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Does the following limit converges to $0$ or not? $$ \lim_{N\to \infty} e^N \int_{N}^{\infty} \frac{e^{-x}}{\log(x)} dx$$ I am interested whether the limit is $0$ or not. How can I get the answer?

I have tried this: $\int_{N}^{\infty} \frac{e^{-x}}{\log(x)} < \frac{1}{\log(N)} \int_{N}^{\infty}e^{-x} = \frac{e^{-N}}{\log(N)}$ hence $$I_N < \frac{1}{\log(N)}$$ therefore $I_N \to 0$ for $N \to \infty$. Is it correct?

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    $\begingroup$ @Frpzzd The OP has added his/her efforts! Yay! $\endgroup$ – Simply Beautiful Art Jun 7 '17 at 22:26
  • $\begingroup$ @CMarius Yes, your work is indeed correct ;) $\endgroup$ – Simply Beautiful Art Jun 7 '17 at 22:27
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HINT:

Rephrase the problem to get $$\lim_{N\to\infty} \frac{\int_N^\infty \frac{e^{-x}}{\ln x}dx}{e^{-N}}$$ Now we have a fraction whose limits on the numerator and denominator are both zero, so perhaps you can use L'Hopital's rule to evaluate it.

Stop reading now if you want to try it yourself.

FULL ANSWER: Let $$f(x)=\frac{e^{-x}}{\ln x}$$ and let $F(x)$ be its antiderivative. Then we have $$\lim_{N\to\infty} \frac{\lim_{k\to\infty} F(k)-F(N)}{e^{-N}}$$ To make a long story short, the integral converges, so $\lim_{k\to\infty} F(k)$ is a constant and when we take the derivative of the top and bottom, we get $$\lim_{N\to\infty} \frac{-f(N)}{-e^{-N}}$$ $$\lim_{N\to\infty} \frac{\frac{e^{-N}}{\ln N}}{e^{-N}}$$ $$\lim_{N\to\infty} \frac{1}{\ln N}$$ Which is zero.

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  • $\begingroup$ Squeeze is still better :-P $\endgroup$ – Simply Beautiful Art Jun 7 '17 at 22:36
  • $\begingroup$ Replace L'Hopital by Stolz-Ces$\mathrm{\grave{a}}$ro, please. Nice answer. $\endgroup$ – Felix Marin Jun 9 '17 at 2:13
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Your approach is correct. The wanted limit is $$ \lim_{N\to +\infty}\int_{N}^{+\infty}\frac{e^{N-x}}{\log x}\,dx = \lim_{N\to +\infty}\int_{0}^{+\infty}\frac{e^{-u}}{\log(u+N)}\,du $$ and for any $N>1$ we have $$ 0\leq \int_{0}^{+\infty}\frac{e^{-u}}{\log(u+N)}\,du \leq \frac{1}{\log N}\int_{0}^{+\infty}e^{-u}\,du = \frac{1}{\log N} $$ hence the limit is trivially $\color{red}{0}$ by squeezing. Indeed $$ \int_{N}^{+\infty}\frac{dx}{e^x \log x} \sim \frac{1}{e^N \log N} $$ by the dominated convergence theorem.

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  • $\begingroup$ I find this to a be a roundabout restatement of the OP's work. $\endgroup$ – Simply Beautiful Art Jun 7 '17 at 22:37
  • $\begingroup$ @SimplyBeautifulArt : I hope it is not anymore with the last addendum. $\endgroup$ – Jack D'Aurizio Jun 7 '17 at 22:42
  • $\begingroup$ Eh, not quite as fabulous as I'd like to hold you up to, so no +1 on this one today :D $\endgroup$ – Simply Beautiful Art Jun 7 '17 at 22:43

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