How can you show
$$\int_x^\infty \frac{e^{-t}}{t} dt\geq \log(1/x)-1 \text{ for }x>0$$?
It's quite straigtforward to prove the estimation in the other direction for log(1/x)+1, but I fail at seeing a lead how to prove this. So any hint is appreciated.
Let $f\colon (0,\infty) \to \mathbb{R}$ be defined by $f(x) = \int_x^\infty \frac{e^{-t}}{t}dt + \log x +1$. We want to show $f$ is non-negative.
It is clearly at least $\mathcal{C}^1$, so differentiating we get $$ f^\prime(x) = \frac{1-e^{-x}}{x} > 0\;; $$ so that is "suffices" to compute the limit of $f$ when $x\searrow 0^+$ (if it exists). Observe first that $\int_1^\infty \frac{e^{-t}}{t}dt = -\mathrm{Ei}(-1)$; we can focus on $g(x) = \int_x^1 \frac{e^{-t}}{t}dt + \log x = \int_x^1 \frac{e^{-t}-1}{t}dt$ (for $0< x < 1$. Then, $$\begin{align} g(x) &= \int_x^1 \sum_{n=1}^\infty \frac{(-1)^n t^{n-1}}{n!} dt = - \int_x^1 \sum_{n=0}^\infty \frac{(-1)^n t^{n}}{(n+1)!} dt = \sum_{n=0}^\infty \frac{(-1)^n}{(n+1)!}\int_{1}^x t^n dt \\ &= \sum_{n=0}^\infty \frac{(-1)^n}{(n+1)!}\frac{x^{n+1}-1}{n+1} \xrightarrow[x\to 0]{} \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(n+1)(n+1)!} = \mathrm{Ei}(-1) - \gamma \end{align}$$ where $\gamma$ is the Euler constant, and the swapping $\int\sum=\sum\int$, the limit and the last equality need justification (but are not too hard). Putting it together, $$ f(x) \xrightarrow[x\to 0]{} -\mathrm{Ei}(-1) + \mathrm{Ei}(-1) - \gamma + 1 = 1-\gamma > 0. $$