# Leading term in exponential integral

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

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.

• Thanks, that's what I was looking for! – Danijel Jun 22 at 23:24

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

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