Improper Integral (comparison test q)

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Homework Help Overview

The discussion revolves around determining the convergence or divergence of the improper integral \(\int \frac{2 + e^{-x}}{x} \, dx\) from 1 to infinity, using the Comparison Theorem.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss potential comparisons for the integral, with some suggesting \(\frac{2 + e^{-x}}{x} \leq \frac{3}{x}\) and others exploring the implications of the behavior of \(e^{-x}\) over the interval. Questions arise about the correctness of inequalities and the approach to integrating the function.

Discussion Status

The discussion is ongoing, with participants providing various comparisons and questioning assumptions about the integral's behavior. Some guidance has been offered regarding the direction of inequalities and the integrability of components of the function.

Contextual Notes

Participants note the challenge of function comparisons and the integral's divergence as indicated by external sources, which influences their exploration of appropriate comparisons.

coals
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Homework Statement



Use the Comparison Theorem to determine whether the integral is convergent or divergent.
[tex] \int \frac{2+ e^{-x} dx}{x} [/tex]
from 1 to infinity

Wolframalpha tells me this integral diverges, now i just need to know what to compare it to.

The Attempt at a Solution



So far I've said over the interval [1,infinity) [tex] \frac{2+ e^{-x}}{x} <= \frac{e^{-x}}{x}[/tex]

When i try to take the limit (t->inf) integral (1,t) of [tex]\frac{e^{-x}}{x}[/tex] i can't integrate this (tried IBP, don't see any obvious trig subs).

Am i just doing it wrong?

Thanks for your help everyone.

That's suposed to be e^(-x) btw i fail at this.
 
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coals said:
So far I've said over the interval [1,infinity) [tex] \frac{2+ e^{-x}}{x} <= \frac{e^{-x}}{x}[/tex]

You might want to recheck the direction of that [itex]\leq[/itex] sign :wink:... Is [itex]e^{-x}[[/itex] ever greater than one on your interval? If not, then surely you can say [itex]\frac{2+ e^{-x}}{x} \leq \frac{3}{x} [/tex][/itex]
 
To expand on gabba's post:
We know that [tex]e^{-x}[/tex] is integrable, so dividing it by isn't likely to change anything. Since we know (from Wolfram) that the integral diverges, we should look at the other half of the function (the constant divided by x)
 
Why not start with the comparison 2 + e-x > 1, then divide by x and compare the integrals?
 
(2+e^(-x))/x > 2/x
 
Thank you all for your help. Very appreciated , i got the answer and i understand what to do now in similar situations. Doing function comparisons seems to be my weak spot I'm alright with everything else but i suck at these.
 

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