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I found in the wolfram site this asymptotic,

$$ \Gamma\left(a,z\right)=\sqrt{2\pi}a^{a-1/2}e^{-a}\left(1+O\left(\frac{1}{a}\right)\right)-\frac{e^{-z}z^{a}}{a}\left(1+O\left(\frac{1}{a}\right)\right), \quad \left|a\right|\rightarrow\infty, $$

and since

$$ \Gamma\left(a\right)=\sqrt{2\pi}a^{a-1/2}e^{-a}\left(1+O\left(\frac{1}{a}\right)\right), \quad \left|a\right|\rightarrow\infty $$

and

$$ \Gamma\left(a\right)=\Gamma\left(a,z\right)+\gamma\left(a,z\right) $$

we have

$$ \gamma\left(a,z\right)=\frac{e^{-z}z^{a}}{a}\left(1+O\left(\frac{1}{a}\right)\right).\tag{1} $$

The question is about this asymptotic: if $(1)$ holds then the implicit constant in the big oh does not depend on $z$. In other words,

Let $z,a\in\mathbb{C},\,\textrm{Re}\left(a\right),\,\textrm{Re}\left(z\right)>0.$ Is it true that $$ \gamma\left(a,z\right)=\frac{e^{-z}z^{a}}{a}\left(1+O\left(\frac{1}{\left|a\right|}\right)\right), \quad \left|a\right|\rightarrow\infty $$ and the constant in the error term does not depend by $z$?

I tried to study the identity

$$ \gamma\left(a,z\right)=e^{-z}z^{a}\sum_{n=0}^{\infty}\frac{z^{n}}{\left(a\right)_{n+1}}, $$

where $\left(a\right)_{n+1}$ is the Pochhammer symbol, but I'm not able to prove it.

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  • $\begingroup$ Can you assume that $|z|<|a|$? It seems to me it does depend on $z$ in general $\endgroup$ – shdp Mar 11 '17 at 11:23
  • $\begingroup$ @LeGrandDODOM, The term "asymptotic expansion" refers to an infinite series, not an approximation with a finite number of terms like the ones in this question. Using the word "asymptotic" as a noun, as in this question, is perfectly acceptable. I'm reverting your edits on this point. $\endgroup$ – Antonio Vargas Mar 11 '17 at 11:34
  • $\begingroup$ Doesn't that identity at the end of your answer give a complete asymptotic series? That's neat. $\endgroup$ – Antonio Vargas Mar 14 '17 at 18:08
  • $\begingroup$ @AntonioVargas Yes but I wanted an asymptotic with a big oh error and an absolute constant. Thank you for your hint. $\endgroup$ – user422009 Mar 14 '17 at 18:43
  • $\begingroup$ @AntonioVargas I wrote another question about the incomplete gamma, if you could see it I would be very grateful. $\endgroup$ – user422009 Mar 15 '17 at 9:05
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I showed in this answer that the first few terms in the asymptotic expansion are

$$ \gamma(a+1,z) \sim \frac{e^{-z}z^{a+1}}{a} \left( 1 + \frac{z-1}{a} + \frac{z^2-3z+1}{a^2} + \cdots\right) $$

or, replacing $a$ by $a-1$,

$$ \gamma(a,z) \sim \frac{e^{-z}z^a}{a} \left( 1 + \frac{z}{a} + \frac{z^2-z}{a^2} + \cdots\right) $$

as $a \to \infty$ with $z$ fixed. This suggests that the constant in the error term does depend on $z$, and that a uniform approximation (where the constant does not depend on either $a$ or $z$) would look like

$$ \gamma(a,z) = \frac{e^{-z}z^a}{a} \left[ 1 + O\!\left(\frac{z}{a}\right)\right] $$

as $a \to \infty$. It should not be too hard to prove a statement like this.

You may be interested in this other answer of mine which discusses uniformity of an asymptotic expansion for the function $\Gamma(a,a-z)$ as $a \to \infty$. The ideas from that answer can be adapted to fit your situation.

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