Is the following integral greater than or equal to zero? I have the following integral: 
$$\int_0^{\infty} (z-1) e^{-a z} f(z) {\rm dz}$$
$a \in [0,1]$. I want to test whether the integral is non negative. I know that $f(z): \mathbb{R^+} \to [0,1]$ is a smooth and an increasing function.
Intuitively I can say yes, because the integrand is negative for small interval, and whether $f(z)$ increases slowly or quickly does not change the result.
I am searching for a bit more rigorous answer. Is my intuition justified, and how to find a counter example if it is not?
Thanks
 A: $\int_{0}^{+\infty}(z-1)e^{-az}f(z)\,dz$ is an integral, integration stands for the process of computing/studying integrals. So we want to prove that such integral is non-negative, under the assumptions that $f$ is smooth, increasing and non-negative. Obviously $f$ cannot grow faster than $e^{az}$, since otherwise the integral would be divergent. We have
$$ \int_{1}^{+\infty}(z-1)e^{-az}f(z)\,dz \geq f(1)\int_{1}^{+\infty}(z-1)e^{-az}\,dz = \frac{f(1)}{a^2 e^a}$$
hence it is enough to prove that
$$ \int_{0}^{1}(1-z) e^{-az}\,f(z)\,dz \leq f(1)\frac{e^{-a}}{a^2}$$
however that is not always granted. We may state that
$$ \int_{0}^{1}(1-z)e^{-az}f(z)\,dz \leq f(1)\int_{0}^{1}(1-z)e^{-az}\,dz = f(1)\frac{e^{-a}+(a-1)}{a^2}$$
but the shown inequalities are sharp if $f(z)$ quickly grows from $0$ to $1-\varepsilon$ on $z\in(0,\varepsilon)$ then stays bounded by $1$ on $z\in(\varepsilon,+\infty)$. An explicit counter-example for $a=2$ is provided by $f(z)=1-e^{-2z}$.
On the other hand, if we further assume $a\in(0,1]$ we have no issues anymore:
$$ \int_{1}^{+\infty}(z-1)e^{-az}f(z)\,dz \geq f(1)\frac{e^{-a}}{a^2} \geq f(1)\frac{e^{-a}+(a-1)}{a^2} \geq \int_{0}^{1}(1-z)e^{-az}f(z)\,dz.$$
A: This is not true. Using integration by parts, we can calculate
$$\int_0^\infty(z-1)e^{-az}dz=\frac1a-\frac1a\int_0^\infty e^{-az}dz=a^{-1}(1-a^{-1})$$
which is negative for all $a\in(0,1)$. So choosing $f=1$ (or, if you require strictly increasing, choose $f$ such that $f>1-\varepsilon$ for appropriately small $\varepsilon>0$) the integral is negative.
