I have the expression $$\frac{a\sinh\left[\frac{p}{c}(L-x)\right]}{p\sinh\left[\frac{pL}{c}\right]}$$ and I want to find the inverse Laplace transform as an infinite series by using the binomial expansion. I tried rewriting the denominator as exponentials and expanding using the binomial theorem, but this gave me an infinite series of terms of the form $\exp\cdot\sinh$ which when inverse Laplace transformed gave me a delta function using the convolution theorem (which doesn't seem right - I don't think my answer should be an infinite sum of delta functions!).
How can I use the binomial theorem here?
Here is an answer to the question but without any use of binomial theorem.
Let us use simplified notations:
$$F(s)=\frac{a}{p}\frac{\sinh(Ap-B)}{\sinh(Ap)}= \frac{a}{p}\left(\frac{\sinh(Ap)\cosh(B)-\sinh(B)\cosh(Ap)}{\sinh(Ap)}\right)$$
$$F(s)=\frac{a}{p}\cosh(B)-a\sinh(B)\frac{1}{p}\operatorname{coth}(Ap)\tag{1}$$
Here, we can use the (ill-known) series representation:
$$\operatorname{coth}(X)=\frac1X + 2\sum_{n=1}^{\infty} \frac{X}{\pi^2n^2+X^2}$$
giving
$$\frac{1}{p}\operatorname{coth}(Ap)=\frac{1}{Ap^2} + 2A \sum_{n=1}^{\infty} \frac{1}{\pi^2n^2+(Ap)^2}\tag{2}$$
From there it is not difficult to obtain the inverse Laplace Transform of (2) due to classical formula:
$$\mathcal{L}^{-1}\frac{1}{k^2+p^2}=\frac1k \sin kt$$
under the form of an infinite series, in fact the Fourier series of a sawtooth function.
But where is the Binomial Theorem hidden ?