Prove that $\int_0^{\infty} \frac{1}{\sqrt{x^3 + x}}dx$ is convergent Could you tell me how to prove that $$\int_0^{\infty} \frac{1}{\sqrt{x^3 + x}}dx$$ is convergent?
 A: HINT:
Split the integral as
$$\int_0^\infty \frac{1}{\sqrt{x^3+x}}\,dx=\int_0^1 \frac{1}{\sqrt{x^3+x}}\,dx+\int_1^\infty \frac{1}{\sqrt{x^3+x}}\,dx$$
For the first integral, note that $$0\le \frac{1}{\sqrt{x^3+x}}\le \frac{1}{\sqrt{x}}$$
For the second integral, note that $$0\le \frac{1}{\sqrt{x^3+x}}\le \frac{1}{ x^{3/2}}$$
A: A simple method is to use the asymptotic equivalence:
$$\frac{1}{\sqrt{x^3+x}}\sim_0 \frac{1}{\sqrt x}$$
and
$$\frac{1}{\sqrt{x^3+x}}\sim_{+\infty}\frac{1}{ x^{3/2}}$$so we conclude the convergence of the given integral since the integrals $\int_0^1 \frac{dx}{\sqrt x}$ and $\int_1^\infty \frac{dx}{ x^{3/2}}$ are convergent.
A: $$\int_0^t\frac{1}{\sqrt{x^3+x}}dx$$
is increasing in $t$.
Besides,
$$\int_0^\infty\frac{1}{\sqrt{x^3+x}}dx=\int_0^1\frac{1}{\sqrt{x^3+x}}dx+\int_1^\infty\frac{1}{\sqrt{x^3+x}}dx$$
$$\le \int_0^1\frac{1}{\sqrt{x}}dx+\int_1^\infty\frac{1}{\sqrt{x^3}}dx$$
$$=2\sqrt{x}\bigg|_0^1+\frac{-2}{\sqrt{x}}\bigg|_1^\infty$$
$$=2+2$$
$$=4$$
A: Through the substitution $x=\tan\theta$ we have:
$$0\leq I=\int_{0}^{+\infty}\frac{dx}{\sqrt{x}\sqrt{x^2+1}}\,dx = \int_{0}^{\pi/2}\frac{d\theta}{\sqrt{\sin\theta\cos\theta}}=\sqrt{2}\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{\sin(2\theta)}}$$
or, by exploiting $\sin(\pi-\theta)=\sin(\theta)$ and the change of variable $\varphi=2\theta$:
$$ I = \sqrt{2}\int_{0}^{\pi/2}\frac{d\varphi}{\sqrt{\sin\varphi}} $$
but $\sin\varphi$ is a concave function on $\left[0,\frac{\pi}{2}\right]$, hence $\sin\varphi \geq \frac{2\varphi}{\pi}$ leads to:
$$ I \leq \sqrt{\pi}\int_{0}^{\pi/2}\frac{d\varphi}{\sqrt{\varphi}}=\pi\sqrt{2}. $$
We may also use Euler's beta function to find the actual value of the integral:
$$ \int_{0}^{+\infty}\frac{dx}{\sqrt{x+x^3}}=\frac{8}{\sqrt{\pi}}\,\Gamma\left(\frac{5}{4}\right)^2. $$
The substitution $x=z^2$ brings such integral into a complete elliptic integral of the first kind: these objects can be computed very fast through the AGM.
$$ I = 2\int_{0}^{+\infty}\frac{dz}{\sqrt{1+z^4}} = 4\int_{0}^{1}\frac{dz}{\sqrt{1+z^4}}<4.$$
A quite tight bound comes from the Cauchy-Schwarz inequality:
$$ I \leq\,4\sqrt{\int_{0}^{1}\frac{dt}{1+t^2}\int_{0}^{1}\frac{1+t^2}{1+t^4}\,dt} = \pi\cdot 2^{1/4}.$$
