# What is the Laplace transform of 1/(1+t)

What is the Laplace transform of $\frac{1}{1+t}$
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benyaep17
It is an exponential integral.
$\mathcal{L}\left\{\frac{1}{1+t}\right\}={\int }_{0}^{\mathrm{\infty }}\frac{{e}^{-st}}{t+1}dt$
We can get the Laplace transform by letting $u=1+t$. Then $du=dt$ and we have
$\begin{array}{rl}{\int }_{0}^{\mathrm{\infty }}\frac{{e}^{-st}}{t+1}dt& ={\int }_{1}^{\mathrm{\infty }}\frac{{e}^{-su+s}}{u}du\\ & ={e}^{s}{\int }_{1}^{\mathrm{\infty }}\frac{{e}^{-su}}{u}du\end{array}$
where ${\int }_{1}^{\mathrm{\infty }}\frac{{e}^{-su}}{u}du=-\text{Ei}\left(-s\right)$. Thus
$\mathcal{L}\left\{\frac{1}{1+t}\right\}=-{e}^{s}\text{Ei}\left(-s\right).$