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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?

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

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