrm{B}\left(\frac1\varphi,1\right)$$ – robjohn Jun 17 '15 at 06:09
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For an answer avoiding the use of special functions: use $w=[x^{\varphi}(1+x^{\varphi})^{-1}]^{\varphi}$ then $$\int_0^1 \mathrm{d}w = 1$$

Zain Patel
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