= Solution
For a full-rank <Euclidean lattice> $\Lambda\subset\mathbb R^n$, its <dual lattice> is
$$
\Lambda'=\{u\in\mathbb R^n:\langle u,\lambda\rangle\in\mathbb Z\text{ for all }\lambda\in\Lambda\}.
$$
If $\Lambda=B\mathbb Z^n$, then $\Lambda'=B^{-\mathsf T}\mathbb Z^n$, $m(\Lambda)=|\det B|$ and $m(\Lambda')=m(\Lambda)^{-1}$. The <characters of a real torus> identify the additive <dual lattice> with the multiplicative character group by
$$
\boxed{u\longmapsto\chi_u,\qquad\chi_u(x+\Lambda)=e^{2\pi i\langle u,x\rangle}.}
$$
The map is well defined precisely because $u\in\Lambda'$, and is injective. For surjectivity, a continuous <group homomorphism> from the compact torus into $\mathbb C^\times$ has compact image. Its modulus has logarithm a homomorphism into $(\mathbb R,+)$ with compact image, hence is zero, so the image lies in the unit circle. Pull the character back to $\mathbb R^n$. Its continuous real lift under $t\mapsto e^{2\pi it}$, normalized to zero at the origin, is additive: its additive defect is an integer-valued continuous function and vanishes at the origin. A continuous additive function is $\langle u,x\rangle$ for a unique $u$. Triviality on $\Lambda$ says $u\in\Lambda'$, proving surjectivity.
Use the <Fourier transform> convention $\widehat f(u)=\int_{\mathbb R^n}f(x)e^{-2\pi i\langle u,x\rangle}\,dx$, and take $f$ to be a <Schwartz function>. The <periodization of a Schwartz function> $P_f(x)=\sum_{\lambda\in\Lambda}f(x+\lambda)$ is smooth and $\Lambda$-periodic. On a fundamental cell $F$, the coefficient of the torus character $\chi_u$ is
$$
\frac1{m(\Lambda)}\int_F P_f(x)e^{-2\pi i\langle u,x\rangle}\,dx
=\frac1{m(\Lambda)}\widehat f(u),\qquad u\in\Lambda'.
$$
The equality follows by translating each cell and using $e^{2\pi i\langle u,\lambda\rangle}=1$. The rapidly convergent <Fourier series> can be evaluated at zero, yielding the <Poisson summation formula for a Euclidean lattice>
$$
\boxed{\sum_{\lambda\in\Lambda}f(\lambda)=\frac1{m(\Lambda)}\sum_{u\in\Lambda'}\widehat f(u).}
$$
This argument keeps track of the covolume factor rather than tacitly assuming a unit-volume lattice.
For $\tau$ in the <complex upper half-plane>, let $g_\tau(x)=e^{\pi i\tau|x|^2}$. Scaling the self-dual real <Gaussian function> gives its <Fourier transform> at $\tau=it$, $t>0$, and holomorphic continuation in $\tau$ gives the <complex Gaussian Fourier transform>
$$
\widehat g_\tau(u)=(\tau/i)^{-n/2}\exp\left(-\frac{\pi i|u|^2}{\tau}\right).
$$
The branch is $\exp[-(n/2)\operatorname{Log}(-i\tau)]$ with the logarithm on the right half-plane; it is positive for $\tau=it$. Both the integrals and the lattice sums are locally normally convergent on the <complex upper half-plane>. Applying the <Poisson summation formula> proves the <lattice theta functional equation>
$$
\boxed{\Theta_\Lambda(\tau)=(\tau/i)^{-n/2}m(\Lambda)^{-1}\Theta_{\Lambda'}(-1/\tau).}
$$
No integrality or self-duality hypothesis on the lattice is needed.
Put $a=n/2$, $m=m(\Lambda)$ and $\theta_\Lambda(t)=\Theta_\Lambda(it)$. The <Epstein zeta function> converges absolutely for $\operatorname{Re}s>a$, and its <Mellin transform> representation is
$$
\mathcal E_\Lambda(s)=\pi^{-s}\Gamma(s)E_\Lambda(s)
=\int_0^\infty(\theta_\Lambda(t)-1)t^{s-1}\,dt.
$$
At infinity, $\theta_\Lambda(t)-1$ decays exponentially. Define the entire function
$$
A_\Lambda(s)=\int_1^\infty(\theta_\Lambda(t)-1)t^{s-1}\,dt.
$$
On $(0,1)$, substitute $\theta_\Lambda(t)=m^{-1}t^{-a}\theta_{\Lambda'}(1/t)$ and then $u=1/t$. Isolate the two elementary terms before integrating; this gives the <pole-subtracted theta integral for an Epstein zeta function>
$$
\boxed{\mathcal E_\Lambda(s)=A_\Lambda(s)+m^{-1}A_{\Lambda'}(a-s)
+\frac{m^{-1}}{s-a}-\frac1s.}
$$
The formula initially holds for $\operatorname{Re}s>a$ and continues the <completed Epstein zeta function> meromorphically to all $\mathbb C$. Apply the same formula to the <dual lattice> at $a-s$, using $m(\Lambda')=m^{-1}$ and $\Lambda''=\Lambda$. The entire terms and the two rational terms match, proving
$$
\boxed{\mathcal E_\Lambda(s)=m(\Lambda)^{-1}\mathcal E_{\Lambda'}(n/2-s).}
$$
Finally $E_\Lambda(s)=\pi^s\mathcal E_\Lambda(s)/\Gamma(s)$ also continues meromorphically. Its only pole is a simple one at $s=n/2$, with residue $\pi^{n/2}/[m(\Lambda)\Gamma(n/2)]$; the pole of the completed function at zero is cancelled by $1/\Gamma(s)$, and $E_\Lambda(0)=-1$. The residues and this cancellation make explicit why subtracting the constant theta term was necessary.
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