I'm trying to find an upper bound for
$f(m, \alpha) := \frac{m\cdot \mathrm{Ei}(m(1-\alpha))}{\exp(m(1-\alpha))}$. Here $m \ge 1$ and $\alpha \in (0, 1)$.
When I compute values of $f$, I see that $f(m, \alpha) \le \frac{2}{1-\alpha}$.
However, I cannot show this rigorously. Could you please help to obtain a tight upper bound? Thank you.
Best Answer
$$f(m, \alpha)=m\frac{ \mathrm{Ei}(m(1-\alpha))}{\exp(m(1-\alpha))}$$ Let $x=m(1-\alpha)$ to make $$g(x, \alpha)=\frac 1{1-\alpha}\, x\,\mathrm{Ei}(x) \,e^{-x}$$ The maximum value of $x\,\mathrm{Ei}(x) \,e^{-x}$ is a bit lower than $1.5$ ($1.4841$ to be precise). So $$f(m, \alpha) \lt \frac {1.4841}{(1-\alpha)}$$