How find this $\int_{1}^{\infty}\frac{1}{x^2\sqrt{x^3-1}}dx$

Question

find The integral $$\int_{1}^{\infty}\dfrac{1}{x^2\sqrt{x^3-1}}dx$$ My try:let $$\sqrt{x^3-1}=t\Longrightarrow x^3=t^2-1$$ Thank I can't,I think this answer maybe use Gamma integral

Answer 1

Sub $x=1/u$, $dx=-du/u^2$. The integral is then equal to

$$\int_0^1 du \left ( \frac{1}{u^3}-1\right )^{-1/2} = \int_0^1 du \, u^{3/2} \, (1-u^3)^{-1/2}$$

Now let $u=v^{1/3}$, $du = \frac13 v^{-2/3} dv$; the integral is then equal to

$$\frac13 \int_0^1 dv \, v^{-2/3} v^{1/2} (1-v)^{-1/2} = \frac13 \int_0^1 dv \, v^{-1/6} (1-v)^{-1/2}$$

This you may recognize as a Beta function:

$$\frac13 \frac{\Gamma(5/6) \Gamma(1/2)}{\Gamma(4/3)} = \frac{\sqrt{\pi} \, \Gamma(5/6)}{\Gamma(1/3)} $$

Note: one may further simplify this by using the duplication and reflection formulae to express this integral in terms of $\Gamma(1/3)^3$, which reduces to a value of the complete elliptic integral of the first kind.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{1}^{\infty}{\dd x \over x^{2}\root{x^{3} - 1}}} = {1 \over 3}\int_{1}^{\infty}{\dd x \over x^{4/3}\root{x - 1}} \\[5mm] = &\ {1 \over 3}\int_{0}^{\infty}x^{\color{red}{1/2} - 1}\,\, \pars{1 + x}^{-4/3}\,\,\dd x \end{align}

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$\ds{\pars{1 + x}^{-4/3} = \sum_{k = 0}^{\infty}{-4/3 \choose k}x^{k} = \sum_{k = 0}^{\infty}\bracks{{k + 1/3 \choose k} \pars{-1}^{k}}x^{k} =}$

$\ds{\sum_{k = 0}^{\infty}\color{red} {\Gamma\pars{4/3 + k} \over \Gamma\pars{4/3}}{\pars{-x}^{k} \over k!}}$.

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Ramanujan's Master Theorem: \begin{align} &\bbox[5px,#ffd]{\int_{1}^{\infty}{\dd x \over x^{2}\root{x^{3} - 1}}} = {1 \over 3}\,\Gamma\pars{\color{red}{1 \over 2}}\,{\Gamma\pars{4/3 - \color{red}{1/2}} \over \Gamma\pars{4/3}} \\[5mm] = &\ {1 \over 3}\,\,{\Gamma\pars{1/2}\Gamma\pars{5/6} \over \Gamma\pars{4/3}} = \bbx{{1 \over 3}\on{B}\pars{{1 \over 2},{5 \over 6}}} \approx 0.7468 \\ & \end{align} $\ds{\on{B}}$: Beta Function.

Answer 3

With your substitution $\sqrt{x^3-1}=t$ the integral becomes $$ \int_1^{\infty} \frac{1}{x^2 \sqrt{x^3-1}} dx =\frac{1}{3} \int_0^{\infty} \frac{t^{-1/2}}{(1+t)^{4/3}} dt=\frac{1}{3}B\left(\frac{1}{2},\frac{5}{6}\right) $$ taking into account that the beta function is $B\left(p,q\right)=\int_0^{\infty} \frac{t^{p-1}}{(1+t)^{p+q}} dt$ (see http://dlmf.nist.gov/5.12#E3).