Rough bounds on the value 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.
 A: Some random thoughts. 


*

*Rewrite the equation for $f'$ as
$$ f'(v,m) = 1 + \left(\frac{m}{v} - 1\right) f(v,m) $$
which implies that $f' = 0 \iff f = \frac{1}{1 - m/v}$. Observe that $f$ by definition is non-negative, and $f(1,m) = 0$. This implies that if $f' \neq 0$ anywhere, $f$ must be strictly increasing. But $(1-m/v)^{-1}$ is strictly decreasing in $v > m$, so if $f'\neq 0$ anywhere, we must have that $f \leq \inf_{v > m} (1-m/v)^{-1} = 1$. 

*Suppose $f'(v^*,m) = 0$. Then 
$$ f''(v^*,m) = \left(\frac{m}{v^*} - 1 \right) f'(v^*,m) - \frac{m}{(v^*)^2} f(v^*,m) < 0 $$
hence any critical point of $f$ must be a maximum. This implies that the maximum, when attained at $v^* < \infty$, is unique. (And in particular answers your second question.) 


Now, for $v > m$, $f'(v,m) < 1$. So we must have 
$$ f(v^*,m) - f(m,m) \leq (v^*-m) = m\left( f(v^*,m)-1\right)^{-1}$$
which can give a crude estimate of $f(v^*,m)$ using the quadratic formula
$$ f(v^*,m) \leq 1 + f(m,m) $$
What remains is to get a good estimate on $f(m,m)$. Observe that for $u < m$, $e^u/u^m$ is decreasing and convex. Just by estimating the integral via a trapezoid, you have that
$$ f(m,m) \leq \frac{m^m}{e^m} \frac{(m-1)}{2} \left( e + \frac{e^m}{m^m}\right) = \frac{m-1}{2} \left( 1 + m^m e^{1-m}\right) $$
which gives that, for $m = 3$
$$ f(v^*,3) \leq 1 + \frac{2}{2} (1 + 3^3 e^{-2}) < 5.66 $$
and for $m = 4$
$$ f(v^*,4) \leq 1 + \frac{3}{2} (1 + 4^4 e^{-3}) < 21.7 $$
which are pretty reasonable estimates, I think. 
