Approximating/estimating values

Question

I am working on a study of epidemic models for heterogeneous networks.

I have the following expression

$$1-e^z (1-z) + z^2(\gamma + \ln z + z + O(z^2))$$

where $\gamma$ is the Euler-Mascheroni constant. This is an expansion of the total epidemic prevalence $R_\infty$ of a network model with power-law degree distribution as presented in this article. Here is a snippet:

$\quad$ The equation for $\,R_\infty,\,$ with the connectivity distribution $(34)$ is $$ R_\infty = 1-2m^2\int_m^\infty k^{-3}e^{-\lambda k\phi_\infty}dx = 1 - 2z^2\int_{z}^\infty x^{-3}e^{-x}dx, \tag{35} $$ where we have defined the new variable $\,z=\lambda m\phi.\,$ Performing the integral, we obtain $$ R_\infty = 1-e^{-z}(1-z) - z^2\Gamma(0,z), \tag{36} $$ where $\,\Gamma(a,z)\,$ is the incomplete Gamma function [37]. For small values of $\,z\,$, the function $\,\Gamma(0,z)\,$ can be expanded in the expression [37] $$ \Gamma(0,z) \approx -(\gamma_E + \ln z) + z + \mathcal{O}(z^2), \tag{37} $$ where $\,\gamma_E\,$ is the Euler's constant. By inserting this expression into the expression for $\,R_\infty,\,$ we obtain for small values of $\,\phi_\infty\,$ $$ R_\infty \approx 2z \equiv \lambda\langle k\rangle \phi_\infty. \tag{38} $$

How will this expression approximately equal $2z$? Is there a way where I could find possible upper bounds to terms in this expression such that I could approximate it to $2z$? I would appreciate any help or any direction I could start with for this problem. Thank you so much for your time.

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snippet image: https://i.stack.imgur.com/Ulv3X.png

Answer 1

With your helpful snippet it is more clear what is going on. From Wikipedia Incomplete gamma function for the upper incomplete gamma function we have

$$\Gamma(0,z) = \underset{s\to 0}\lim \big(\Gamma(s) - \frac1s - (\gamma(s,z) - \frac1s)\big) = -\gamma - \ln(z) - \sum_{k=1}^\infty \frac{(-z)^k}{k(k!)} $$ is the limiting function to the upper incomplete gamma function as $\,s\to 0,\,$ also known as the exponential integral $\,E_1(z).^{[5]}$

Multiply both sides of equation $(37)$ by $\,-z^2\,$ to get

$$ -z^2 \Gamma(0,z) = z^2(\gamma + \ln(z)) - z^3 + \mathcal{O}(z^4). $$

Now expand in power series the expression

$$ 1-e^{-z}(1-z) = 1 - (1 - z + z^2/2 + \mathcal{O}(z^3))(1-z) = \\ 1 -(1-2z+3z^2/2 +\mathcal{O}(z^3)) = 2z-3z^2/2 +\mathcal{O}(z^3). $$

Adding these two equations gives

$$ 1-e^{-z}(1-z) -z^2 \Gamma(0,z) = 2z + z^2(\gamma -3/2 + \ln(z)) +\mathcal{O}(z^3). $$