I am trying analyze an integral of the form

$$I(\varepsilon)=\int_0^\infty f(t,\varepsilon) \,dt$$

where $\varepsilon$ is a small real parameter. The function $f(t,\varepsilon)$ is very complicated, so the integral cannot be evaluated analytically, but it is enough for me to understand its behavior around $\varepsilon=0$. The function $f(t,\varepsilon)$ has a Taylor expansion $f(t,\varepsilon) =\sum_{n=0}^\infty f_n(t) \varepsilon^n$, so the naive answer would be $I(\varepsilon)=\sum_{n=0}^\infty \varepsilon^n\int_0^\infty f_n(t) \, dt$.

However, some of the integrals $\int_0^\infty f_n(t) \, dt$ in the series diverge (to be more specific, the integrands behave as $e^{-t}t^{-m}$, so the integrals diverge around 0). Is there any way to compute the expansion in such case? I have tried various substitutions, division of the integral into several parts, etc., but I always get some kind of divergent or ambiguous result.

As a simple example, consider the integral

$$I(\varepsilon)=\int_0^\infty \frac{\varepsilon\, e^{-t}}{t^2+\varepsilon^2} \, dt$$

This integral can be evaluated explicitly using special functions and its expansion is

$$I(\varepsilon)=\frac{\pi }{2}+\varepsilon (\log(\varepsilon ) + \gamma -1)+\cdots$$

The expansion of the integrand is $$\frac{\varepsilon\, e^{-t}}{t^2+\varepsilon^2}=\frac{e^{-t} \varepsilon}{t^2} -\frac{e^{-t} \varepsilon^3}{t^4}+\frac{e^{-t} \varepsilon ^5}{t^6}+\cdots,$$ which means that integrals of all terms in the series diverge. I would like to know how to get the expansion of $I(\varepsilon)$ without computing $I(\varepsilon)$ explicitly.

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