In one book the author defines the following value: $ F(m) = \sup\limits_{v\geq 1}\frac{v^m}{e^v}\int\limits_1^v\frac{e^u}{u^m}\mathrm du $ for $m\in [3,4]$. Further he puts an upper bound for this value: $ F(m)\leq 1+\frac m2+\frac m2(m+1)^{m+1}e^{-m}, $ so e.g. for $m = 3$ on has $F(M)\leq 21.67$. This bound seems to be too rough: I do not know the actual value for $F(3)$ but at lease Mathematica provides the answer $\approx3.58$.
It's clear that I cannot rely upon this approximation and I would like to find more nice bounds for $F(m)$ especially for $m\in [3,4]$.
If one define $ f(v,m) = \frac{v^m}{e^v}\int\limits_1^v\frac{e^u}{u^m}\mathrm du $ then f'(v,m) = 1+\frac{v^{m-1}(m-v)}{e^v}\int\limits_1^v\frac{e^u}{u^m}\mathrm du where the deirvative is taken w.r.t. $v$. Since we are interested only in the case $v\geq 1$ then for $v\in[1,m]$ clearly f'\geq 1>0. So there are two questions:
if there $v^*$ such that f'(v^*,m) = 0?
if it is unique?
These questions are strongly related to the behaviour of $ \frac{v^{m-1}(m-v)}{e^v}\int\limits_1^v\frac{e^u}{u^m}\mathrm du $ which I cannot understand well. E.g. calculation of f'' does not help much to me (again, the derivative is w.r.t $v$).
Edited: thanks to Willie Wong, I've fixed the bound for $F(3)\leq 21.67$. This bound is still to rough since I need to operate thereafter with values like $\exp(F(m))$ where the difference between numbers $3.58$ and $21.67$ is significant.