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I need to determine if the following integral converges/diverges.

$$ \int_{0}^{\infty}\frac{\arctan x}{\sqrt{x^3+x}}dx $$

What i tried:

We can write the integral as:

$$ \int_{0}^{\infty}\frac{\arctan x}{\sqrt{x^3+x}}dx = \int_{0}^{\infty}\frac{\arctan x}{\sqrt{x(x^2+1)}}dx $$

Because the $x^2+1$ i thought about defining $x = \tan u$, therefore i will be able to use the identity: $$ \tan^2u + 1 = \frac{1}{\cos^2u} $$

Define: $$ x = \tan u \Rightarrow u = \arctan x $$ $$ x = 0 \Rightarrow u = 0 $$ $$ x \to \infty \Rightarrow u = \pi/2 $$

Therefore we can write the integral as:

$$ \int_{0}^{\infty}\frac{\arctan x}{\sqrt{x(x^2+1)}}dx = \int_{0}^{\pi/2}\frac{u\cdot du}{\sqrt{\tan u(\tan^2u+1)}} $$

$$ = \int_{0}^{\pi/2}\frac{u\cdot du}{\sqrt{\tan u\frac{1}{\cos^2u}}} = \int_{0}^{\pi/2}\frac{u\cos u}{\sqrt{\tan u}}du $$

And here i am stuck.

Another way:

I thought maybe to devide the integral into two, not sure even how, maybe:

$$ \int_{0}^{\infty}\frac{\arctan x}{\sqrt{x^3+x}}dx = \int_{0}^{\pi/2}\frac{\arctan x}{\sqrt{x^3+x}}dx + \int_{\pi/2}^{\infty}\frac{\arctan x}{\sqrt{x^3+x}}dx $$

But i dont see how it gives me something.

Can i have a hint?

Thank you.

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  • $\begingroup$ Can't you also use the comparison test for this? $\endgroup$ Apr 6 '20 at 3:42
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The long term behavior of the function is $\frac{\pi/2}{x^{3/2}}.$ The convergence of $\int_1^{\infty} x^{-3/2} \, dx$ should now tell you something.

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  • $\begingroup$ I see what you say, when $x \to \infty$ we can write $arctanx$ as $\pi/2$ and the dominant variable at the denominator is $x^{3/2}$ but im not sure how to explain this so i can write my term as $\frac{\pi/2}{x^{3/2}}$ If i get to this, surly my integral converges $\endgroup$
    – Alon
    Apr 6 '20 at 3:43
  • $\begingroup$ Oh maybe i can write: $\frac{arctanx}{\sqrt{x^3+x}} < \frac{\pi/2}{\sqrt{x^3}}$ $\endgroup$
    – Alon
    Apr 6 '20 at 3:45
  • $\begingroup$ Then use the comparison test $\endgroup$
    – Alon
    Apr 6 '20 at 3:45
  • $\begingroup$ @Alon Yes, that is a correct estimate, but that only solves half your problem; it ensures that $\displaystyle\int_1^{\infty} \dfrac{\arctan x}{\sqrt{x^3 + x}} \, dx \leq \displaystyle\int_1^{\infty} \dfrac{\pi/2}{x^{3/2}}\, dx < \infty$. But you also have to worry about $\displaystyle\int_0^1 \dfrac{\arctan x}{\sqrt{x^3 + x}}\, dx$ being finite. The answer by herb steinberg addresses this case. $\endgroup$
    – peek-a-boo
    Apr 6 '20 at 5:27
  • $\begingroup$ Thank you, i dont quite understand how i can say $\arctan x$ less or more equal $x$... We are in real analysis, can i say such things? its not... strict enough, isnt it? $\endgroup$
    – Alon
    Apr 6 '20 at 5:30
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As long as you are using the principal value for arctan$x(\le \frac{\pi}{2}$), then the integrand behaves like $x^{-\frac{3}{2}}$ as $x\to \infty$.

For $x\to 0$, arctan$x \approx x$, so the integrand behaves like $\sqrt{x}$.

Therefore you have convergence at both ends

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  • $\begingroup$ Thank you, I dont understand how you formalaize behave like. for the deminator, you just say $\lim_{x \to 0}\sqrt{x^3+x} = \lim_{x \to 0}\sqrt{x^3}$ Is it a correct statment? $\endgroup$
    – Alon
    Apr 6 '20 at 5:34
  • $\begingroup$ For $x\to \infty$, $x^3\gg x$. For $x\to 0$, $x^3+x=x(x^2+1) \approx x$. $\endgroup$ Apr 6 '20 at 17:42
  • $\begingroup$ But as far as i know this is an approximation you can write in something like physics, not, as much as i see, in real analysis, are you sure you can write such things in real analysis? If you do so great, i will take it $\endgroup$
    – Alon
    Apr 6 '20 at 17:47
  • $\begingroup$ The approximations are perfectly valid in real analysis. If you are taking a course where they first come up, you may have to formalize them. For example $x\lt x(x^2+1)\lt x(1+\epsilon)$ for $x\lt \sqrt{\epsilon}$ for any $\epsilon \gt 0$. $\endgroup$ Apr 6 '20 at 18:19
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Note that for $x>0$, $\arctan x < x$ and $\sqrt {x^3 + x} > \sqrt x$, whence $\frac{{\arctan x}}{\sqrt {x^3 + x}} < \sqrt x$ which shows the convergence near $0$. Also for $x>0$, $\arctan x < \frac{\pi }{2}$ and $\sqrt {x^3 + x} > \sqrt {x^3 } = x^{3/2}$, whence $\frac{{\arctan x}}{\sqrt {x^3 + x}} < \frac{\pi }{2}\frac{1}{{x^{3/2} }}$ which shows the convergence at $+\infty$.

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A quick way to show convergence is the limit comparison test.

For $x \to 0^+$ we have

$$\lim_{x\to 0^+}\frac{\arctan x}{x}=1 \text{ and } \lim_{x\to 0^+}\frac{\sqrt x}{\sqrt{x^3+x}} = 1 \Rightarrow \lim_{x\to 0^+}\frac{\frac{\arctan x}{\sqrt{x^3+x}}}{\sqrt x}= 1$$

Since $\int_0^1 \sqrt x\; dx$ is convergent, $\int_0^1 \frac{\arctan x}{\sqrt{x^3+x}}dx$ is convergent, as well.

For $x \to +\infty$ we have

$$\lim_{x\to +\infty}\left(\arctan x\cdot \frac{x^{\frac 32}}{\sqrt{x^3+x}}\right)=\frac{\pi}2\cdot 1\Rightarrow \lim_{x\to }\frac{\frac{\arctan x}{\sqrt{x^3+x}}}{\frac 1{x^{\frac 32}}}= \frac{\pi}2$$

Since $\int_1^{+\infty} \frac{dx}{x^{\frac 32}}$ is convergent, $\int_1^{+\infty} \frac{\arctan x}{\sqrt{x^3+x}}dx$ is convergent, as well.

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