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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!

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    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)...2012-01-26
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    @N.s: but $\int_{0}^{a} \frac{dx}{\sqrt{x^3}}$ doesn't converge..2012-01-26
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    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...2012-01-26
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    Thanks! :) so this kind of an answer (after the correction) would be enough in your opinion?2012-01-26
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    Well, it also matters a lot how you write the answer ;) But basically this is the idea, split it in two and compare each of them with the right thing....2012-01-26
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    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 $1/2$.2012-01-26
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    Thanks a lot. Can I conclude that an integral such as $\int_{0}^{a} f(x)$ for example when it is improper only at 0, converges if the limit of $f(x)$ when x tends to 0 is not 0?2012-01-26
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    and what is actual problem when its limit it is 0?2012-01-26
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    If $\lim_{x\rightarrow 0}f(x)=0$, the integral $I=\int_{0}^{a}f(x)dx$ is not improper at $x=0$. $I$ is an improper integral at $x=0$ only if $\lim_{x\rightarrow 0}f(x)=\infty $. And it may converge or not.2012-01-26
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    @AméricoTavares: thank you very much. This comment and your answer were very helpful.2012-01-27
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    @AméricoTavares: Regarding to what you said in your last comment. here's an example for a function which $\lim f(x)_{x \to 0} =0$ and it is still counted as an improper integral at $x=0$: $\int_{0}^{1}\frac{x}{\sqrt{1-x^4}}$- Why is that? Thanks!2012-01-27
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    The integral $\int_{0}^{1}\frac{x}{\sqrt{1-x^{4}}}dx$ is improper at $x=1$ but not at $x=0$.2012-01-27
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    BTW, $\int_0^\infty\frac{dx}{\sqrt{x^3+x}}=\frac{8}{\sqrt{\pi}}\Gamma\left(\frac{1}{4}\right)^2$2014-03-03

2 Answers 2

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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.

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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 $1/ x^{0.5}$)