I have the function
$$
f(x)=\cases{\frac{1}{\sqrt{\sinh^2(a)-\sinh^2(ax)}} & $-1 \leq x \leq1$ \\ 0 & otherwise}
$$
Where $a \in \mathbb{R}^+$. Here is a plot of $f$:
It would be very convenient if I could justify (in any way: maybe distributionally?) that
$$
f(x)=A\left[ \delta(x-1)+\delta(x+1) \right]+B+\mathcal{O}(?) \quad ; \quad a \to \infty \tag{1}
$$
For some $x$ independent constants $A$ and $B$. I am uncertain of how to do this, or how to estimate the error.
What I've tried
My thought was to look at the Fourier series of $f$, and associate terms with the Fourier series of (2). By expanding (1) around $x= \pm1$, I can find a Laurent series in $(x \pm 1)$, and perform the Fourier integral term by term. The result is messy, and worse: does not allow easy comparison to the Fourier series of (2). This may not be a fruitful approach.
By looking at the wiki page about mollifiers, it seems I need to claim something like: $a f(x) \to \delta (x \pm1), \ a \to \infty$. Superficially, this appears to work, but I'm not sure how to make it concrete, or if there should be any other factors next to $af(x)$ (other than a factor to ensure normalization to unity).
Questions
- How can I estimate the error and find $A$ and $B$ in (1)?
- If $A$ and $B$ cannot be found, how can I find the ratio $A/B$?
Background
$f(x)$ will eventually appear beneath an integral with a Green's function $G$. I'd like to formulate this for arbitrary $G$. We can take some liberties about the 'niceness' of $G$, as necessary. In particular, it has a well behaved Fourier transform.
Best Answer
@Sal: Not the answer you're looking for I'm sure, and I anticipate the MSE gods will delete this in short order, but perhaps it helps you in the meantime as you continue your research.
After messing about with things a bit, analytically and numerically, a similar function has the delta-like property. Take $\phi_a(x) = \frac{2}{\pi}\frac{a\sinh(ax)}{(\sinh^2 a - \sinh^2(ax))^{1/2}}$ on $0 < x < 1$. Then completely experimentally and without proof (or perhaps fodder for another post!) it appears $\int_0^1\, g(x)\phi_a(x)\, dx \rightarrow g(1)$ as $a \rightarrow \infty$ for a smooth function $g$.
== ADDED ==
If you want to approximate $f$ by $A\delta(x + 1) + A\delta(x - 1) + B$ in some sense, you'll probably want something like \begin{equation*} \int_{-1}^{1}\, g(x)f(x; a)\, dx - A g(-1) - Ag(1) - B\int_{-1}^1\, g(x)\, dx \rightarrow 0 \end{equation*} as $a \rightarrow \infty$ to hold. Use a few examples, $g(x) = 1, x^2, x^4, \dots$ and work out what $A$ and $B$ would need to look like. For example with $g = 1$ I get $\int_{-1}^1\, f(x; 20)\, dx = 4.12252\times 10^{-9}$ and with $g(x) = x^2$, $\int_{-1}^1\, x^2 f(x; 20)\, dx = 1.37424 \times 10^{-9}$. To me this path doesn't look promising. The area under the singular portions simply isn't big enough to mimic a delta function.