The Upper Incomplete Gamma function, for $t \in \mathbb{R}$, is defined as:
\begin{equation} \Gamma(α,β)=\int_{β}^{\infty}t^{α-1}e^{-t}dt \end{equation}
For the problem which I am studying it takes the form:
\begin{equation} \Gamma(1+d,A-c\ln x)=\int_{(A-c\ln x)}^{\infty}t^{d}e^{-t}dt \end{equation}
where
\begin{equation} A=\frac{cdr}{1-(1-c)r} \end{equation}
For the parameters it holds that $c\in (0,1]$, $d \in \mathbb{R}$ and $r$ is an arbitary real constant which plays no significant role at this particular stage.
My task is to perturbate the Incomplete Gamma function above by a parameter $\epsilon$, such as $0<\epsilon <<1$. To do so, I have to expand the $Γ(1+d+ε, A-c\ln x)$ function into a Taylor series around the point $0$. But to do a Taylor's expansion I need, by definition:
\begin{equation} Γ(1+d+ε, A-c\ln x)=\sum_{n=0}^{\infty}\frac{\partial^n}{\partialε^n}Γ(1+d+ε, A-c \lim_{x\to 0}\ln x)\frac{ε^n}{n!} \end{equation}
which can't be done since $\ln x \to -\infty$ as $x \to 0$.
I am really looking for reason here. Am I doing something wrong? Should I not take $x \to 0$ but $ε \to 0$ instead?
Another thought which has crossed my mind is perhaps take $Γ(0,0)$? This is how I would have done it if I would like to perturbate an ODE. Therefore, perhaps it needs a Taylor expansion of two variables. I dont know, I have totally got something wrong here..
How am I going to perturbate this function given all the above? If anyone could point out to me how to proceed with the first derivative and the right way of thinking I am pretty sure that I can manage the rest.
Any help would be greatly appreciated. Thank you!
