Show $\int_{0}^{\infty}\frac{dx}{\sqrt{x^{4}+x}} = \frac{2 \Gamma(1/3) \Gamma(7/6)}{\sqrtπ}≈2.80436$

Question

How the following integral can be calculated? $$\int_{0}^{\infty}\frac{dx}{\sqrt{x^{4}+x}}$$

I tried to substitute $\sqrt{x^{3}+1}=u$, but that would not help.

Another way I used was $$\int_{0}^{\infty}\frac{dx}{\sqrt{x^{4}+x}}\le\int_{0}^{\infty}\frac{dx}{x^{2}}$$

which again is not useful, since I'm looking for a convergent integral. Also integral-calculator.com could not solve the integral and WolframAlpha gave the following solution:

$$\int_{0}^{\infty}\frac{dx}{\sqrt{x^{4}+x}} = \frac{2 \Gamma(1/3) \Gamma(7/6)}{\sqrtπ}≈2.80436$$

I don't know where this comes from, so can someone solve this integral or at least show me that the integral is convergent?

Answer 1

Notice that $$\int_0^\infty\frac{{\rm d}x}{\sqrt{x^4+x}}\stackrel{x\mapsto\frac1x}=-\int_\infty^0\frac1{x^2}\frac{{\rm d}x}{\sqrt{\frac1x+\frac1{x^4}}}=\int_0^\infty\frac{{\rm d}x}{\sqrt{1+x^3}}$$ From here it is even clearer why to apply the substitution hinted by Zacky in the comments. I will follow a different track. Recall Ramanujan's Master Theorem (RMT) and reshape the integral accordingly to obtain \begin{align*} \int_0^\infty\frac{{\rm d}x}{\sqrt{1+x^3}}&=\int_0^\infty(1+x^3)^{-1/2}{\rm d}x&&;x^3\mapsto x\\ &=\frac13\int_0^\infty x^{1/3-1}(1+x)^{-1/2}{\rm d}x\\ &=\frac13\int_0^\infty x^{1/3-1}\left[\sum_{n\geq0}\binom{2n}n\frac{(-x)^n}{4^n}\right]{\rm d}x\\ &=\frac13\int_0^\infty x^{1/3-1}\left[\sum_{n\geq0}\frac{\phi(n)}{n!}(-x)^n\right]&&;\phi(n)=\frac{\Gamma(2n+1)}{4^n\Gamma(n+1)}\\ &=\frac13\Gamma\left(\frac13\right)\phi\left(-\frac13\right)&&;\text{RMT}\\ &=\frac13\Gamma\left(\frac13\right)\frac{\Gamma\left(-\frac23+1\right)}{4^{-1/3}\Gamma\left(-\frac13+1\right)}\\ &=\frac{2^{2/3}}3\frac{\Gamma^2\left(\frac13\right)}{\Gamma\left(\frac23\right)} \end{align*} This expression evaluates to $2.803~642\dots$, matching WolframAlpha's overall result numerically. We can do better by using Legendre's Duplication formula for $z=1/6$ to see \begin{align*} \Gamma\left(2\frac16\right)&=\frac1{\sqrt\pi}2^{2(1/6)-1}\Gamma\left(\frac16\right)\Gamma\left(\frac16+\frac12\right)\\ \Gamma\left(\frac13\right)&=\frac{2^{-2/3}}{\sqrt\pi}\Gamma\left(\frac16\right)\Gamma\left(\frac23\right)\\ \therefore~\Gamma\left(\frac23\right)&=\sqrt\pi2^{2/3}\Gamma\left(\frac13\right)\Gamma\left(\frac16\right)^{-1} \end{align*} Using this result we further obtain \begin{align*} \frac{2^{2/3}}3\frac{\Gamma^2\left(\frac13\right)}{\Gamma\left(\frac23\right)}&=\frac{2^{2/3}}3\frac{\Gamma^2\left(\frac13\right)}{\sqrt\pi2^{2/3}\Gamma\left(\frac13\right)\Gamma\left(\frac16\right)^{-1}}\\ &=\frac1{3\sqrt\pi}\Gamma\left(\frac13\right)\Gamma\left(\frac16\right)\\ &=\frac2{\sqrt\pi}\Gamma\left(\frac13\right)\Gamma\left(\frac76\right) \end{align*}

$$\therefore~\int_0^\infty\frac{{\rm d}x}{\sqrt{x^4+x}}~=~\frac2{\sqrt\pi}\Gamma\left(\frac13\right)\Gamma\left(\frac76\right)$$

I do not think there is an completely elementary way to obtain this result (arguably, using the Beta Function is more elementary than RMT) as the final expression is in terms of the Gamma Function.

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Regarding convergence. Split the integral as $$\int_0^\infty\frac{{\rm d}x}{\sqrt{x^4+x}}=\int_0^1\frac{{\rm d}x}{\sqrt{x^4+x}}+\int_1^\infty\frac{{\rm d}x}{\sqrt{x^4+x}}$$ Now, we have $x^4+x\geq x^4$ and for $x\in[1;\infty)$ this shows that the latter integral converges. Regarding the first one, apply $x\mapsto\frac1x$ and then using that $x^3+1\geq x^3$ shows that the first integral converges aswell. Continue as above to obtain its value.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}{\dd x \over \root{x^{4} + x}}} \,\,\,\stackrel{x\ \mapsto\ 1/x}{=}\,\,\, \int_{\infty}^{0}{-\,\dd x/x^{2} \over \root{1/x^{4} + 1/x}} \\[5mm] = & \int_{0}^{\infty}{\dd x \over \root{x^{3} + 1}} \\[5mm] \stackrel{\large x\ =\ \pars{1/t - 1}^{1/3}}{=}\,\,\,& {1 \over 3}\int_{0}^{1}t^{-5/6}\pars{1 - t}^{-2/3}\,\dd t = {1 \over 3}\,{\Gamma\pars{1/6}\Gamma\pars{1/3} \over \Gamma\pars{1/2}} \\[6mm] = &\ {1 \over 3}\,{\bracks{\vphantom{\Large A} \pars{1/6}\Gamma\pars{1/6}}\Gamma\pars{1/3} \over \root{\pi}}\, 6 \\[5mm] = &\ \bbx{\large{2\,\Gamma\pars{1/3}\Gamma\pars{7/6} \over \root{\pi}}}\ \approx 2.8044 \\ & \end{align}