The inverse Laplace transform of $e^{-z}\textrm{Ei}(z)z^{-1}\log(z).$ For a work I need to evaluate the following integral: $$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}e^{uz}e^{-z}\textrm{Ei}\left(z\right)z^{-1}\log\left(z\right)dz,\ c>0,\,u>2$$ where $\mathrm{Ei}\left(z\right)$ is the exponential integral function. I know that $$\mathfrak{L}^{-1}\left(z^{-1}\log\left(z\right)\right)\left(u\right)=(-\log\left(u\right)-\gamma)1_{u>0}$$ where $\gamma$ is the Euler-Mascheroni constant, so my idea was to find the inverse Laplace transform of $e^{-z}\textrm{Ei}\left(z\right)$ and then to use the convolution theorem. I found that $$e^{-z}\textrm{Ei}\left(z\right)=PV\int_{0}^{\infty}\frac{e^{-uz}}{1-u}du\tag{1}$$ so my question is: 

Can I use the convolution theorem even if the integral in $(1)$ exists only in the sense of the Cauchy Principal Value? In other words, can I write $$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}e^{uz}e^{-z}\textrm{Ei}\left(z\right)z^{-1}\log\left(z\right)dz=PV\int_{0}^{u}\frac{\log\left(t\right)+\gamma}{u-t-1}dt?$$

I don't know if it is a standard property or is a idiocy. I searched in my textbooks but I didn't find anything like that. Thank you for your time.
 A: $\because\mathcal{L}_{z\to t}^{-1}\left\{\dfrac{\textrm{Ei}(z)\ln z}{z}\right\}$
$=\mathcal{L}_{z\to t}^{-1}\left\{\dfrac{\gamma\ln z}{z}+\dfrac{\ln^2z}{z}+\sum\limits_{n=1}^\infty\dfrac{z^{n-1}\ln z}{n!n}\right\}$ (according to https://en.wikipedia.org/wiki/Exponential_integral#Convergent_series)
$=\mathcal{L}_{z\to t}^{-1}\left\{\dfrac{\gamma\ln z}{z}+\dfrac{\ln^2z}{z}+\sum\limits_{n=0}^\infty\dfrac{z^n\ln z}{(n+1)!(n+1)}\right\}$
Contains the term $\ln z$
A: Your derivation is correct and can be made rigorous through the use of distributions. If
$$(f, \phi) = \int_0^\infty (-\ln u - \gamma) \phi(u) du, \\
(g, \phi) = \operatorname{v.\!p.} \int_0^\infty \frac {\phi(u)} {1 - u} du$$
($f$ is just a regular distribution), then
$$\mathcal L[f] = \frac {\ln z} z, \\
\mathcal L[g] = e^{-z} \operatorname{Ei}(z), \\
\mathcal L[f * g] = \frac {\ln z} z e^{-z} \operatorname{Ei}(z).$$
The convolution is defined as
$$(f * g, \phi) = (f \otimes g, (x,y) \mapsto \phi(x + y)) = \\
(f, x \mapsto (g, y \mapsto \phi(x + y))).$$
$f*g$ is a functional that can be identified with the ordinary function $h(u)$ defined by your formula on any interval not containing $u = 1$.
Further, the inverse transform exists in the sense of ordinary functions, which means that $f*g$ doesn't have a singular component, and this also holds:
$$\mathcal L[h] = \frac {\ln z} z e^{-z} \operatorname{Ei}(z).$$
$h(u)$ has a singularity of the order $\ln^2$ at $u = 1$.
