How to approximate a sharply peaked function with deltas? 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.
 A: @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.
A: Too long for comments.
Starting from @Han de Bruijn's answer, computing
$$N(a) = \int_{-1}^{+1} \left[\frac{1}{\sqrt{\sinh^2(a)-\sinh^2(ax)}}-\frac{1}{\sinh(a)}\right] dx$$ is not so bad using elliptic integrals
$$N(a) =-\frac{2  }{a}\,\text{csch}(a)\left(a+i F\left(i
   a\left|-\text{csch}^2(a)\right.\right)\right)$$
$$N^*(a) = N(a)\,a\,e^a=-2 e^a \text{csch}(a) \left(a+i F\left(i
   a\left|-\text{csch}^2(a)\right.\right)\right)$$ which approach very fast it horizontal asymptote. For $a=100$, the value is
$$2.772588722239781237668928485832706272302\cdots$$ This number differs from $\log(16)$ by $5.5 \times 10^{-85}$.
A: One way to interpret the idea behind the question is to ask how much mass concentrates in the boundaries. A clean argument interprets $f$ as density or respectively as a measure and embeds those measures in an appropriate space with a norm to define "closeness" or approximation. Accordingly, I show that the integral of $f$ converges to zero. To me this is strong indication that any definition of the limit of $f$ should not contain $\delta$ distributions. In terms of the original questions this means: $A=0.$
To avoid any worries about the singularity of the integrand we will integrate from $-1+\epsilon$ to $1-\epsilon$ and then argue about the limit $\epsilon\rightarrow 0$ and due to the symmetry of the integrand we only need to deal with the positive part.
$$ \int_{0}^{1-\epsilon}\frac{dx}{\sqrt{\sinh^2 a - \sinh ^2 ax}}=\frac{1}{a}\int_0^{a(1-\epsilon)}\frac{dz}{\sinh a\sqrt{1 - (\frac{\sinh z}{\sinh a})^2}}.$$
Now substitute $y = \frac{\sinh z}{\sinh a}$ with $z=\text{arsinh}(y \sinh a)$ and $dz = \frac{\sinh a dy}{\sqrt{1 + (y \sinh a)^2}}$ and use $\sqrt{1 + (y \sinh a )^2} \geq 1$ to obtain
$$ 0
\leq \frac{1}{a} \int_0^{\frac{\sinh a(1-\epsilon)}{\sinh a}} \frac{dy}{\sqrt{1 + (y \sinh a )^2}\sqrt{1 - y^2}}
\leq \frac{1}{a} \int_0^{\frac{\sinh a(1-\epsilon)}{\sinh a}} \frac{dy}{\sqrt{1 - y^2}} = \frac{1}{a}\arcsin\left(\frac{\sinh a(1-\epsilon)}{\sinh a}\right).$$
Ultimately $$\lim_{\epsilon\rightarrow 0} \lim_{a\rightarrow\infty}\frac{1}{a}\arcsin\left(\frac{\sinh a(1-\epsilon)}{\sinh a}\right) =  \lim_{a\rightarrow\infty}\lim_{\epsilon\rightarrow 0}\frac{1}{a}\arcsin\left(\frac{\sinh a(1-\epsilon)}{\sinh a}\right)=0.$$
A: This answer is intended to provide a more complete mathematical account of the heuristic approach used by @Han de Bruijn in a previous answer.
The function as prescribed above does not converge to a delta-function in the limit $a\to\infty$. However it seems to converge to one with some appropriate modifications. Consider the modification
$$f_a(x)=N(a)\left(\frac{1}{\sqrt{\sinh^2 a-\sinh^2ax}}-\frac{1}{\sinh a}\right)$$
So far the normalization factor in front is undetermined but we will see that it is necessary for the convergence towards a distribution with the desired properties. Consider now an integral over a test function and perform the change of variables $\sinh a \tanh t=\sinh ax$
$$\int_{-1}^{1}dx\phi(x)f_a(x)=\frac{N(a)}{a}\int_{-\infty}^{\infty}\phi\left(\frac{1}{a}\sinh^{-1}(\sinh a\tanh t)\right)\frac{\cosh t-1}{\cosh^2 t\sqrt{1+\sinh^2a\tanh^2 t}}$$
Note that $\lim_{a\to\infty}\frac{1}{a}\sinh^{-1}(\sinh a\tanh t)=\text{sgn}(t)$. In order for the limit of the integrand to exist we need to assume that
$$\lim_{a\to\infty}\frac{N(a)}{a\sinh a}=L\in\mathbb{R}$$
With this determined, it is easy to see that we can use the dominated convergence theorem since the integrand is dominated by an integrable function
$$|\phi|\frac{\sinh a}{\sqrt{1+\sinh^2 a\tanh^2 t}}\frac{\cosh t -1}{\cosh^2 t}\leq M_\phi\frac{\cosh t -1}{\cosh t|\sinh t|} $$
where $M_\phi$ is the maximum of $|\phi(x)|,x\in(-1,1)$  (here we have assumed that $\phi$ is an appropriate test function).
Whenever the condition on the normalization factor is satisfied we readily see that
$$\lim_{a\to\infty}\int_{-1}^1dx \phi(x)f_a(x)=L(\phi(1)+\phi(-1))\int_0^{\infty}dt\frac{\cosh t-1}{\cosh t\sinh t}$$
where the integral can be exactly evaluated to be
$$\int_0^{\infty}dt\frac{\cosh t-1}{\cosh t\sinh t}=\log2$$
It can be shown that normalizing $f_a(x)$ by the area under it's curve naturally chooses $L=(2\log 2)^{-1}$ and is enough to guarantee the convergence to exactly $\frac{\delta(x-1)+\delta(x+1)}{2}$, as required by the fact that the area of the normalized curve is $1$.
$$$$
