Find the inverse Laplace transform of $f(t) = \int_t^\infty \frac{e^{ - u}}{u}du$ Find the inverse Laplace transform of the integral:$$f(t) = \int_t^\infty  {\frac{{{e^{ - u}}}}{u}du} $$
If the integral: $$f(t) = \int_0^\infty  {\frac{{{e^{ - u}}}}{u}du} $$
I had done. However the integral from t (not ZERO) to infinity???
 A: I think you mean the Laplace transform, not its inverse.  Rewrite $f$ as
$$f(t) = \int_1^{\infty} dy \frac{e^{-t y}}{y} $$
The the LT of $f$, $\hat{f}(s)$, is
$$\begin{align}\hat{f}(s) &= \int_0^{\infty} dt \, f(t) e^{-s t}\\ &= \int_0^{\infty} dt \,\int_1^{\infty} dy \frac{e^{-t y}}{y} e^{-s t} \\ &=\int_1^{\infty} \frac{dy}{y} \:  \int_0^{\infty} dt \, e^{-(y+s) t} \\ &=\int_1^{\infty} \frac{dy}{y (y+s)}\\&= \frac{1}{s} \lim_{Y \to \infty} \int_1^Y dy \, \left ( \frac{1}{y} - \frac{1}{y+s}\right )\\ &= \frac{1}{s}\lim_{Y \to \infty}\left [\log{Y} - \log{(Y+s)} + \log{(1+s)} \right ] \end{align}$$
Thus,
$$\hat{f}(s) = \frac{\log{(1+s)}}{s}$$
A: One wants to write $f$ as the Laplace transform of some function $g$, that is, to find some function $g$ such that $f(t)=\int\limits_0^{+\infty}\mathrm e^{-tx}g(x)\mathrm dx$ for every $t\geqslant0$. The change of variable $u=tx$ indicates that, for every $t\geqslant0$, $f(t)=\int\limits_1^{+\infty}\mathrm e^{-tx}x^{-1}\mathrm dx$.
Thus, the function $g$ defined by $g(x)=x^{-1}\mathbf 1_{x\geqslant1}$ for every $x$ fits the bill, that is, $f=\mathcal L(g)$.
A: $\mathcal{L}^{-1}_{t\to x}\left\{\int_t^\infty\dfrac{e^{-u}}{u}du\right\}=\dfrac{\mathcal{L}^{-1}_{t\to x}\left\{\dfrac{e^{-u}}{u}\right\}}{x}=\dfrac{H(x-1)}{x}$
