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