How to simplify this integral?

Question

I have managed to solve this integral by using Taylor series expansion to approximate the $e^x=1+x$. However, I am not successful due to the integral is not converged.

Could you please give me a hint to simplify this one or express it in other function or terms. \begin{align} I= \int_0^{\infty } \frac{\text{exp}[-a x]}{(1+b x)^{p+2}(1+c x)^q} \, dx \end{align} where $a>0$, $b>0$, $c>0$ $p\ge 0$, $q \ge 0$.

Thank you very much!

Answer 1

You can't approximate $e^t\approx1+t$, since this is only valid for $t\approx0$. But, in this case, $t\in(0,\infty)$, since the upper integration limit is $\infty$. If the interval of integration would've been $(u,v)\subset(-1,1)$ instead of $(0,\infty)$, then indeed the approximation could have proven useful, but, as it stands, it is useless.

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Now as to the actual integral: Begin first by evaluating $I(B,C)=\displaystyle\int_0^\infty\frac{e^{-ax}}{(B+bx)(C+cx)}~dx$,

which can be done by decomposing the integrand into partial fractions, and appealing to the definitions of certain special functions, such as the incomplete $\Gamma$ function, or the exponential integral. Then notice that for integer values of p and q, the integral can be expressed in terms of $I^{(p+1,~q-1)}(1,1)$. Alternately, split the interval of integration into $(0,1)$ and $(1,\infty)$, let $t=1/x$ on the latter, and expand the integrand into its binomial series, then reverse the order of summation and integration, and rewrite the new expression on terms of hypergeometric functions.

Answer 2

Hint:

$\int_0^\infty\dfrac{e^{-ax}}{(1+bx)^{p+2}(1+cx)^q}dx$

$=\int_0^\infty\dfrac{e^{-\frac{ax}{b}}}{(1+x)^{p+2}\left(1+\dfrac{cx}{b}\right)^q}d\left(\dfrac{x}{b}\right)$

$=\dfrac{b^{q-1}}{c^q}\int_0^\infty\dfrac{e^{-\frac{ax}{b}}}{(1+x)^{p+2}\left(\dfrac{b}{c}+x\right)^q}dx$

$=\dfrac{b^{q-1}}{c^q}\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^na^nx^n}{b^nn!(1+x)^{p+2}\left(\dfrac{b}{c}+x\right)^q}dx$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^na^nB(p-q-n+1,n+1){_2}F_1\left(q,p-q-n+1;p-q+2;\dfrac{c-b}{c}\right)}{b^{n-q+1}c^qn!}$ (according to http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/07/01/01)

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^na^nB(p-q-n+1,n+1){_2}F_1\left(q,n+1;p-q+2;\dfrac{b-c}{b}\right)}{b^{n+1}n!}$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^na^nc^qB(p-q-n+1,n+1){_2}F_1\left(q,n+1;p+2;\dfrac{c-b}{c}\right)}{b^{n+q+1}n!}$