TITLE: How to approximate a sharply peaked function with deltas?
QUESTION [8 upvotes]: 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.

REPLY [3 votes]: 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 $5.5 \times 10^{-85}$.