Asymptotics of the incomplete gamma function

Question

Consider the following integral
$$I=\Gamma\left(1-\alpha,\beta\right)=\int\limits _{\beta}^{+\infty}t^{-\alpha}e^{-t}dt,$$
where $\alpha,\beta>0$, and $\Gamma$ is the incomplete gamma function. Is it possible to find an asymptotic estimate of $I$ for $\beta\ll\alpha$?

Answer 1

Let $\lambda> 0$ and $\beta = \lambda (\alpha - 1)$. Then $$ \Gamma (1 - \alpha ,\beta ) \sim \frac{{\beta ^{1 - \alpha } \mathrm{e}^{ - \beta } }}{{\alpha - 1}}\sum\limits_{n = 0}^\infty {\frac{{p_n (\lambda )}}{{(\lambda + 1)^{2n + 1} }}\frac{1}{{(\alpha - 1)^n }}} $$ as $\alpha\to+\infty$ uniformly in $\lambda > 0$. Here $p_0(\lambda)=1$ and $$ p_n (\lambda ) = \lambda (1 + \lambda )p'_{n - 1} (\lambda ) - (2n - 1)\lambda p_{n - 1} (\lambda ) $$ for $n\geq 1$. These are polynomials in $\lambda$ of degree $n$.

To achive the best numerical accuracy, stop the series after about $\left\lfloor {\alpha \sqrt {(\lambda + \log \lambda + 1)^2 + \pi ^2 } } \right\rfloor$ terms. See this paper for more details.