Both the summation routine I coded to verify the hypothesis and the hypothesis itself was wrong.
Firstly, the summation that I reach from (1) to (2) is correct. However, transition from (2) to (3) is not. $\vartheta_3(.)$ is defined as
$$
\vartheta_3(q,z)=1+2\sum\limits_{i=1}^\infty q^{i^2}\cos(2iz).
$$
Rearranging the terms here, we have
$$
\sum\limits_{i=1}^\infty q^{i^2}\cos(2iz)=\frac{\vartheta_3(q,z)-1}{2}
$$
The summation limit starts at $i=1$. Thus, (1) becomes
$$
\sum\limits_{i=0}^\infty \frac{1}{A^x}\left[\exp\left(-\frac{(id-a)^2}{A}\right)+\exp\left(-\frac{(id+a)^2}{A}\right)\right]=\frac{2}{A^x}\exp\left(-\frac{a^2}{A}\right)\left[\frac{1}{2}\left(\vartheta_3\left(q,pj\right)-1 \right)+1\right],
$$
where
$$
q=\exp\left(-\frac{d^2}{A}\right),\phantom{0000} p=\frac{ad}{A}.
$$
Note that the last term comes from $i=0$ of the summation.
The verification code can be corrected as follows
def vartheta(A,d,a,pow):
p=(a*d)/A
q=np.exp(-(d**2)/A)
a1=np.exp(-(a**2)/A)
r=mpmath.jtheta(3, p*1j, q)
qr=a1*(r+1)
qr=float(str(qr.real))
return qr/(A**pow)
As a final note, the first term of the series dominates the summation for small $A$. This is why the result looks perfect for small $A$. As $A$ increases, the contribution of $i=0$ term in the total sum drops, and the result gets more off.
Here's how I was able to get a seemingly correct answer. I, being a physicist by training, have played very fast and loose with somewhat unfamiliar mathematics here. I am sure, on some level, this is illegitimate. However, we have...
$$A(\chi)=-\sum_{i=0}^{\infty}(-1)^{i}(2i+1)\chi\int_{0}^{\infty}t^{-\frac{1}{2}}e^{-\frac{(2i+1)^2\chi^2}{t}}dt$$
applying the substitution $u=\frac{(2i+1)^2\chi^2}{t}$ yields;
\begin{align}
\tag{1}
A(\chi) &=-\sum_{i=0}^{\infty}(-1)^{i}(2i+1)^2\chi^2\int_{0}^{\infty}u^{-\frac{3}{2}}e^{-u}du\\
\tag{2}
&=-\sum_{i=0}^{\infty}(-1)^{i}(2i+1)^2\chi^2\Gamma\big(-\frac{1}{2}\big)\\
\tag{3}
&=2\chi^2\sqrt{\pi}\sum_{i=0}^{\infty}(-1)^{i}(2i+1)^2
\end{align}
Now, we can write the remaining sum thusly...
\begin{align}
\tag{4} &\space\space\space\space\space \sum_{i=0}^{\infty}(-1)^{i}(2i+1)^2\\
\tag{5}
&=\sum_{i=0}^{\infty}(-1)^{i}(4i^2+4i+1)
\end{align}
Which gives us;
\begin{align}
\tag{6}
&= 4\sum_{i=0}^{\infty}(-1)^{i}i^2+4\sum_{i=0}^{\infty}(-1)^{i}i+\sum_{i=0}^{\infty}(-1)^{i}\\
\tag{7}
&=-4\eta(-2)-4\eta(-1)+\eta(0)=0-1+\frac{1}{2}=-\frac{1}{2}
\end{align}
So our final answer, which I have investigated and confirmed numerically, appears to be...
$$\tag{8} A(\chi)=2\chi^2\sqrt{\pi}(-\frac{1}{2})=-\chi^2\sqrt{\pi}$$
Again, I have not been careful about interchanging the integral and sum at the start nor have I been careful in my application of known Dirichlet eta function values to a classically non-convergent sum. So, whatever outrage this may cause among the actual mathematicians is well deserved.
Best Answer
While not definitive, both Mathematica and Wolfram Alpha entirely refuse to integrate this. Moreover, I can't find any such results in Gradshteyn and Ryzhik's tables of integrals (indefinite or definite). So that suggests that the series expansions or approximations seem the best hope. Conveniently the series will converge rapidly for most $-1<q<1$, slowing down only as $q$ approaches $1$ from the left.
In particular, the series converges as $q\to -1^+$. Moreover, while Mathematica fails to compute these integrals directly, it actually can resum this series when $q=-1$. Given that the series evaluates to zero when $q=0$, one deduces the definite integral
\begin{align} \int_{-1}^0 \vartheta_3(q;0)\,dq &= 0-\left[-1+2\sum_{n=1}\frac{(-1)^{n^2+1}}{2n+1}\right]\\ &= 1+2\sum_{n=1}^\infty \frac{(-1)^{n}}{n^2+1}\\ &=1-1 +\pi \operatorname{csch} \pi\\ &=\pi \operatorname{csch}\pi \end{align} which is remarkably simple and agrees in value with Mathematica's own numerical integration.