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Is it generally true that $$\int_t^\infty \exp(-x+o(x))dx = \exp(-t+o(t))$$ as $t \to \infty$.

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For a rigorous argument, let $\eta(x)$ be any measurable function such that $\eta(x)/x \to 0$ as $x\to\infty$. In particular, for each $\epsilon \in (0, 1)$, there exists $R > 0$ such that $|\eta(x)/x| < \epsilon$ whenever $x \geq R$. So, if the function $\delta(t)$ is implicitly defined by the relation

$$ \int_{t}^{\infty} e^{-x+\eta(x)} \, \mathrm{d}x = e^{-t+\delta(t)}, $$

then for $ t \geq R$,

$$ \frac{1}{1+\epsilon} e^{-t - \epsilon t} =\int_{t}^{\infty} e^{-x-\epsilon x} \, \mathrm{d}x \leq \int_{t}^{\infty} e^{-x+\eta(x)} \, \mathrm{d}x \leq \int_{t}^{\infty} e^{-x+\epsilon x} \, \mathrm{d}x = \frac{1}{1-\epsilon} e^{-t + \epsilon t} $$

and thus

$$ -\epsilon -\frac{\log(1+\epsilon)}{t} \leq \frac{\delta(t)}{t} \leq \epsilon -\frac{\log(1-\epsilon)}{t}. $$

From this, it is easy to see that $\limsup_{t\to\infty} |\delta(t)/t| \leq \epsilon$ for any $\epsilon \in (0, 1)$, and so, we get $\delta(t)/t \to 0$ as required.

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  • $\begingroup$ Thanks, that's what I was looking for! $\endgroup$ – Danijel Jun 22 at 23:24
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near $+\infty$,

$$e^{-x+o(x)}\sim e^{-x} $$

and $$\int_0^{+\infty}e^{-x}dx \text{ converges}$$

thus

$$\int_t^{+\infty}e^{x+o(x)}dx \sim \int_t^{+\infty}e^{-x}dx$$

or $$\int_t^{+\infty}e^{x+o(x)}dx \sim e^{-t}\sim e^{-t+o(t)}.$$

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  • $\begingroup$ Just one thing that still puzzles me. Why is $\exp(-x+o(x)) \sim \exp(-x)$. In which sense do you mean $\sim$? $\endgroup$ – Danijel Jun 22 at 21:59
  • $\begingroup$ $e^{o(x)}$ goes to $1$ when $x$ goes to $+\infty$. $\endgroup$ – hamam_Abdallah Jun 22 at 22:04
  • $\begingroup$ Hmm not necessarily... What if $o(x)=\sqrt{x}$. Then $\exp(o(x)) \to \infty$. $\endgroup$ – Danijel Jun 22 at 22:08

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