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}


\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!

  • $\begingroup$ This may be terribly naive, but can't you take one of the standard expansions for the upper incomplete gamma function $\Gamma(a,z)$ and substitute $a$ with $1+d+\epsilon$? $\endgroup$
    – njuffa
    Feb 21 '16 at 1:58
  • $\begingroup$ @njuffa Well, thats exactly what I would like to do but there are some issues. First of all, are those expansions around $0$? And moreover, are you aware of where could I find the proof of these? I need to go through the whole procedure, since my concern is to understand the method of pertrbation for functions like this. $\endgroup$
    – Bazinga
    Feb 21 '16 at 12:08
  • $\begingroup$ @njuffa Moreover, the integration limit, does depend on the variable of differentiation, given that this is $ε$, therefore it will not be the same result. $\endgroup$
    – Bazinga
    Feb 21 '16 at 14:03
  • 1
    $\begingroup$ You're doing an expansion around $\epsilon=0$, in which case $x$ is a constant just like $d$, $A$, and $c$. $\endgroup$
    – David
    Feb 25 '16 at 23:20

To continue from @David's comment: let's consider the situation. You've got a function $f(y;\epsilon)$, where $y = (x,d,c,A,r)$, and you would like to find out more about $f(y;\epsilon)$ assuming that $0<\epsilon \ll 1$. That is, you would like to perturb $f(y;\epsilon)$ around $\epsilon = 0$. Assuming $f(y;\epsilon)$ depends on $\epsilon$ in a sufficiently smooth way, we can write \begin{equation} f(y;\epsilon) = f(y;0) + \epsilon \frac{\partial f}{\partial \epsilon}(y;0) + \mathcal{O}(\epsilon^2). \end{equation} To specify things some more, you have in particular \begin{equation} f(y;\epsilon) = f(x,d,c,A,r;\epsilon) = \Gamma(1+d+\epsilon,A-c \ln x). \end{equation} That means that \begin{equation} \frac{\partial f}{\partial \epsilon}(y;0) = \frac{\partial \Gamma}{\partial \alpha}(1+d,A-c \ln x), \end{equation} where $\frac{\partial \Gamma}{\partial \alpha}$ is the derivative of the incomplete Gamma function to the first argument. You can find the expression for that derivative here; the $n$-th derivative is given here, if you would like to expand $f$ to higher orders in $\epsilon$.

  • $\begingroup$ Hello and please excuse me for my late reply. Seems pretty correct to me, I think that this is what I should have done. I am now sure that this is the way and I have already tried to expand until terms of order 2. Thank you. $\endgroup$
    – Bazinga
    Mar 21 '16 at 19:58

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