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

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?

• Hint: Substitute $\frac{1}{x^3+1}=t$ then use the beta function. Feb 16, 2020 at 18:25
• @thank you so much, but I still don't know about beta function, do you know a way to show that the integral is convergent?
– user715522
Feb 16, 2020 at 18:34
• @user715522 Do you want to evaluate the integral or do you want to show that it converges? Feb 16, 2020 at 18:37
• well I prefer the first one, but it seems to be a little complicated, so I just want to show that the integral is convergent
– user715522
Feb 16, 2020 at 18:38
• @Zacky,thank you so much
– user715522
Feb 16, 2020 at 19:20

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.

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.

• Your answer is really awesome, also can we determine the convergence of this integral?if yes then how, because I really don't know how to do that
– user715522
Feb 16, 2020 at 18:59
• @user715522 I've added the proof of convergence as Zacky suggested in the comments. Feb 16, 2020 at 19:11
• Also, there is a chance for the following shortcut in the computation: Indeed for $x,y>0$ we have by using the subsitution $u=\frac{t}{1-t}$: $$\int_0^\infty u^{x-1} (1+u)^{-(x+y)}\,\mathrm du = \int_0^1 \left(\frac{t}{1-t}\right)^{x-1} \cdot (1-t)^{-(x+y)} \cdot (1-t)^{-2} \, \mathrm dt = \operatorname B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ (in your particular case, $x=y=\frac13$). Feb 16, 2020 at 19:28
• @MaximilianJanisch I always forget about this particular integral representation of the Beta Function... Thank you for pointing out! Feb 16, 2020 at 19:31
• @mrtaurho I didn't know about it before writing the above comment but when I saw $\frac{\Gamma(1/3)^2}{\Gamma(2/3)}$ I was like "there must be a Beta function somewhere" 😃 Feb 16, 2020 at 19:34

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \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}