I am trying to find a series representation for $E_n(z)$, and I have gotten thus far: $$E_n(z)=\int_{1}^{\infty}\frac{e^{-zt}}{t^n}dt$$ $$E_n(z)=\int_{1}^{\infty}t^{-n}\sum_{k\geq0}\frac{(-z)^k}{k!}t^kdt$$ $$E_n(z)=\sum_{k\geq0}\frac{(-z)^k}{k!}\int_{1}^{\infty}t^{k-n}dt$$ $$E_n(z)=\sum_{k\geq0}\frac{(-z)^k}{k!}\biggl(\frac{t^{k-n+1}}{k-n+1}\biggr)\biggr|_{1}^{\infty}$$ $$E_n(z)=\sum_{k\geq0}\frac{(-z)^k}{k!}\biggl(\frac{1}{n-k-1}+\frac{1}{k-n+1}\lim_{t\to\infty}t^{k-n+1}\biggr)$$ I should know how to compute that limit, but I don't. This site gives $$\int_{1}^{\infty}t^{k-n}dt=\frac{1}{n-k-1}$$ for $n-k>0$, and $n-k-1>0$ which legitimately makes no sense. Did I do something wrong?
EDIT: I am not asking for a series representation for $E_n(z)$, as I am trying to find it myself, but I am trying to find out why the integral calculator I linked gave me the integral in the title, and if I could apply it to my search.
$\newcommand{\d}{\mathrm{d}}$ $\newcommand{\e}{\mathrm{e}}$ Let $n>1$ be an integer. Then
$$\e^s=\sum_{k=0}^{n-2}\frac{s^k}{k!}+\frac{s^{n-1}}{(n-2)!}\int_0^1\e^{su}(1-u)^{n-2}\d u$$ so $$\begin{align} E_n(z)&=\sum_{k=0}^{n-2}\frac{(-z)^k}{k!(n-k+1)}+\frac{(-z)^{n-1}}{(n-2)!}\int_0^1 E_1(uz)(1-u)^{n-2}\d u& (\Re z &> 0)\end{align}\text{.}$$
The trouble is that $k=n-2$ is as far as the Maclaurin expansion can go: $$\begin{align} E_1(z)&\sim -\ln z& (z&\to 0)\text{.}\end{align}$$ But you can subtract out this logarithmic singularity to get $$E_{n}\left(z\right)=\frac{(-z)^{n-1}}{(n-1)!}(H_{n-1}-\ln z-\gamma)-\sum_{% \begin{subarray}{c}k=0\\ k\neq n-1\end{subarray}}^{\infty}\frac{(-z)^{k}}{k!(1-n+k)}$$ (DLMF 8.19.8) where $$H_{n}=\sum_{k=1}^n\frac{1}{k}$$ denotes the harmonic numbers and $\gamma$ is the Euler–Mascheroni constant.