I would like to know the Laplace Transform and/or Fourier Transform of this function

Question

Let $ c(t) = A \cdot\left( 1 + 2\sum_{n=1}^{\infty} (-1)^n e^{-B\cdot t \cdot n^2} \right).$

Also known as

$A \vartheta _4 (0, e^{-Bt})$

(Jacobi's elliptic theta-function).

Answer 1

As a reminder : $$\vartheta_4(0,x)=\sum_{n=-\infty}^{\infty}(-1)^n x^{n^2}=1+2\sum_{n=1}^{\infty}(-1)^n x^{n^2}$$ $$1+\sum_{n=1}^{\infty}(-1)^n x^{n^2}=\frac{1}{2}\left(\vartheta_4(0,x)+1\right)$$ There is something wrong in the wording of the question :

$ c(t) = A \cdot\left( 1 + \sum_{n=1}^{\infty} (-1)^n e^{-B\cdot t \cdot n^2} \right) \quad$ Also known as $\quad A \vartheta _4 (0, e^{-Bt})$

First interpretation : $$ c(t) = A \cdot\left( 1 + \sum_{n=1}^{\infty} (-1)^n e^{-B\cdot t \cdot n^2} \right) =\frac{A}{2} \cdot\left(\vartheta_4\left(0,e^{-B\cdot t}\right)+1\right) $$ $$\text{which is not }\quad A \vartheta _4 (0, e^{-Bt})\quad\text{as stated.}$$

Second interpretation : $$c(t)=A \vartheta _4 (0, e^{-Bt}) = A \sum_{n=-\infty}^{\infty} (-1)^n e^{-B\cdot t \cdot n^2} $$ $$\text{which is not }\quad A \cdot\left( 1 + \sum_{n=1}^{\infty} (-1)^n e^{-B\cdot t \cdot n^2} \right)\quad\text{as stated.}$$

Supposing that the second is the right one :

$\mathcal{L}_t\big[e^{-B\cdot t \cdot n^2}\big](p)=\frac{1}{B\:n^2+p}\quad$ where $\mathcal{L}$ denotes the Laplace transform.

So, the Laplace transform of $c(t)$ is : $$\mathcal{L}_t\big[c(t)\big](p)=A \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{B\:n^2+p}$$

$$\mathcal{L}_t\big[c(t)\big](p)=\frac{\pi\:A}{\sqrt{B\:p}\:\sinh\left(\pi\sqrt{\frac{p}{B} }\right)} $$

Answer 2

I think this might be a good one. Since the solution is available for composite slabs, let's say this is a composite of two equal membranes, so in the image

https://i.stack.imgur.com/l60sR.jpg

everything _1 = _2

therefore

$Q (s, C , D) = \dfrac{C} {s l \sqrt{s/D} \cdot \cosh \left( \dfrac{l}{2} \sqrt{\dfrac{s}{D}} \right) \cdot \sinh\left( \dfrac{l}{2} \sqrt{\dfrac{s}{D}} \right) }$