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Find the Laplace of 1

Working

Definition of Laplace Transform:

{$$F(s)=\int _0^\infty f(t)e^{-st}\, dt\$$}

To find the Laplace transform of {$f(t) = 1$}

{\begin{align} F(s)=\int _0^\infty 1e^{-st}\, dt\ \cr \end{align}}

Solving by parts:

{\large \begin{align} F(s)=\int _0^\infty 1e^{-st}\, dt = \int _0^\infty uv' \, dt\ &= \left. uv \right\rvert ^\infty _0 - \int _0^\infty u^\prime v \, dt\ \cr \int _0^\infty 1e^{-st} \, dt\ &= \left. 1\frac{e^{-st}}{-s} \right\rvert ^\infty _0 - \int _0^\infty 1^\prime e^{-st} \, dt\ \cr &= \left. \frac{e^{-st}}{-s} \right\rvert ^\infty _0 - 0\,,\quad s \ne 0 \cr &= \lim_{R\rightarrow\infty} \left. \frac{e^{-sR}}{-s} \right\rvert ^R _0 \cr &= \lim_{R\rightarrow\infty} \frac{e^{-sR}}{-s} - \frac{e^{-s0}}{-s} \cr &= \lim_{R\rightarrow\infty} \frac{e^{-sR}}{-s} - \frac{1}{-s} \cr &= \lim_{R\rightarrow\infty} \frac{e^{-sR} - 1}{-s} \cr &= \frac{ - 1}{-s}\,,\quad s > 0 \cr \therefore F(s) &= \frac{1}{s}\,,\quad s > 0 \end{align}}

Solving by parts is an unnecessary complication here, but was done for exercise.

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