$$\displaystyle \int_{0}^\infty \left(e^{\large-x}- e^{\large -Ax-Bx^{c}}\right) \cdot x^{-1} \mathrm dx,$$ with $A>0,\,\, B>0,\,\, 1>C>0$
I have attempted to apply the Taylor expansion to the exponential term to eliminate the denominator x and the constant.
I believe that after applying the Taylor expansion, the summation and integration can be interchanged, and some parameters can be transformed into gamma functions to obtain the final expression. However, I am not sure how to handle this.
=$\int_{0}^{\infty}e^{-x}(1-e^{-(A-1)x-Bx^{c}})\cdot x^{-1}dx$
=$\int_{0}^{\infty}e^{-x}(1-e^{[-(A-1)-Bx^{c-1}]x})\cdot x^{-1}dx$
=$\int_{0}^{\infty}e^{-x}(1-\sum_{k=0}^{\infty} \frac {[-(A-1)-Bx^{c-1}]^kx^k} {k!})\cdot x^{-1}dx$
=-$\int_{0}^{\infty}e^{-x}(\sum_{k=1}^{\infty} \frac {[-(A-1)-Bx^{c-1}]^kx^{k-1}} {k!})\cdot dx$
If I split the integral, it can be represented using the Ei function and the Fox-H function (How are Fox-H functions useful in math?) as shown below. However, this expression tends to infinity for both terms, making it impossible to verify through simulation.
=$\int_{0}^{\infty}e^{-x}\cdot x^{-1}dx -\int_{0}^{\infty}e^{-(A-1)x-Bx^{c}}\cdot x^{-1}dx$
=$-Ei(0) -\int_{0}^{\infty}\sum{\frac{b^kx^{c-1}e^{-(A-1)x}}{k!}}dx$
=$-Ei(0) -\sum{\frac{b^k\Gamma(ck)}{(A-1)^{ck}k!}}$
=$-Ei(0)-_1\Psi_0[\frac{(0,c)}{};\frac{b}{(A-1)^{c}}]$
