Using Watson's lemma to derive asymptotic expansion

Question

Let $f(\epsilon)=\int_0^{\infty}\frac{e^{-t}}{(1+\epsilon t)^{1/2}}dt$. Derive an asymptotic expansion of the form $f(\epsilon)\approx \sum_{c=0}^{\infty}c_n\epsilon^n$ for $\epsilon\to 0+$.

Idea: This looks a bit like Watson's Lemma, which says that if $f(t)$ is continuous on $(0,\infty)$ with $f(t)=O(e^{\mu t})$ for some $\mu>0$ as $t\to\infty$ and have an asymptotic expansion of the form $f(t)\approx\sum_{n=0}^{\infty}c_nt^{\lambda_n}$ for $\lambda_n>-1$ a sequence of strictly increasing real numbers. Then \begin{equation}\int_0^{\infty}e^{-xt}f(t)dt\approx\sum_{n=0}^{\infty}c_n\frac{\Gamma(\lambda_n+1)}{x^{\lambda_n+1}}\end{equation} as $x\to\infty$. But I am not sure how to transform the original integral to the required form - we have an $e^{-t}$ term but not $e^{-xt}$. So we could, for instance, make the substitution $t=xs$, but I don't see how to proceed here.

Answer 1

Make the change of variables $t\mapsto t/\varepsilon$ so that your integral becomes $$\frac{1}{\varepsilon}\int_0^\infty \frac{e^{-t/\varepsilon}}{(1+t)^{1/2}}\,dt$$ Apply Watson's lemma with $x=1/\varepsilon\to\infty$ as $\varepsilon\to 0$.

We have that $(1+t)^{-\frac{1}{2}}=1-\frac{1}{2}t+O(t^2)$ so that your integral becomes $$1-\frac{1}{2}\varepsilon+O(\varepsilon^2).$$

Answer 2

Starting from @Zachary's solution $$\frac 1{\epsilon}\int \frac{e^{-\frac{t}{\epsilon }}}{\sqrt{1+t}}\,dt=-\frac{e^{\frac{1}{\epsilon }}}{\sqrt{\epsilon }} \Gamma \left(\frac{1}{2},\frac{t+1}{\epsilon }\right)$$ $$\frac 1{\epsilon}\int_0^\infty \frac{e^{-\frac{t}{\epsilon }}}{\sqrt{1+t}}\,dt=\sqrt{\frac{\pi}{\epsilon }}\,\,e^{\frac{1}{\epsilon }}\,\, \text{erfc}\left(\frac{1}{\sqrt{\epsilon }}\right)$$ which gives $$\color{blue}{f(\epsilon)=\sum_{n=0}^\infty (-1)^n\,\frac{(2n-1)!!}{2^n}\,\epsilon^n=\frac{1}{\sqrt{\pi }}\sum_{n=0}^\infty (-1)^n\,\Gamma \left(n+\frac{1}{2}\right)\epsilon^n}$$