Does the following improper integral converge ?
$$\int _1^{\infty}\:\cfrac{e^{1/x}-1}{x}dx$$
I have tried to compare it to some known improper integrals but with no luck.
Thanks for helping.
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HINT:
Note that as $x\to \infty$, $e^{1/x}=1+\frac1x+O\left(\frac1{x^2}\right)$.
Can you finish now?
Mark Viola
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$$\int _1^{\infty}:\cfrac{1+1/x-1}{x}dx =\int _1^{\infty}:\cfrac{1}{x^2}dx$$ that converges, is that correct? – Richy65 Jan 24 '21 at 20:56
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2@Richy65 Yes, it converges. Well done! – Mark Viola Jan 24 '21 at 21:01
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Thank you very much – Richy65 Jan 24 '21 at 21:02
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2You're welcome. My pleasure. ;-) – Mark Viola Jan 24 '21 at 21:05