TITLE: Estimate for differences of exponential integral
QUESTION [5 upvotes]: Since the function $\exp(-t)/t$ is monotone decreasing for $t>0$ it is immediate to get
$$f(x) := \int_x^{2x} \frac{\exp(-t)}{t} dt \le \exp(-x)$$
for $x>0$. Using
$$E_1(x) = -\gamma - \log(x) - \sum_{k=1}^{\infty} \frac{(-x)^k}{k\; k!}$$
for $x>0$ (see Wikipedia for that formular) and realizing
$$f(x)=E_1(x)-E_1(2x)$$
we even get
$$\lim_{x \to 0} f(x) = \log(2).$$
Question:
Is there a $C \in (0,\log(2)]$ such that
$$C \exp(-x) \le f(x) \le \exp(-x)$$
for all $x>0$?

REPLY [2 votes]: We can also show $f(x)/e^{-x}\to 0$ at $\infty$ using L'Hopital. Let $F$ be an antiderivative for $e^{-t}/t.$ Then
$$\frac{f(x)}{e^{-x}} = \frac{F(2x)-F(x)}{e^{-x}}.$$
Applying LHR, we consider
$$\frac{2F'(2x)-F'(x)}{-e^{-x}} = \frac{e^{-2x}/x-e^{-x}/x}{-e^{-x}}.$$
This $\to 0$ at $\infty.$