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.
$\because\mathcal{L}_{z\to t}^{-1}\left\{\dfrac{\textrm{Ei}(z)\ln z}{z}\right\}$
$=\mathcal{L}_{z\to t}^{-1}\left\{\dfrac{\gamma\ln z}{z}+\dfrac{\ln^2z}{z}+\sum\limits_{n=1}^\infty\dfrac{z^{n-1}\ln z}{n!n}\right\}$ (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 $\ln z$