Determining Whether An Improper Integral Converges Or Not

Question

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.

Answer 1

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.

Answer 2

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: $$\boxed{\; I = \frac{B\left(\frac{1}{10},\frac{2}{5}\right)}{5}=\color{red}{\frac{\Gamma\left(\frac{1}{10}\right)\Gamma\left(\frac{2}{5}\right)}{5\sqrt{\pi}}}\approx 2.38115964324. }$$