Improper Integral Sinx/x^2 and similar with sinx

[mekise]

Homework Statement

[/B]
This is the improper integral of which I have to study the convergence.

∫[1,+∞] sinx/x² dx

The Attempt at a Solution

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I have tried to use the absolute convergence.

∫f(x)dx converges ⇔ ∫|f(x)|dx converges

but after i have observed that x^2 is always positive and the absolute value is only for the sinx, I do not how to move!

How to see the convergence or the divergence of sinx when you can not to approximate to x for x->0?

Thank you for the help!

 

[bhobba]

Try the p test, then the comparison test for absolute convergence.

Thanks
Bill

 

[ecastro]

Try starting with the Squeeze Theorem.

 

[ecastro]

Try starting with the Squeeze Theorem.

 

[mekise]

I tried the Squeeze Theorem.

-1≤sinx≤+1 ⇒ -1/x²≤sinx/x²≤+1/x²

the two functions converge so my function converge!

Thank you very much!

 

[bhobba]

I tried the Squeeze Theorem.

-1≤sinx≤+1 ⇒ -1/x²≤sinx/x²≤+1/x²

the two functions converge so my function converge!

I gave you a like because you are on the right track - but that is wrong - the squeeze test applied to functions that converge to the same thing.

Like I said use the p test on 1/x^2, note that the absolute value of your function is less than or equal to 1/x^2. Thus its absolute value converges by the comparison test, then apply the absolute convergence test.

Thanks
Bill

 

[STEMucator]

You have already pointed this out:

$$\int_{1}^{\infty} \left| \frac{\text{sin}(x)}{x^2} \right| \space dx \leq \int_{1}^{\infty} \frac{1}{x^2} \space dx$$