# Question regarding $\mathcal{L}\{f'(t)\}$.

Based on the Laplace Transform page in Wikipedia, we can find the function $$\mathcal{L}\{f'(t)\}$$ by solving $$\mathcal{L}\{f(t)\}$$ using Integration by Parts:

\begin{align*} \mathcal{L}\{f(t)\} &= \int_{0^-}^{\infty} e^{-st} f(t) \ dt \\ \\ &= \left[\dfrac{f(t)e^{-st}}{-s} \right]_{0^-}^{\infty} - \int_{0^-}^{\infty} \dfrac{e^{-st}}{-s} f'(t) \ dt \\ \\ &= \left[-\dfrac{f(0^{-})}{-s} \right] + \dfrac{1}{s}\mathcal{L}\{f'(t)\} \ , \end{align*}

and isolating $$\mathcal{L}\{f'(t)\}$$. But on the second line, we evaluate $$\left[\dfrac{f(t)e^{-st}}{-s} \right]_{0^-}^{\infty}$$ as

\begin{align*} \left[\dfrac{f(t)e^{-st}}{-s} \right]_{0^-}^{\infty} &= \lim_{N\rightarrow\infty} \left[\dfrac{f(t)e^{-st}}{-s} \right]_{0^-}^{N} \\ \\ &= \lim_{N\rightarrow\infty} \left[\dfrac{f(N)e^{-sN}}{-s} - \dfrac{f(0^-)e^{-s(0^-)}}{-s} \right] \\ \\ &= \left[0 - \dfrac{f(0^-)}{-s} \right] \\ \\ &= \left[ - \dfrac{f(0^-)}{-s} \right] \ , \end{align*}

assuming that $$s > 0$$. My question is, how is it true that $$\lim_{N \rightarrow \infty}\dfrac{f(N)e^{-sn}}{-s} = 0$$ for all real functions $$f(N)$$? What if $$f(N) = N^N$$? Wouldn't the first term diverge if that's the case?

• Presumably we are assuming ${\cal L}\{f\}$ exists, so its integrand tends to $0$.
– anon
Apr 29, 2021 at 7:50