Asymptotic Expansion for a Function involving a Weird Integral

So I'm trying to find the asymptotic expansion as $x \to \infty$ of $$f(x)=\frac{1}{\bigg[A-\int \frac{\lambda^x}{\Gamma(x+1)}dx\bigg]^\frac{1}{\alpha}}$$

Note that $\lambda>0$ and $\alpha>0$. We can also see that $x \to \infty$ means that we can basically treat A as some arbitrary constant of integration and ignore the constant factor that comes out of the integral, but I have no clue on where to go from here. I'm basically stuck. Note here that $f(x) \to \infty$ is a constraint as well.

• Your question is ill-posed. You can't use $x$ as both the argument of a function $f(x)$ and a dummy integration variable. – Pierpaolo Vivo Mar 7 '16 at 14:55
• It's an indefinite integral. It has to be a function of x. You can rewrite the integral in the question as $$\int_{x_0}^x \frac{\lambda^y}{\Gamma(y+1)}dy$$ where $x_0>0$ fixed. – Suvy Mar 7 '16 at 14:57
• it's not 'if I want'. The way you wrote the question is meaningless. And it is not equivalent to what you wrote in your comment, which instead makes perfect sense. Please edit your question. – Pierpaolo Vivo Mar 7 '16 at 15:00
• How is my question unclear? An indefinite integral of a function of x has to be a function of x. It's not unclear at all. If I'm looking for an asymptotic expansion as a function of x with an indefinite integral that's a function of x, the question is obviously well-defined. What I wrote there is just another way of defining what amounts to the same thing. – Suvy Mar 7 '16 at 15:01
• That's particularly true because A is arbitrary, which means you have a constant of integration that fits right into the outer part of the integral. So both ways of defining the integral must necessarily be equivalent. – Suvy Mar 7 '16 at 15:02