Leading term in exponential integral

Question

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

Answer 1

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.

Answer 2

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