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How can I show that$$\sqrt[3]6>\int_1^\infty\frac{(1+x)^{1/3}}{x^2}\mathrm dx?$$

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Perhaps you can try to use a Taylor expansion on $(1+x)^{1/3}$ and use a suitable convergence theorem (e.g. Monotone convergence theorem, Fatou's lemma or the Dominated convergence theorem) –  gifty Aug 3 '12 at 16:11

5 Answers 5

Let $p=3$, $q=3/2$. Then \begin{align} \int_1^{+\infty}\frac{(1+x)^{1/3}}{x^2}dx&=\int_1^{+\infty}\color{green}{\frac{(1+x)^{1/3}}{x}}\color{red}{\frac 1x}dx\\ &< \color{green}{\left(\int_1^{+\infty}\frac{1+x}{x^3}dx\right)^{1/3} }\color{red}{\left(\int_1^{+\infty}x^{—3/2}dx\right)^{2/3}}\\ &=(1+\int_1^{+\infty}x^{-3}dx)^{1/3}\left(\left[-2x^{-1/2}\right]_1^{+\infty}\right)^{2/3}\\ &=(1+\left[-\frac 12x^{-2}\right]_1^{+\infty})^{1/3}2^{2/3}\\ &=\sqrt[3]{3/2\cdot 4}\\ &=6^{1/3}. \end{align} Note that the inequality is strict as we are not in equality case in Hölder's inequality.

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Nice use of Hölder's inequality. –  nbubis Aug 3 '12 at 16:25
    
Indeed. First, I tried using Jensen applied to the measure $\mu=\frac 1{t^2}\lambda$ but it didn't worked. Maybe gifty's approach gives a better bound. –  Davide Giraudo Aug 3 '12 at 16:37

In the interval from $1$ to $2$, $\frac{(1+x)^{1/3}}{x^2}\le \frac{3^{1/3}}{x^2}$. Thus $$\int_1^2 \frac{(1+x)^{1/3}}{x^2}\,dx\le \int_1^2\frac{3^{1/3}}{x^2}\,dx= \frac{1}{2}(3^{1/3}).$$ In the interval from $2$ to $\infty$, $(1+x)^{1/3}\le \left(\frac{x}{2}+x\right)^{1/3}$. So $$\int_2^\infty \frac{(1+x)^{1/3}}{x^2}\,dx\le \int_2^\infty(3/2)^{1/3}x^{-5/3}\,dx= \frac{3}{4}(3^{1/3}).$$ So our full integral is $\le \frac{5}{4}(3^{1/3})$. Since $(5/4)^3 \lt 2$, the result follows.

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$$ \int_1^\infty \frac{(1+x)^{1/3}}{x^2} \mathrm{d} x \stackrel{x=\frac{2-u}{u}}{=} 2^{4/3} \int_0^1 u^{-1/3} \left(2-u\right)^{-2} \mathrm{d} u = 2^{-2/3} \int_0^1 u^{-1/3} \left(1-\frac{1}{2} u\right)^{-2} \mathrm{d} u $$ This is just the Euler type integral for Gauss's hypergeometric function: $$ \frac{\Gamma(a) \Gamma(c-a)} {\Gamma(c)} {}_2F_1\left(a,b;c;x\right) = \int_0^1 u^{a-1} (1-u)^{c-a-1} (1-x u)^{-b} \mathrm{d} u $$ where $a=\frac{2}{3}$, $c=a+1 = \frac{5}{3}$, $b=2$ and $x=\frac{1}{2}$. Since $c=a+1$ the ratio of $\Gamma$-functions simplifies to $\frac{1}{a} = \frac{3}{2}$. Now $$ \int_1^\infty \frac{(1+x)^{1/3}}{x^2} \mathrm{d} x = \frac{3}{2^{5/3}} {}_2F_1\left(\frac{2}{3}, 2; \frac{5}{3}; \frac{1}{2} \right) $$ The said hypergeometric can be computed in (not very simple, but elementary) closed form: $$ \int_1^\infty \frac{(1+x)^{1/3}}{x^2} \mathrm{d} x = \sqrt[3]{2}+\frac{1}{24} \log \left(3505753+2782518 \sqrt[3]{2}+2208486\ 2^{2/3}\right)-\frac{1}{2 \sqrt{3}}\arcsin\left(\frac{\sqrt{3}}{2} \sqrt{\left(7-3 \sqrt[3]{2}\right) \left(\sqrt[3]{2}-1\right)}\right) \approx 1.6695914 \approx 4.6540453^{1/3} $$ So $6^{1/3}$ strikes me as a pretty tight bound.

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Please, give the corresponding Maple, Mathematica, etc commands were used to simplify the hypergeometric function. –  vesszabo Aug 3 '12 at 20:17

The given integral ($=:Q$) can be evaluated in elementary terms. Using the substitution $u:=(1+x)^{1/3}$ one obtains $$Q=\int_{2^{1/3}}^\infty {3u^3\over (u^3-1)^2}\ du\ .$$ Mathematica produces the output

enter image description here

which numerically is $\doteq 1.66959$ (cf. Sasha's answer), whereas $6^{1/3}\doteq1.81712$.

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The integral can be written as $$\int_1^{\infty} {1 \over x^{5 \over 3}} \bigg(1 + {1 \over x}\bigg)^{1 \over 3}\,dx$$ By Taylor expanding, $(1 + {1 \over x})^{1 \over 3} = 1 + {1 \over 3x} +$ error term, where the error term is negative by the Lagrange form of the remainder. So the integral is less than $$\int_1^{\infty} {1 \over x^{5 \over 3}} + {1 \over 3 x^{8 \over 3}}\,dx$$ $$= {3 \over 2} + {1 \over 3}\cdot{3 \over 5}$$ $$= 1.7$$ $$ < \sqrt[3]6$$

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Nice, it's almost the exact value. –  vesszabo Aug 3 '12 at 20:30

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