towels
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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?
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davros
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(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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towels
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(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.
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BabyMaths
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(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.
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Implication
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Evaluation aside, the integral converges if the limit defining it exists and is finite.

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1/x should work

dx=-x2du
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davros
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(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?
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Hasufel
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An alternative way of looking at this is (I`m just saying) to note what the series for e^{1/x} is, which is:

\displaystyle 1+\frac{1}{x}+\frac{1}{2x^{2}}+ \frac{1}{6x^{3}} +\frac{1}{24x^{4}}+\frac{1}{120x  ^5}+.....

multiplying this by the x^{2} we get:

\displaystyle x^{2}+x+\frac{1}{2}+ \frac{1}{6x}+\frac{1}{24x^{3}}+ \frac{1}{120x^{4}}+...

the last line in turn is greater than or equal to: x^{2}+x+ \frac{1}{2} \equiv (x+\frac{1}{2})^{2}+\frac{1}{4}

this means that:

\displaystyle \frac{1}{x^{2}e^{1/x}} \leq \frac{1}{ (x+\frac{1}{2})^{2}+\frac{1}{4}}

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
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ztibor
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(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?
\displaystyle \int_{0}^{\infty} \frac{1}{x^2}\cdot e^{\frac{1}{x}} dx =\lim_ {b \to \infty \\ a\to 0} \int_{a}^{b} \frac{1}{x^2}\cdot e^{\frac{1}{x}} dx =\lim_ {b \to \infty \\ a\to 0} \left [-e^{\frac{1}{x}} \right ]_a^b =\lim_ {b \to \infty \\ a\to 0} (-e^{\frac{1}{b}}+e^{\frac{1}{a}})  =\infty
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