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.

  • $\begingroup$ Can you not re-write the integral as $\int_0^\infty e^{-M y} [f(x+y)+f(x-y)] dy$ and from here you apply standard asymptotic expansion? $\endgroup$ – Chip Apr 3 at 1:49
  • $\begingroup$ @Chip I think in general $f(x+y)+f(x-y)$ may still have a corner close to $y=0$. $\endgroup$ – Ian Apr 3 at 2:15
  • $\begingroup$ can you define more precisely 'corner singularity' : are the derivatives discontinues there, or only the first derivative, $\it etc$... $\endgroup$ – Chip Apr 3 at 2:41
  • $\begingroup$ If one writes starting with my comments above $f(x)$ as its back Fourier transform of $\tilde f(\omega)$ on $[0,1]$ (the period of $f(x)$), one gets: $2 \int_0^1 d\omega e^{I \omega x} \frac{M}{M^2+\omega^2} \tilde{f}(\omega)$. Does this help you? (see mathworld.wolfram.com/DampedExponentialCosineIntegral.html and extra.research.philips.com/hera/people/aarts/…) First term may be $2 f(x)/M + {\cal O}(1/M^3)$? $\endgroup$ – Chip Apr 3 at 3:02
  • $\begingroup$ @Chip In my actual setting (which is a bit different, this is a simplification for ease of brainstorming) all the odd derivatives are discontinuous and the even ones are continuous. $\endgroup$ – Ian Apr 3 at 3:20

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