# Uniform asymptotic expansion of integral with corner singularities

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.

• 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? – Chip Apr 3 at 1:49
• @Chip I think in general $f(x+y)+f(x-y)$ may still have a corner close to $y=0$. – Ian Apr 3 at 2:15
• can you define more precisely 'corner singularity' : are the derivatives discontinues there, or only the first derivative, $\it etc$... – Chip Apr 3 at 2:41
• 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 $2 f(x)/M + {\cal O}(1/M^3)$? – Chip Apr 3 at 3:02
• @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. – Ian Apr 3 at 3:20