Inverse Laplace Transform of (1)/(e^s-1) and (s)/(e^s-1)

Inverse Laplace Transform of $\frac{1}{{e}^{s}-1}$ and $\frac{s}{{e}^{s}-1}$
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Giancarlo Callahan
It's not hard to check that
${\int }_{0}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\delta \left(t-n\right){e}^{-st}=\sum _{n=1}^{\mathrm{\infty }}{e}^{-ns}=\frac{1}{{e}^{s}-1}$
and
${\int }_{0}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{\delta }^{\prime }\left(t-n\right){e}^{-st}=-\sum _{n=1}^{\mathrm{\infty }}s{e}^{-ns}=-\frac{s}{{e}^{s}-1}$
and hence from the Laplace uniqueness theorem
${\mathcal{L}}^{-1}\left[\frac{1}{{e}^{s}-1}\right]\left(t\right)=\sum _{n=1}^{\mathrm{\infty }}\delta \left(t-n\right)$
${\mathcal{L}}^{-1}\left[\frac{s}{{e}^{s}-1}\right]\left(t\right)=-\sum _{n=1}^{\mathrm{\infty }}{\delta }^{\prime }\left(t-n\right)$
which is your original train of delta functions.