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This integral converges because $$ \int_{1}^{\infty} \frac{1}{\sqrt{x^3 -1}}\,dx < \int_{1}^{\infty} \frac{1}{x\sqrt{x -1}}\,dx $$ $$\int_{1}^{\infty} \frac{1}{x\sqrt{x -1}}\,dx = 2\int_{0}^{\infty} \frac{1}{u^2+1}\,dx =\pi$$ But, how to solve it?

Guilherme Namen
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  • I don't think that there is a nice closed form which you'll be happy with. – Zuy Feb 24 '23 at 17:11
  • You could plug it into wolframalpha: https://www.wolframalpha.com/input?i=integrate+%28x%5E3-1%29%5E%28-1%2F2%29+dx+from+1+to+infinity

    It comes up with the reasonably nice $\frac{2\sqrt\pi\ \Gamma(\frac76)}{\Gamma(\frac23)}$, but I don't think you will be very happy with that.

     – student91 Feb 24 '23 at 17:13
  • In general, there are no closed expressions for integrals involving roots and exponents - rather, this is how a lot of "special functions" arise. Have a look at this similar question: https://math.stackexchange.com/questions/33412/how-to-integrate-int-frac1-sqrt1x3-mathrm-dx – Nuke_Gunray Feb 24 '23 at 17:15

2 Answers2

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Substitute $y=x^{-3}$ to recover a beta integral.

$$\int_1^\infty \frac{dx}{\sqrt{x^3-1}} = \int_1^\infty \frac{dx}{x^{3/2} \sqrt{1-x^{-3}}} \\ = \frac13 \int_0^1 \frac{y^{-4/3}\,dy}{y^{-1/2}\sqrt{1-y}}=\frac13 \int_0^1 y^{-5/6} (1-y)^{-1/2} \, dy$$

user170231
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1

Using Mellin transform: $$ \int_{1}^{\infty}\frac{dx}{\sqrt{x^3-1}} = \frac{1}{3} \int_{0}^{\infty}\frac{dx}{\sqrt{u}(u+1)^{\frac{2}{3}}} \quad x^3=u+1 $$ $$\frac{1}{3} \int_{0}^{\infty}\frac{dx}{\sqrt{u}(u+1)^{\frac{2}{3}}} = \frac{1}{3}\mathcal{M}\left (\frac{1}{2}\right)\left[ \frac{1}{(u+1)^\frac{2}{3}}\right ] = \frac{\Gamma \left( \frac{1}{2} \right)\Gamma \left( \frac{1}{6} \right)}{3\Gamma \left( \frac{2}{3} \right)}$$

Guilherme Namen
  • 315