# Converging integrals?

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#1
If I have the integral of 1/[(x^2)*(e^(1/x))], with integration limits of 0 and infinity, how do I find out if it converges or not?
0
6 years ago
#2
(Original post by towels)
If I have the integral of 1/[(x^2)*(e^(1/x))], with integration limits of 0 and infinity, how do I find out if it converges or not?
Just a suggestion - will the sub u = 1/x help you out?
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#3
(Original post by davros)
Just a suggestion - will the sub u = 1/x help you out?
I don't think so. That would rearrange the integrand to be 1/[(x^2)(e^u)] or u/[x*(e^u)]. I don't see how that would help with finding out more about the convergence of the integral.
0
6 years ago
#4
(Original post by towels)
I don't think so. That would rearrange the integrand to be 1/[(x^2)(e^u)] or u/[x*(e^u)]. I don't see how that would help with finding out more about the convergence of the integral.
Try that again. You will find that it simplifies nicely.
0
6 years ago
#5
Evaluation aside, the integral converges if the limit defining it exists and is finite.

Spoiler:
Show
1/x should work

dx=-x2du
0
6 years ago
#6
(Original post by towels)
I don't think so. That would rearrange the integrand to be 1/[(x^2)(e^u)] or u/[x*(e^u)]. I don't see how that would help with finding out more about the convergence of the integral.
It should work straight away - what is du in terms of dx?
0
6 years ago
#7
An alternative way of looking at this is (I`m just saying) to note what the series for is, which is:

multiplying this by the we get:

the last line in turn is greater than or equal to:

this means that:

The right-hand side of this is an ArcTan expression which converges , between the 0 and infinity limits.

(just an alternative - if somewhat long-winded) way of doing this
0
6 years ago
#8
(Original post by towels)
If I have the integral of 1/[(x^2)*(e^(1/x))], with integration limits of 0 and infinity, how do I find out if it converges or not?
0
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