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

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In summary, the professor's proof is better because it uses a limit comparison test which is specific to positive functions. Additionally, the comparison integral is negative, which is significant.
  • #1
CGandC
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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.
 
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  • #2
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.
 
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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.
 
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  • #4
The same theorem gives you comparison criteria for non-positive case, as well.
 
  • #5
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.
 
  • #6
CGandC said:
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.
 
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1. What is the meaning of "diverges" in this context?

In mathematics, "diverges" means that a sequence or series does not have a finite limit. In this case, the integral does not have a finite value and therefore does not converge.

2. What is the significance of the interval (0,∞)?

The interval (0,∞) represents all positive real numbers. In this context, it indicates that the integral is being evaluated from 0 to infinity.

3. Why does the integral of (sin x + arctan x)/x^2 diverge?

The integral of (sin x + arctan x)/x^2 diverges because the function is not integrable at x=0. This means that the function has a singularity at x=0 and therefore the integral cannot be evaluated.

4. Can the integral be evaluated using other methods?

No, the integral of (sin x + arctan x)/x^2 cannot be evaluated using other methods. The divergence at x=0 is a fundamental issue that cannot be resolved by any other mathematical techniques.

5. Are there any practical applications for this concept?

Yes, the concept of divergence in integrals has practical applications in physics and engineering. It can be used to analyze systems that have infinite or singular solutions, such as in the study of electric fields or fluid dynamics.

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