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?
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2$\begingroup$ Is there a reason to believe it can be done in some closed form? $\endgroup$– Thomas AndrewsJun 17, 2015 at 5:05
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1$\begingroup$ It can apparently be shown using the gamma function that this integral comes out to $1$. $\endgroup$– Ben GrossmannJun 17, 2015 at 5:11
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2$\begingroup$ What a nice question! $\endgroup$– tiredJun 17, 2015 at 19:14
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4$\begingroup$ 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 $1/\varphi=\varphi-1$. $\endgroup$– Vladimir ReshetnikovJun 24, 2015 at 17:03
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5 Answers
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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 $\varphi(\varphi-1)=1$
$(2)$: $\int_0^\infty\frac{x^{\alpha-1}}{(1+x)^\beta}\,\mathrm{d}x=\mathrm{B}(\alpha,\beta-\alpha)$
$(3)$: $\mathrm{B}(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$
$(4)$: $\alpha\,\Gamma(\alpha)=\Gamma(\alpha+1)$
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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 $1$.
$$\int_0^{\infty} \frac{1}{(1+x^{\phi})^{\phi}}\,dx = \int_0^{\infty} \frac{x^{\phi^2}}{(1+x^{\phi})^{\phi}}\frac{dx}{x^2} = \int_0^{\infty} \frac{x^{\phi - 1}}{(1+x^{\phi})^{\phi}}\,dx = \, \cdots$$
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1$\begingroup$ @Winther when you put it that way it's obvious too :-) after all $\displaystyle \Gamma (\phi) = \int_0^{\infty} x^{\phi -1}e^{-x}\,dx = \int_0^{\infty} xe^{-x^{\phi}}\,dx^{\phi} = \Gamma \left(1+\frac{1}{\phi}\right)$ .. !! :-) $\endgroup$– r9mJun 17, 2015 at 5:44
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1$\begingroup$ Well, there you go. Now you have two proofs for the statement. $\endgroup$– WintherJun 17, 2015 at 5:46
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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 $I(\infty)=1$.
answered Jun 17, 2015 at 5:48
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3$\begingroup$ Filling in the missing integers: $I(3)=\frac{10\pi\sqrt3}{81}$ and $I(4)=\frac{77\pi\sqrt2}{512}$ and $I(5)=\frac{399\pi\sqrt{50+10\sqrt5}}{15625}$ $\endgroup$– robjohn ♦Apr 2, 2016 at 8:34
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Hint: let $x^{\varphi}=u$ then $$\dfrac{1}{\varphi}\int_{0}^{+\infty}\dfrac{u^{1/\varphi-1}}{(1+u)^{\varphi}}du=\dfrac{1}{\varphi}B(\dfrac{1}{\varphi},\varphi-\dfrac{1}{\varphi})=\dfrac{1}{\varphi}B(\dfrac{1}{\varphi},1)$$
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$\begingroup$ 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 $2 \alpha = 1 + \sqrt{5}$. $\endgroup$ Jun 17, 2015 at 5:31
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$\begingroup$ I believe the last two terms should be $$\frac1\varphi\mathrm{B}\left(\frac1\varphi,\varphi-\frac1\varphi\right) =\frac1\varphi\mathrm{B}\left(\frac1\varphi,1\right)$$ $\endgroup$– robjohn ♦Jun 17, 2015 at 6:09
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For an answer avoiding the use of special functions: use $w=[x^{\varphi}(1+x^{\varphi})^{-1}]^{\varphi}$ then $$\int_0^1 \mathrm{d}w = 1$$
answered Dec 17, 2016 at 4:50