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I want to know if the following integral converges:

$\int_0^\infty \dfrac{xdx}{\sqrt{x^5+1}}$

I assume I'd have to study $lim_a\rightarrow \infty \int_0^a \dfrac{xdx}{\sqrt{x^5+1}} $ But I don't know how to go about.

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  • $\begingroup$ What tools do you know to show convergence? $\endgroup$ – mickep Aug 12 '16 at 14:13
  • $\begingroup$ Your numerator is of degree $1$. What "degree" would you assign to the denominator? I hope it is more than $2$ because then..... $\endgroup$ – imranfat Aug 12 '16 at 14:17
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Note that near $x = 0$ the integrand stays bounded so there is no issue there. The simple answer observes the asymptotic behavior of the integrand as $x \to \infty$. For large $x$, we see $$\frac{x}{\sqrt{x^5 + 1}} \approx \frac x {\sqrt{x^5}} = \frac{1}{x^{1.5}}$$ so your integral should converge since $\int_\epsilon^\infty \frac{1}{x^{1.5}} dx$ converges for any positive $\epsilon$. Of course in this particular case, it is very easy to make this rigorous since $$\frac{x}{\sqrt{x^5 + 1}} \le \frac x {\sqrt{x^5}} = \frac{1}{x^{1.5}},$$ so the integral converges by comparison.

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  • $\begingroup$ isn't $\dfrac{x}{x^\dfrac{1}{5}}= x^\dfrac{4}{5}$? $\endgroup$ – aribaldi Aug 12 '16 at 14:48
  • $\begingroup$ we have $$\frac x {\sqrt{x^5}} = \frac{x}{x^{5/2}} = \frac{1}{x^{3/2}}.$$ There shouldn't be any $x^{1/5}$. $\endgroup$ – User8128 Aug 12 '16 at 14:50
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The integrand function is continuous and behaves like $\frac{1}{x^{3/2}}$ for large $x$, hence the integral is convergent. By setting $x=y^{2/5}$, then $y=\tan t$, we get:

$$ I = \frac{2}{5}\int_{0}^{+\infty}\frac{y^{-1/5}}{\sqrt{1+y^2}}\,dy=\frac{2}{5}\int_{0}^{\pi/2}\cos^{-4/5}(t)\sin^{-1/5}(t)\,dt $$ and by Euler's beta function we get: $$\boxed{\; I = \frac{B\left(\frac{1}{10},\frac{2}{5}\right)}{5}=\color{red}{\frac{\Gamma\left(\frac{1}{10}\right)\Gamma\left(\frac{2}{5}\right)}{5\sqrt{\pi}}}\approx 2.38115964324. }$$

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