Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 infinty 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!

share|improve this question
2  
Your integral is improper at $0$ to so you need to split it. Your argument then works well for the second integral ("acts like" formally means a type of comparison test)... –  N. S. Jan 26 '12 at 16:31
    
@N.s: but $\int_{0}^{a} \frac{dx}{\sqrt{x^3}}$ doesn't converge.. –  Jozef Jan 26 '12 at 16:34
2  
True, but when $x$ goes to zero which is dominant (larger)? $x$ or $x^3$? ;) The integral doesn't "act" like $\int \frac{dx}{\sqrt{x^3}}$ near zero... –  N. S. Jan 26 '12 at 16:36
2  
The improper integral potentially goes bad at two places, $0$ where the function blows up, and $\infty$ of course. It is usually a good idea to split up such an integral into pieces that only have one bad feature. So for example if we want to study $\int_0^1 \frac{dx}{\sqrt{x(1-x)}}$, it is a good idea to split at some place between $0$ and $1$, say $1/2$. –  André Nicolas Jan 26 '12 at 17:40
1  
@AméricoTavares: thank you very much. This comment and your answer were very helpful. –  Jozef Jan 27 '12 at 8:04

2 Answers 2

up vote 7 down vote accepted

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.

share|improve this answer

You must separate in two integrals for example $0$ to $1$ and $1$ to infinite and both converge then the integral converges. (from $0$ to $1$ the integral is of the same class of $1/ x^{0.5}$)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.