Integration divided by the function How do I guarantee that
$
\frac{\int_0^v f(x) dx}{f(v)}
$
is increasing? Under which assumptions is this true? Or, what types of properties would such a function have? Thanks.
 A: Suppose $f$ is a $C^1$ function on some open interval $I$. Then applying the quotient rule we find that
$$
\frac{d}{dx}\left(\displaystyle\frac{\int_0^v f(x)\,dx}{f(v)}\right) = \frac{[f(v)]^2 - f'(v)\int_0^v f(x)\,dx}{[f(v)]^2}
$$
Therefore, this function is increasing on $I$ if
$$
[f(v)]^2 - f'(v)\int_0^v f(x)\,dx \geq 0 ~~\forall v\in I
$$
This inequality is satisfied, for instance, if $f$ is additionally a decreasing function on $I$. Since the integral
$$
F(v) = \int_0^v f(x)\,dx
$$
represents the area under $f$ over the interval $[0,v]$ (assuming $f$ is positive), it is clear that $F$ can only increase as $v \rightarrow\infty$. Therefore, we may conclude that $F(v)/f(v)$ is increasing on an interval $I$ if $f$ is a decreasing, strictly positive, continuous function on $I$. Of course, there may be much more general conditions under which $F(v)/f(v)$ is increasing. 
As an example, consider the function $f(v) = 1/(v + 1)$. It follows that
$$
\frac{F(v)}{f(v)} = (v + 1)\ln(v + 1)
$$
which is clearly increasing on $[0,\infty)$.
