Showing $\rho = -1 + \int_0^\infty \frac{e^{-y}}{1+\delta y} \ dy = \sum_{k=1}^\infty (-1)^k \delta^k k!$

Question

Trying to understand the expansion in the title, i.e. $$ \rho = -1 + \int_0^\infty \frac{e^{-y}}{1+\delta y} \ dy = \sum_{k=1}^\infty (-1)^k \delta^k k!. $$ The paper claims to "expand in increasing powers of $\delta y$"; we have $0 \leq \delta \leq 1$.

Any hints are greatly appreciated.

Background information (can skip): $\rho$ is the coefficient of correlation between two random variables $X$ and $Y$, where $(X, Y)$ follows a specific bivariate distribution, with dependence given through $\delta$. Above equivalence serves the purpose to proof that the limit for $\delta\rightarrow 0$ of the correlation ratio $$\eta^2 = \frac{\delta}{3} - \frac{1}{6}-\frac{\rho}{6\delta}$$ is $0$, since from $\rho = \sum_{k=1}^\infty (-1)^k \delta^k k!$ we have for $\delta=0$ that $\frac{\rho}{\delta} = -1$.

Answer 1

As stated the first equation is simply wrong: it holds only for $\delta=0$, otherwise the right-hand side is a divergent series. The correct thing to do to examine the behaviour for small $\delta$ in these situations is to integrate by parts: we have $$ -1+ \int_0^{\infty} \frac{e^{-y}}{1+\delta y} \, dy = \left[ \frac{-e^{-y}}{1+\delta y} \right]_0^{\infty} - \delta \int_0^{\infty} \frac{e^{-y}}{(1+\delta y)^2} \, dy \\ = -1+1 - \delta \left( \left[ \frac{-e^{-y}}{(1+\delta y)^2} \right]_0^{\infty} - \delta \int_0^{\infty} \frac{2e^{-y}}{(1+\delta y)^3} \, dy \right) \\ = -\delta +\delta^2 \int_0^{\infty} \frac{2e^{-y}}{(1+\delta y)^3} \, dy $$ This last integral is finite, and for $\delta>0$, the denominator is larger than $1$, so it is bounded above by $\int_0^{\infty} 2e^{-y} \, dy = 2$, so as $\delta \downarrow 0$, $$ \rho = -\delta + O(\delta^2), $$ which is enough to take the limit you need.

Answer 2

What could be useful (I hope) is to consider (simple substitution $y=t-\frac{1}{\delta }$) $$I=\int \frac{e^{-y}}{1+\delta y} dy=\frac{e^{\frac{1}{\delta }} }{\delta }\int \frac{e^{-t}}{t} dt=\frac{e^{\frac{1}{\delta }} }{\delta }\text{Ei}\left(-t\right)=\frac{e^{\frac{1}{\delta }} }{\delta }\text{Ei}\left(-y-\frac{1}{\delta }\right)$$ where appears the exponential integral function. $$J=\int_0^\infty \frac{e^{-y}}{1+\delta y} dy=\frac{e^{\frac{1}{\delta }} }{\delta }\Gamma \left(0,\frac{1}{\delta }\right)$$ where appears the incomplete gamma function.

Now, using the asymptotics of $\Gamma \left(0,z\right)$ when $z \to \infty$ $$J=\sum_{k=0}^\infty (-1)^k\, k!\, \delta ^k$$ and then the result.