Let $f : \mathbb{R} \to \mathbb{R}$ be a given function which is $1$-periodic and is smooth except for finitely many corner singularities in each period. An example is $-|\sin(2\pi x)|$, which has a corner singularity at $x=k/2$ for each integer $k$.
I'm now looking at
$$I(x,M)=\int_{-\infty}^\infty e^{-M|y|} f(x-y) dy$$
where $M$ is a large parameter. I would like to obtain an asymptotic expansion for $I$ as $M \to \infty$ which is uniform in $x$ when $f$ is fixed.
The difficulty in doing so is that approximating $f$ on an interval of length $O(M^{-1})$ near $x$ cannot be done by a polynomial when $x$ is within $O(M^{-1})$ of a singularity of $f$. This causes the Laplace method to fail to provide a uniform expansion, essentially because it provides two different expansions when $x$ is a singularity or not, and these do not agree with one another as $x$ passes through a singularity.
Generally when I see methods for uniform asymptotic expansions of integrals in the literature, they still rely on local analyticity assumptions in order to convert the problem into a complex analysis problem. Or they simply identify the problem with some special function and then perhaps perform further asymptotics directly on the special function. In any case, I don't see how this can work here. Is there some alternative method?
An alternative that comes to mind is to use integration by parts. Putting aside remainder estimation, a first step of integration by parts for the $y \in [0,\infty)$ integral can be done as follows. Identify $\delta>0$ where the first singularity is located in $y$, then split the integral there. Now the first step of integration by parts gives only the "local" term $\frac{f(x)}{M}$ which is exactly what the first step of the Laplace method would give. The next step gives more interesting terms:
$$I=\frac{f(x)}{M} - \frac{1}{M} \int_0^\infty e^{-My} f'(x-y) dy$$
Now when we split the second integral and integrate by parts, the boundary terms provide a term corresponding to $f'(x)$ from the left endpoint of integration and additionally provide a non-cancelling pair of terms from the jump in $f'$ at $y=\delta$.
Can we continue, or is there some breakdown further along in the procedure? It seems to work, providing a term corresponding to the jump (if present) in each derivative at $y=\delta$. And the expansion itself appears to be a nice function of $x$ as long as singularity-to-singularity is a full period (as in the example, which is actually best characterized as being $1/2$-periodic). If singularity-to-singularity is not a full period then we need to retain more terms in order to maintain continuity at the midpoint between two singularities, but that's a technical detail.