Calculate integral $\int_0^\infty e^{-x} (e^{-\frac a b x} - 1)^{b} dx$ for $b>0$ and any $a \in \mathbb{R}$

Question

i am working on following task:

Choose any nonzero $a \in \mathbb{R}$ so the integral converges and for a given $b > 0$ compute $\int_t^\infty e^{-x} (e^{-\frac a b x} - 1)^{b} dx$.

I am looking for reference material or some tips how to approach such problems (improper integrals of rational functions of $e^x$).

It seems such integrals are computable, as Wolfram gives back some answer when I don't give limits of integration. The answer I get back from Wolfram is given using hypergeometric function and I need this integral as a function of $t$.

Answer 1

Enforce the substitution $e^{-\frac{a}{b}x}\to x$. Then,

$$\begin{align} I(t;a,b)&=\int_t^{\infty}e^{-x}\left(e^{-\frac{a}{b}x}-1\right)^{b}dx\\\\ &=\frac{b}{a}\int_{0}^{e^{-(a/b)t}}x^{b/a-1}\left(x-1\right)^{b}du\\\\ &=\frac{b}{a}\text{B}\left(e^{-(a/b)t};\frac{b}{a},b+1\right) \end{align}$$

where $B(x;\alpha,\beta)$ is the Incomplete Beta Function. For $t=0$, we have

$$I(0;a,b)=\frac{b}{a}\text{B}\left(\frac{b}{a},b+1\right)$$

where $B(\alpha,\beta)$ is the Beta Function.

Answer 2

Hint: Letting $t=e^{-x}$, we have the $($ incomplete $)$ beta function.