= Solution
{c}
For fixed nonzero $x$, expand the denominator geometrically:
$$
f(x)=\frac{e^{-x}}{x^2}-\epsilon^2\frac{e^{-x}}{x^4}+O(\epsilon^4).
$$
The effective expansion parameter is $\epsilon^2/x^2$, so the expansion fails when $x=O(\epsilon)$. In the <inner variable> $X=x/\epsilon$,
$$
F(X)=\frac{e^{-\epsilon X}}{\epsilon^2(1+X^2)}
=\frac{\epsilon^{-2}}{1+X^2}-\frac{\epsilon^{-1}X}{1+X^2}+O(1)
$$
for fixed $X$. Expanding the leading outer term toward $x=0$ gives $x^{-2}$ at leading order; expanding the leading inner term for large $X$ and returning to $x$ gives $\epsilon^{-2}X^{-2}=x^{-2}$. Thus the two one-term re-expansions match in $\epsilon\ll x\ll1$. The next common contribution is $-x^{-1}$ when the corresponding next inner order is retained.
On $x\geq\delta$, the relative denominator error is bounded by $\epsilon^2/\delta^2$, hence
$$
I_2=\left[1+O\left(\frac{\epsilon^2}{\delta^2}\right)\right]
\int_\delta^\infty\frac{e^{-x}}{x^2}dx.
$$
<Integration by parts> yields
$$
\int_\delta^\infty\frac{e^{-x}}{x^2}dx
=\frac{e^{-\delta}}{\delta}-\int_\delta^\infty\frac{e^{-x}}x\,dx
=\frac1\delta+\log\delta+O(1).
$$
For the last step split the remaining <integral> at one: $\int_\delta^1 dx/x=-\log\delta$, and replacing $e^{-x}$ by one on that interval has a bounded error.
In the inner interval, $|e^{-x}-1|\leq x$ gives an explicit error estimate:
$$
I_1=\frac1\epsilon\arctan\frac\delta\epsilon
+O\left(\log\left(1+\frac{\delta^2}{\epsilon^2}\right)\right)
=\frac{\pi}{2\epsilon}-\frac1\delta
+O\left(\frac{\epsilon^2}{\delta^3}+
\log\left(1+\frac{\delta^2}{\epsilon^2}\right)\right).
$$
The $1/\delta$ terms cancel between regions. Every displayed remainder is $o(\epsilon^{-1})$ whenever $\epsilon/\delta\to0$ and $\delta\to0$; for example $\epsilon(\epsilon^2/\delta^3)=(\epsilon/\delta)^3$. Therefore the leading answer is independent of the arbitrary intermediate cutoff:
$$
\boxed{I\sim\frac{\pi}{2\epsilon}.}
$$
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