I am asking if there is a closed form or tight upper or lower bound for the general exponential integral defined as:
$\int_{x=x_{0}}^{\infty}x^{-n}e^{-x}dx$
where $n\geq2$ (my original problem is for $n=3$) and $x_{0}$ is a real positive number
For $n=1$, the integral is called the exponential integral and bounded as follows: Exponential Integral
$\frac{1}{2}e^{-x_{0}}\ln\left(1+\frac{2}{x_{0}}\right)<\int_{x=x_{0}}^{\infty}x^{-1}e^{-x}dx\leq e^{-x_{0}}\ln\left(1+\frac{1}{x_{0}}\right)$
I am looking for similar bounds for general $n\geq2$.
I appreciate your help. Thank you
Using integration by parts, we have $\int_{x=x_{0}}^{\infty}x^{-n}e^{-x}dx=\frac{1}{\left(n-1\right)!}\left(\frac{e^{-x_{0}}}{x_{0}^{n-1}}\sum_{k=0}^{n-2}\left(-1\right)^{k}\left(n-k-2\right)!x_{0}^{k}+\left(-1\right)^{n-1}\int_{x=x_{0}}^{\infty}x^{-1}e^{-x}dx\right)$
Using the upper and lower bounds of first ($n=1$) exponential integral, we get
$\frac{1}{\left(n-1\right)!}\left(\frac{e^{-x_{0}}}{x_{0}^{n-1}}\sum_{k=0}^{n-2}\left(-1\right)^{k}\left(n-k-2\right)!x_{0}^{k}+\left(-1\right)^{n-1}\frac{1}{2}e^{-x_{0}}\ln\left(1+\frac{2}{x_{0}}\right)\right)<\int_{x=x_{0}}^{\infty}x^{-n}e^{-x}dx\leq\frac{1}{\left(n-1\right)!}\left(\frac{e^{-x_{0}}}{x_{0}^{n-1}}\sum_{k=0}^{n-2}\left(-1\right)^{k}\left(n-k-2\right)!x_{0}^{k}+\left(-1\right)^{n-1}e^{-x_{0}}\ln\left(1+\frac{1}{x_{0}}\right)\right)$