Finding out if $\int_{0}^{\infty} \frac{x \arctan{x}}{x^{2}+2x+3} dx$ converges or diverges.

Question

It seems so that the integral diverges but to show that I want to find an $f(x)$ which meets the requirements $f(x) \leq $ the original equation for all $x \in (0,\infty)$ which can be reasonably integrated but im having trouble finding that function. Any help?

Thanks in advance.

Answer 1

hint

For any $x>0$,

$$\arctan(x)=\frac{\pi}{2}-\arctan(\frac 1x)$$

Your integral is then a sum of a divergent and a convergent integral.

You will use $$\arctan(\frac 1x) \sim \frac 1x \; (x\to +\infty)$$

Answer 2

Hint Find $$\lim_{x \to \infty} \frac{ \frac{x \arctan{x}}{x^{2}+2x+3}}{\frac{1}{x}}$$

Answer 3

You don't need to have $f(x)\le$ the original equation for all $x\in(0,\infty)$ - you only need that to be valid for sufficiently large $x$.

Note that:

$$\lim_{x\to\infty}\frac{\frac{x\arctan x}{x^2+2x+3}}{\frac{1}{x}}=\frac{\pi}{2}$$

or:

$$\lim_{x\to\infty}\frac{\frac{x\arctan x}{x^2+2x+3}}{\frac{\pi}{2x}}=1$$

This already means that $\frac{x\arctan x}{x^2+2x+3}\sim\frac{\pi}{2x}$ when $x\to\infty$, which (provided you have access to the corresponding theorem - "limit comparison theorem" for integrals) already implies that the integrals of the left and the right side both converge or both diverge on intervals of the type $(N,\infty)$ for $N$ sufficiently large.

Even if you don't have access to that theorem, recall the $\epsilon$-definition of the limit, which means that, for any $\epsilon>0$ you can find large enough $N$ such that the quotient within the limit is in $(1-\epsilon, 1+\epsilon)$. Take, for example, $\epsilon=\frac{1}{2}$ and there will be large enough $N$ such that, for $x>N$ you have:

$$\frac{\frac{x\arctan x}{x^2+2x+3}}{\frac{\pi}{2x}}>1-\frac{1}{2}=\frac{1}{2}$$

i.e.

$$\frac{x\arctan x}{x^2+2x+3}>\frac{\pi}{4x}$$

for $x>N$. Obviously, the integral $\int_N^\infty\frac{\pi}{4x}dx$ diverges, which proves that the original integral diverges as well.

Answer 4

It diverges because:

• $\lim_{x\to\infty}\arctan x=\frac\pi 2$, so $\arctan x\sim_\infty\frac\pi 2;
• a polynomial is asymptotically equivalent to its leading term, therefore $$\frac{x \arctan{x}}{x^{2}+2x+3}\sim_\infty\frac{\frac\pi2 x}{x^2}=\frac\pi{2x},$$ and the integral of the latter function diverges.

Answer 5

HINT.-Try to show, for example, that for $x\gt 10$ you do have $f(x)\gt \dfrac 1x$ and using the harmonic series deduce the integral does not converge.