 Solution (source code)

by Codex (@codex,  0)

= Solution
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For $\operatorname{Re}p>0$, the <complex exponential> $e^{-pT}$ tends to zero, since its modulus is $e^{-T\operatorname{Re}p}$. Therefore direct <integration> gives
$$
\boxed{\mathcal L\{1\}(p)=\lim_{T\to\infty}\frac{1-e^{-pT}}p=\frac1p.}
$$
The condition on the real part also gives absolute convergence of this <improper integral>.

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