Prove that $\int_0^{\infty} \frac{1}{\sqrt{x^3 + x}}dx$ is convergent

Question

Could you tell me how to prove that $$\int_0^{\infty} \frac{1}{\sqrt{x^3 + x}}dx$$ is convergent?

Answer 1

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 $$0\le \frac{1}{\sqrt{x^3+x}}\le \frac{1}{ x^{3/2}}$$

Answer 2

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.

Answer 3

$$\int_0^t\frac{1}{\sqrt{x^3+x}}dx$$ is increasing in $t$.

Besides, $$\int_0^\infty\frac{1}{\sqrt{x^3+x}}dx=\int_0^1\frac{1}{\sqrt{x^3+x}}dx+\int_1^\infty\frac{1}{\sqrt{x^3+x}}dx$$ $$\le \int_0^1\frac{1}{\sqrt{x}}dx+\int_1^\infty\frac{1}{\sqrt{x^3}}dx$$ $$=2\sqrt{x}\bigg|_0^1+\frac{-2}{\sqrt{x}}\bigg|_1^\infty$$ $$=2+2$$ $$=4$$

Answer 4

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: $$ I \leq\,4\sqrt{\int_{0}^{1}\frac{dt}{1+t^2}\int_{0}^{1}\frac{1+t^2}{1+t^4}\,dt} = \pi\cdot 2^{1/4}.$$