How can I asymptotically expand this integral? How do you asymptotically expand the following to linear order:
$$\frac{1}{\theta}\int_{-\infty}^{t}\exp\left(\frac{s-t}{\theta}\right)\frac{d}{ds}D(s)\mathrm{d}s?$$
I need to show that in the limit $\theta\to 0$, the above becomes $\frac{dD}{dt}$.
My instinct is to Taylor expand the integrand and use IBP, but surely this would give a term involving $D(t)$ and not $\frac{dD}{dt}$. Any help is much appreciated.
 A: Is there a specific reason you need an asymptotic expansion?
If you only care about showing that the integral tends to $dD/dt$ in the limit, then you just need to formulate the integral as an approximate identity. Let $H(s)$ denote the unit step function, i.e.
$$
H(s) = \begin{cases}
1 & s\geq 0\\
0 & s < 0
\end{cases}
$$
and define
$$
f(s) = \exp(-s)H(s).
$$
Note that for all $\theta>0$, $H(s/\theta) = H(s)$. Setting $g(s) = \frac{dD}{ds}(s)$, your integral is equal to
$$
\frac{1}{\theta}\int_{-\infty}^\infty f\left(\frac{t-s}{\theta}\right)g(s)~ds.
$$
Writing $f_\theta(s) = (1/\theta)f(s/\theta)$, we can therefore rewrite the integral as the convolution $(f_\theta * g)(t)$. It then remains to show that $f_\theta$ is an approximate identity (or mollifier, or good kernel, whatever your preferred language is). This is simple as $f$ is an $L^1$ function, $\int_\mathbb{R} f(s)~ds = 1$, and $f_\theta$ are just the 1-dimensional rescalings of $f$. Therefore $(f_\theta * g)(t) \to g(t)$ pointwise almost everywhere as $\theta\to\infty$ for any reasonably nice function $g$. The convergence can be upgraded to stronger than pointwise a.e. depending on how nice $g$ is.
