Looking at the definition of EllipticTheta
from the documentation:
$$\vartheta _3(u,q)=1+2 \sum _n^{\infty } q^{n^2} \cos (2 n u)$$
it is clear that $\vartheta_3(0, 1)$ diverges. Your problematic term is:
term = EllipticTheta[3, -((d0 g π t)/(2 l)), E^(-((d0 π^2 t)/l^2))]
EllipticTheta[3, -((d0 g π t)/(2 l)), E^(-((d0 π^2 t)/l^2))]
Let $l = 1/s$ and replace the arguments with the small $s$ approximations:
term /. l->1/Series[s, {s, 0, 1}] //TeXForm
$\vartheta _3\left(-\frac{1}{2} s (\pi \operatorname{d0} g t)+O\left(s^2\right),1-\pi ^2 \operatorname{d0} s^2
t+O\left(s^3\right)\right)$
which is of the form:
EllipticTheta[3, -b s, 1 - a s^2]
When $b=0$, Mathematica knows about the asymptotic behavior of this function as $s \to 0$:
Assuming[a>0, Series[EllipticTheta[3, 0, 1 - a s^2], {s, 0, 0}]] //TeXForm
$\frac{\sqrt{\pi } (-1)^{\left\lfloor -\frac{\arg (s)}{\pi }\right\rfloor } e^{-i \pi \left\lfloor
\frac{3}{4}-\frac{\arg \left(-a s^2\right)}{2 \pi }\right\rfloor }}{\sqrt{a} s}$
However, when the second argument also depends on $s$, there appears to be a bug:
Series[EllipticTheta[3, -b s, 1 - a s^2], {s, 0, 0}, Assumptions->a>0] //TeXForm
$\vartheta _3(0,1)+O\left(s^1\right)$
There is an internal function that is designed to handle EllipticTheta
series, but it doesn't seem to be used. We can work around the above issue by using it directly. First, let's reproduce the known result above:
Assuming[a>0, System`SeriesDump`ThetaFunction[3, 0, 1-a s^2, s, 0, 0]] //TeXForm
$\frac{\sqrt{\pi } (-1)^{\left\lfloor -\frac{\arg (s)}{\pi }\right\rfloor } e^{-i \pi \left\lfloor
\frac{3}{4}-\frac{\arg \left(-a s^2\right)}{2 \pi }\right\rfloor }}{\sqrt{a} s}$
Now, let's try it on the second example:
series = Assuming[
a>0 && b>0,
System`SeriesDump`ThetaFunction[3, -b s, 1-a s^2, s, 0, 0]
];
series //TeXForm
$\frac{\sqrt{\pi } (-1)^{\left\lfloor -\frac{\arg (s)}{\pi }\right\rfloor } \exp \left(b^2 s^2
\left(\frac{1}{2}-\frac{1}{a s^2}\right)-i \pi \left\lfloor \frac{3}{4}-\frac{\arg \left(-a s^2\right)}{2
\pi }\right\rfloor \right)}{\sqrt{a} s}$
Much better! Now, recall that the above EllipticTheta
term comes with a prefactor of $\sqrt{\frac{1}{l^2}}$, or equivalently $\sqrt{s^2}$, so we want the limit of s series
as $s \to 0$. Hence:
lim[a_, b_] = Assuming[
a>0 && b>0 && s>0,
Limit[Simplify[s series], s -> 0]
];
lim[a, b] //TeXForm
$\frac{\sqrt{\pi } e^{-\frac{b^2}{a}}}{\sqrt{a}}$
Summarizing, we have the limit:
$$\underset{s\to 0}{\text{lim}}\,s \vartheta _3(-b s,1-a s^2)=\frac{e^{-\frac{b^2}{a}} \sqrt{\pi
}}{\sqrt{a}}$$
Now, the only nonzero term in the limit $s \to 0$ is then:
((a1 Sqrt[(d0 t)/l^2] EllipticTheta[3,-((d0 g π t)/(2 l)),E^(-((d0 π^2 t)/l^2))])/Sqrt[d0 t])
which takes the form:
c s EllipticTheta[3, -b s, 1 - a s^2]
where:
c = a1;
a = d0 π^2 t
b = d0 g π t/2
Hence, the symbolic limit is:
res = a1 lim[d0 π^2 t, d0 g π t/2];
res //TeXForm
$\frac{\operatorname{a1} e^{-\frac{1}{4} \operatorname{d0} g^2 t}}{\sqrt{\pi } \sqrt{\operatorname{d0} t}}$
Let's check:
v = res /. {a1 -> 1, d0 -> 1, g -> 3/2, t -> 1/1000};
v //TeXForm
N[v, 20]
$\frac{10 \sqrt{\frac{10}{\pi }}}{e^{9/16000}}$
17.831208285385305584
in agreement with Bob's answer.
EllipticTheta
function arguments, you see all values ofEllipticTheta
in your equation converging to zero. (ConsiderLimit[EllipticTheta[3, Pi/2, x], x -> 1, Direction -> 1]
and friends.) Numerics ofPlot
behaving the way they do is an entirely different question. $\endgroup$Limit
should generate conditions for its output if you addGenerateConditions -> True
option to it - but it doesn't. $\endgroup$