How to rewrite $\int\limits_A^B \frac{x^n \exp(-\alpha x)}{\small(x + \beta\small)^m} \, dx$?

Question

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!

Answer 1

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.

Answer 2

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

$$ \operatorname{Ei}_m(x)=\frac{1}{(m-1)!}\left[(-x)^{m-1}\operatorname{Ei}_1(x)+e^{-x}\sum\limits_{s=0}^{m-2}(m-s-2)!(-x)^s \right] $$