How to rewrite $\int\limits_A^B \frac{x^n \exp(-\alpha x)}{\small(x + \beta\small)^m} \, dx$? Currently, I am a post-graduate researcher in Telecommunications. During the process of evaluating the transmission error probability, I need to evaluate the following integral $I = \int\limits_A^B \frac{x^n \exp(-\alpha x)}{\small(x + \beta\small)^m} \, dx$?
How to rewrite this improper integral in terms of special function (for example $Ei(x)$, Bessel,...)?
Notice that $A, B, \alpha,\beta > 0$ (positive real number) and $m,n$ are two positive integers.
I have tried to compute this integral with different values of $A, B, \alpha,\beta > 0$ and $m,n$  by using Wolfram Mathematica.
It seems that the results of this integral have a form of the exponential integral function $Ei\left( x \right) = \int\limits_{t =  - x}^\infty  {\frac{{{e^{ - t}}}}{t}dt}  = \int\limits_{t =  - \infty }^{t = x} {\frac{{{e^t}}}{t}dt}$ as:
$I = C_1\bigg[C_2 + C_3\big[ {\rm Ei}\big(- \alpha(\beta+ A)\big) - {\rm Ei}\big(- \alpha(\beta+ B)\big) \big] \bigg]$.
Are there any way to find out the correct values of $C_1$, $C_2$, and $C_3$.
Thank you for your enthusiasm!
 A: Consider the integral
$$
I(\alpha)= \int\limits_{A}^{B} dx \ \frac{e^{-\alpha x}}{(x+\beta)^m}
$$
By changing variables $y=x+\beta$, then scaling: $t=y/(A+\beta),$ this is in the form of two 'En-Functions' , defined as $\operatorname{Ei}_n(x)=\int\limits_1^\infty dt \ t^{-n}e^{-xt} $  so we have
$$
I(\alpha)=e^{\alpha \beta}(\beta+A)^{1-m}\operatorname{Ei}_m(\alpha(\beta+A))-e^{\alpha \beta}(\beta+B)^{1-m}\operatorname{Ei}_m(\alpha(\beta+B))
$$
The original integral is given by the $n$th derivative of $I(\alpha)$
$$
\int\limits_{A}^{B} dx \ \frac{x^n e^{-\alpha x}}{(x+\beta)^m}=(-1)^n\frac{d^n }{d\alpha^n}I(\alpha)
$$
You may write $\operatorname{Ei}_m$, and consequently $I(\alpha)$, in terms of only $\operatorname{Ei}_1$, the exponential integral. The formula is given here
$$
\operatorname{Ei}_m(x)=\frac{1}{(m-1)!}\left[(-x)^{m-1}\operatorname{Ei}_1(x)+e^{-x}\sum\limits_{s=0}^{m-2}(m-s-2)!(-x)^s \right]
$$
A: Thank you for considering my concern.
I have found out myself.
For the given integral above, I firstly change variable $y = x + \beta \to dx = dy$. Now, $I$ can be rewritten as
$I = \int_{y_{\min}}^{y_{\max}}\frac{\small(y - \beta\small)^n \exp\big(- \alpha\small(y-\beta\small)\big)}{y^m}dy$, where $y_{\min} = A +\beta$ and $y_{\max} = B + \beta$.
By using the binomial theorem for $\small(y - \beta\small)^n$, we can be modified $I$ as
$I = \exp(\alpha\beta)\sum_\limits{k=0}^n\binom{k}{n}\small(-\beta\small)^{n-k}
\int_{y_{\min}}^{y_{\max}}\frac{\exp\big(- \alpha y\big)}{y^{m-k}}dy$.
Based on the condition of $(m-k)$ and thanks to the help of Wolfram Mathematica, I can achieve the final result of $I$ as follows:
$\begin{array}{l}
I = \exp \left( {\alpha \beta } \right)\sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}}
n\\
k
\end{array}} \right){{\left( { - \beta } \right)}^{n - k}}} \\
\,\,\,\,\,\,\, \times \left\{ \begin{array}{l}
{\alpha ^{ - \left( {m - k} \right) - 1}}\left[ {\Gamma \left( {\left( {m - k} \right) + 1,{y_{\min }}\alpha } \right) - \Gamma \left( {\left( {m - k} \right) + 1,{y_{\max }}\alpha } \right)} \right],m - k = 0\\
{\alpha ^{\left( {m - k} \right) - 1}}\left[ {\Gamma \left( {1 - \left( {m - k} \right),{y_{\min }}\alpha } \right) - \Gamma \left( {1 - \left( {m - k} \right),{y_{\max }}\alpha } \right)} \right],m - k \ne 0
\end{array} \right.
\end{array}$,
where $\Gamma \left(\cdot,\cdot\right)$ is upper bound incomplete Gamma function.
