Why does the constant term of the antiderivative disappear when deriving the Laplace transformation of f(t) = t

Question

I am trying to derive the Laplace transformation

\begin{align} f(t)&=t\\ F(s)&=\int_0^\infty t\cdot e^{-st}\;\mathrm{d}t.\\ \end{align}

Using the partial integration rule

\begin{align} \int f'g = \left[fg\right]_0^\infty - \int fg', \end{align}

with $f' = e^{-st}, f = \tfrac{-1}{s}e^{-st} + C$ and $g = t, g' =1$. In every derivation I come across the constant $C$ is gone.

My question is: where does the constant $C$ go? What step am I missing?

Answer 1

When evaluating definite integrals, we ignore constants of integration. We do this because we'd be writing $C-C$ which is zero. As $C$ is independent of $x$ or whatever independent variable is chosen, we will always end up getting zero in the evaluation step.