Uniqueness of a function that satisfies a condition

Let $F:[0, \infty)\to\mathbb R$ such that \begin{align} \int_0^\infty F(t)e^{-st}(-1+st)dt=\frac{1}{s^2}. \end{align} One possibility to solve the previous problem is to have $F(t)=t$ (I can see the integral as an expectation respect to an exponential distribution). There exists another function $F$ that satisfies that condition?

Solution 1:

Note that we can write

$$\begin{align} \int_0^\infty F(t)(-1+st)e^{-st}\,dt&=-\int_0^\infty F(t)e^{-st}\,dt-s\frac{d}{ds}\int_0^\infty F(t)e^{-st}\,dt\\\\ &=-sf'(s)-f(s)\\\\ &=-(sf(s))'\tag1 \end{align}$$

where $f(s) =\int_0^\infty F(t)e^{-st}\,dt$ is the Laplace Transform of $F$.

Setting the right-hand side of $(1)$ equal to $\frac{1}{s^2}$, we find by integrating that

$$f(s)=\frac1{s^2}+\frac{C}{s}\tag2$$

where $C$ is an integration constant.

Taking the inverse Laplace Transform of $(2)$, we find that

$$F(t) =t+C$$

for any constant $C$.