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By definition incomplete Gamma function is:$$\Gamma(0,x)=\int_{x}^{\infty}t^{-1}e^{-t}dt $$ I have an expression which includes

$$\Gamma(0,r(A)e^{i\phi(A)}),$$ where $A>0$ is a parameter, and $r(A)$ is almost linear, $\displaystyle \phi(A)\rightarrow\pm\frac{\pi}{2},\ as\ A\rightarrow\infty $

I want to have more intuition about the influence of parameter $A$. I will be very appreciate any comments or a good reference.

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You could use the Chain Rule, Product Rule, and Fundamental Theorem of Calculus to examine $\frac{d}{dA}\Gamma$. –  alex.jordan May 23 '11 at 0:07

3 Answers 3

up vote 2 down vote accepted

When you say $r(A)$ is almost linear, do you mean that for large $A$ that $r(A)\approx\rho A+c$ for some constants $\rho$ and $c$? Also, what do you mean by $\phi(A)\rightarrow\pm\frac{\pi}{2}$?

Without the $\pm$, and with the above assumption about $r(A)$:

$$\frac{d}{dA}\Gamma(0,r(A)e^{i\phi(A)})=-\left[r(A)e^{i\phi(A)}\right]^{-1}e^{-\left[r(A)e^{i\phi(A)}\right]}\left(r'(A)e^{i\phi(A)}+r(A)i\phi'(A)e^{i\phi(A)}\right)$$

For large $A$, $r(A)\approx\rho A+c$, $r'(A)\approx\rho$, $\phi(A)\approx\frac{\pi}{2}$, and $\phi'(A)\approx0$. And the above reduces to:

$$\frac{d}{dA}\Gamma(0,r(A)e^{i\phi(A)})\approx-\left[(\rho A+c) i\right]^{-1}e^{-\left[(\rho A+c)i\right]}\left(\rho i +0\right)$$

$$\frac{d}{dA}\Gamma(0,r(A)e^{i\phi(A)})\approx-\frac{1}{A+\frac{c}{\rho}}e^{-(\rho A+c)i}$$

So for large $A$, increasing $A$ by a differential $\Delta A$ will "add" about $\frac{\Delta A}{A+\frac{c}{\rho}}$ to your quantity, but in the direction of $-e^{-(\rho A+c)i}$; that is it will add a complex differential of about $\frac{\Delta A}{A+\frac{c}{\rho}}$ in magnitude at an angle of $-\rho A-c+\pi$.

Imagine an angle that spins around the origin (clockwise if $\rho$ is positive) at a linear rate with respect to $A$. Now add to your quantity in that rotating direction with a magnitudes that decrease roughly like $\frac{1}{A}$.

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The incomplete $\Gamma$ function behaves like $$\Gamma(0,z) \sim \frac{e^{-z}}{z} \, \sum_{k=0} \frac{(-1)^k k!} {z^{k}},$$ valid for $|\text{arg} z| < 3\pi/2$, see here. If you substitute $$z = r(A) e^{i\phi(A)}$$ and take a few terms (maybe even only one), you will know how $\Gamma(0,z)$ behaves as a function of $A$.

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I've done an (amateurish) discussion on the gamma-function some monthes ago. I arrived at the incomplete gamma by some naive approach. Possibly this gives some additional intuition/insight to you. See "uncompleting the gamma"

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