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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

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    When $x_0 = 0$ you've written down $\Gamma(1-n)$. – 2017-01-01
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    Thank you @Mark. I am looking for the case $x_{0}\gg0$ – 2017-01-01
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    I believe what you have written is an incomplete gamma function. There may be some helpful information here: https://en.wikipedia.org/wiki/Incomplete_gamma_function – 2017-01-01
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    @user71352 appears to be [correct](https://en.wikipedia.org/wiki/Incomplete_gamma_function). – 2017-01-01
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    Thank you @user71352 and Mark. So the integration is the upper incomplete gamma function $\Gamma\left(1-n,x_{0}\right)$. – 2017-01-01
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    Let $f_n(x)=x^{-n}e^{-x}.$ For $n\geq 2$ we have $f'_{n_1}+f_{n-1}=(1-n)f_n.$ Therefore $\int_a^{\infty}f_n(x)dx=$ $(n-1)^{-1}(a^{1-n}e^{-a}-\int_a^{\infty}f_{n-1}(x)dx).$ – 2017-01-01
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    Thank you @user254665. I noticed this result. So we can write the general exponential integral in terms of the first ($n=1$) exponential integral. Therefore, I can use the upper and lower bounds after some modifications. – 2017-01-01
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    Similar methods apply to other integrals. E.g. $\int_A^{\infty}f(x)dx$ with $f(x)=e^{-x^2}$ and $A>0.$ Put $f(x)=-f'(x)/2x.$ – 2017-01-02

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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

$\frac{1}{\left(n-1\right)!}\left(\frac{e^{-x_{0}}}{x_{0}^{n-1}}\sum_{k=0}^{n-2}\left(-1\right)^{k}\left(n-k-2\right)!x_{0}^{k}+\left(-1\right)^{n-1}\frac{1}{2}e^{-x_{0}}\ln\left(1+\frac{2}{x_{0}}\right)\right)<\int_{x=x_{0}}^{\infty}x^{-n}e^{-x}dx\leq\frac{1}{\left(n-1\right)!}\left(\frac{e^{-x_{0}}}{x_{0}^{n-1}}\sum_{k=0}^{n-2}\left(-1\right)^{k}\left(n-k-2\right)!x_{0}^{k}+\left(-1\right)^{n-1}e^{-x_{0}}\ln\left(1+\frac{1}{x_{0}}\right)\right)$