# Is $\int_0^\infty\frac{\arctan x}{\sqrt{x^3+x}}dx$ converges/diverges

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.

• Can't you also use the comparison test for this? Apr 6 '20 at 3:42

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.

• 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
– Alon
Apr 6 '20 at 3:43
• Oh maybe i can write: $\frac{arctanx}{\sqrt{x^3+x}} < \frac{\pi/2}{\sqrt{x^3}}$
– Alon
Apr 6 '20 at 3:45
• Then use the comparison test
– Alon
Apr 6 '20 at 3:45
• @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. Apr 6 '20 at 5:27
• 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?
– Alon
Apr 6 '20 at 5:30

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

• 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?
– Alon
Apr 6 '20 at 5:34
• For $x\to \infty$, $x^3\gg x$. For $x\to 0$, $x^3+x=x(x^2+1) \approx x$. Apr 6 '20 at 17:42
• 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
– Alon
Apr 6 '20 at 17:47
• 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 $\epsilon \gt 0$. Apr 6 '20 at 18:19

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 $$+\infty$$.

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.