The integral of (sin x + arctan x)/x^2 diverges over (0,∞)

[CGandC]

Homework Statement

Prove/Disprove: the integral $J=\int_{0}^{\infty} \frac{\sin x+\arctan x}{x^{2}} d x$ converges.

Relevant Equations

Using integral comparison test in the sense of comparing between integrals, like here https://web.njit.edu/~bg263/Lecture%20notes%20and%20supplements/L19.pdf ( and not in the context of sums )

My attempt:
Disprove. Note that $\int_{0}^{\infty} \frac{-1 - \frac{\pi}{2} }{x^2} d x \leq \int_{0}^{\infty} \frac{\sin x+\arctan x}{x^{2}} d x$ and that $\int_{0}^{\infty} \frac{-1 - \frac{\pi}{2} }{x^2} d x$ diverges, hence by the integral Direct Comparison Test, $J$ diverges.

Was the above proof specious? My professor gave the following proof:

Disprove. In the interval $(0, \frac{\pi}{2}]$ the integrand is positive and also
$\frac{\sin x+\arctan x}{x^{2}} \geq \frac{ \sin x }{ x^2}$
And $\lim _{x \rightarrow 0} \frac{\frac{\sin x}{x^{2}}}{\frac{1}{x}}=\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
Hence by the Limit Comparison Test, $\frac{ \sin x }{ x^2}$ , $\frac{1}{x}$ converge/diverge together.
Since $\int_{0}^{\infty} \frac{1}{x} d x$ diverges, thus $\frac{ \sin x }{ x^2}$ diverges, hence by the integral Direct Comparison Test, the integral in $(0, \frac{\pi}{2}]$ over the integrand $\frac{\sin x+\arctan x}{x^{2}}$ diverges which means $J$ diverges.

 

[Office_Shredder]

I like the professors proof better. Your comparison integral is negative, so the fact that it diverges doesn't say much. The actual integral could be closer to 0 in magnitude. Notice your attempt "proves" that every integral with a positive integrand diverges.

 

[CGandC]

Thanks, I've realized my mistake. The Direct Comparison Test works for positive functions, that is: if $0 \leq f \leq g$ then:
1. if $\int f$ diverges then $\int g$ diverges.
2. if $\int g$ converges then $\int f$ converges.
where the integral is on the desired interval

My mistake was that I took a negative function and tried to apply the theorem which works only for positive functions.

 

[nuuskur]

The same theorem gives you comparison criteria for non-positive case, as well.

 

[CGandC]

Yes, for example if we look at $--f$ and $--g$ where $0 \leq f \leq g$
the theorem would still apply ( we'd have $0 \geq -f \geq -g$ ) but the results would require me to look at integrals over positive integrals.

In general, I think the comparison criteria for negative functions could be stated symmetrically as:
if $0 \geq f \geq g$ then:
1. if $\int g$ diverges ( to $-\infty$ ) then $\int f$ diverges.
2. if $\int f$ converges then $\int g$ converges.
where the integration is over the desired interval.

or the other way around ( I think this one is the correct version):
1. if $\int f$ diverges ( to $-\infty$ ) then $\int g$ diverges.
2. if $\int g$ converges then $\int f$ converges.

 

[Office_Shredder]

or the other way around ( I think this one is the correct version):
1. if $\int f$ diverges ( to $-\infty$ ) then $\int g$ diverges.
2. if $\int g$ converges then $\int f$ converges.

Yes, this is right.

A common way to refer to this is as an absolutely convergent condition. The integral of f converges absolutely if the integral of |f| converges. If $|g|\leq |f|$ and f converges absolutely, then g converges absolutely.