I'd like your help with checking whether $\int_{0}^{\infty} \frac{dx}{\sqrt{x^3+x}}$ converges or not. Here are the steps which led me to conclude that the integral does converge, but I'm not really sure.
First, I saw that I can't easily compute $\int \frac{dx}{\sqrt{x^3+x}}$ so I wanted to see what happens when x tends to infinity. It's not really formal but I can see that the integral "acts" as $\int_{0}^{\infty} \frac{dx}{\sqrt{x^3}}$ so in the infinity it is almost as $\int_{0}^{\infty} \frac{dx}{x^{3/2}}$ and since it's in the format $\int_{0}^{\infty}\frac{1}{x^p}$ where $p>1$ I concluded that it converges. (Is it true? or do I must separate it to two integrals $\int_{0}^{a}\frac{1}{x^p}$ + $\int_{a}^{\infty}\frac{1}{x^p}$ and check both?).
What do you think? what is the correct way to check convergence for this integral?
Thanks a lot!
Best Answer
Since the integrand has a singularity at $x=0$, you must split the integral into two, such as $$ \int_{0}^{\infty }\frac{1}{\sqrt{x^{3}+x}}dx=\int_{0}^{b}\frac{1}{\sqrt{x^{3}+x}}dx+\int_{b}^{\infty }\frac{1}{\sqrt{x^{3}+x}}dx\qquad b>0. $$
As for the first integral use the limit test
$$\lim_{x\rightarrow 0}\frac{\frac{1}{\sqrt{x^{3}+x}}}{\frac{1}{\sqrt{x}}} =\lim_{x\rightarrow 0}\frac{1}{\sqrt{x^{2}+1}}=1$$
to conclude that $\int_{0}^{b}\frac{1}{\sqrt{x^{3}+x}}dx$ is convergent, because so is $\int_{0}^{b}\frac{1}{\sqrt{x}}dx$. The second integral is convergent by the argument you indicated.
Comment: How to choose the function $g(x)=x^{-1/2}$ to compare with $$f(x)=\frac{1}{\sqrt{x^{3}+x}}=x^{-1/2}\frac{1}{\sqrt{1+x^{2}}}\ ?$$ The binomial series $\left( 1+x\right) ^{\alpha }=\sum_{k=0}^{\infty }\binom{\alpha }{k}x^{k}$ yields for $\alpha =-1/2$
$$\frac{1}{\sqrt{1+x}}=1-\frac{1}{2}x+\frac{3}{8}x^{2}+\ldots .$$
Substituting $x^{2}$ for $x$, we get
$$\frac{1}{\sqrt{1+x^{2}}}=1-\frac{1}{2}x^{2}+\frac{3}{8}x^{4}+\ldots $$
Consequently,
$$f(x)=\frac{1}{\sqrt{x^{3}+x}}=x^{-1/2}-\frac{1}{2}x^{3/2}+\frac{3}{8} x^{7/2}+\ldots .$$
The function $g(x)=x^{-1/2}$ is the first term of this expansion.