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Use comparison theorem to show if integral is convergent or divergent

  • Thread starter zero13428
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  • #1
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Homework Statement



int (e^-x)/(x)dx from 0 to infinity

Determine if integral is convergent or divergent


2. The attempt at a solution
I assume because the bottom limit is 0 and there is an x in the bottom of the integral that this is going to be divergent but I still have to use the comparison theorem. I'm trying to find a function less then (e^-x)/(x) that also diverges to show that (e^-x)/(x) will diverge but I'm drawing a blank. I've tried messing around with 1/x^2 or just e^-x but both of those are still larger then the original (e^-x)/(x). Any suggestions?
 

Answers and Replies

  • #2
Office_Shredder
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1/x is an interesting function in that it diverges both ways, the integral from 0 to 1 of 1/x is divergent, as is the integral from 1 to infinity (compare this to [tex]x^{1/2}[/tex] and [tex] \frac{1}{x^2}[/tex]). So when trying to do comparison with a 1/x term, it often helps to break up the integral into one from 0 to 1 and one from 1 to infinity, and see on each of those whether it converges or diverges
 
  • #3
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Ok, so if I can break the function into two parts, as long as one part (in this case 1/x) converges or diverges then the entire function does?

I thought I could only do these type of integrals by picking a function that was larger or smaller and then comparing.
 
  • #4
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Since this is an improper integral at both limits of integration, that's why you should consider the two intervals (0, 1] and [1, ∞) separately. Try finding whether the integral from (0, 1] converges first.
 

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