$$\displaystyle I = \int_{0}^{1} \left (( \ln u)^{-z} - \sum_{k=0}^{n-1} \frac {(\ln u)^{k-z}}{u \cdot k!} \right ) \mathrm {d}u$$
My attempt :
Distributing the integration over the two parts as follows:
$$\displaystyle \int_{0}^{1} ( \ln u)^{-z} \mathrm {d}u - \int_{0}^{1} \left ( \sum_{k=0}^{n-1} \frac {(\ln u)^{k-z}}{u \cdot k!} \right ) \mathrm {d}u.$$
Solving for first part : let $ \ \displaystyle \ln u = -y$
Then $\displaystyle \int_{0}^{1} ( \ln u)^{-z} \mathrm {d}u = (-1)^z \int_{0}^{\infty} ( y)^{-z} e^{-y} \mathrm {d}y = (-1)^z \Gamma (-z+1) .$
Solving for Second part :
Let
$\displaystyle f(u) = \sum_{k=0}^{n-1} \frac {(\ln u)^{k-z}}{u \cdot k!} $
Then $\displaystyle \int_{0}^{1} f(u) \mathrm {d}u = F(1) - F(0),$ where $\displaystyle F'(u) = f(u) $
Using same substitution,
$ \ \displaystyle F(u) = \sum_{k=0}^{n-1} \frac {1}{k!} \int \frac {(\ln u)^{k-z}}{u} \mathrm {d}u \implies F(y) = - (-1)^z \sum_{k=0}^{n-1} \frac {(-1)^k(y)^{k+1-z}}{k!(k+1-z)} \\ \{ \ y \ ; \ 0 < y < \infty \}$
Now $\displaystyle I = (-1)^z \Gamma (-z+1) + F(y)|_{0}^{\infty} $
OR
$$\displaystyle I = (-1)^z \left ( \Gamma (-z+1) + \lim _{y\rightarrow \infty} \sum_{k=0}^{n-1} \frac {(-1)^k(y)^{k+1-z}}{k!(k+1-z)} - \lim _{y\rightarrow 0} \sum_{k=0}^{n-1} \frac {(-1)^k(y)^{k+1-z}}{k!(k+1-z)} \right )$$
The summation $\displaystyle \ \ \lim _{y\rightarrow 0} \sum_{k=0}^{n-1} \frac {(-1)^k(y)^{k+1-z}}{k!(k+1-z)} $ can be finite if $ 0 < z < 1 $ or I may be wrong.
My Doubt : What I did is correct or not and is a there a closed form for this integral ?
Thank you for reading.