I'm am trying to solve an integral of the form $$I=\lim_{t\to-\infty}\int_{0}^{4\sin^2(\pi t/N)}\delta(x-a)f(x)\hspace{1mm}dx.$$
Since, (I'm probably wrong to assume this, since the limit of this function does not exist)$\lim_{t\to-\infty}4\sin^2(\pi t/N)$ oscillates between $[0,4]$. Could I say that $$I=f(a)$$ when $0<a<4$?
EDIT: I simplified the actual problem, but I will write it out in full now. I have to solve the following: $$I=\displaystyle\sum_{\boldsymbol{k}\in\mathbb{Z}^4}\int_{0}^{\infty}d^4n\hspace{1mm}\delta\left(\lambda-E+c^2\left(\sin^2(\pi n_1/N)+\sin^2(\pi n_2/N)+sin^2(\pi n_3/N)+\sin^2(\pi n_4/N)\right)\right)\mathrm{e}^{2\pi i\boldsymbol{k}\cdot\boldsymbol{n}}.$$ Where $c$ and $N$ are constants. Basically, what I've tried is to turn the sine terms into a vector $\boldsymbol{a}=(\sin(\pi n_1/N),\sin(\pi n_2/N),\sin(\pi n_3/N),\sin(\pi n_4/N))$, and re-parameterise the exponent e.g., $$\mu_i=\frac{k_i\cdot n_i}{\sin(\pi n_i/N)},$$ there the exponential term is $\mathrm{e}^{2\pi i\boldsymbol{a}\cdot\boldsymbol{\mu}}$ and $i=1,2,3,4$. But when I change the variables to $\boldsymbol{a}$, the top limit of integration becomes undefined... The reason for the re-parameterisation of the exponent is so then the solution is a Bessel function. I hope this helps.
Discussion rather than answer (but too big for a comment): I sort of see the origin of the problem. It still looks to me as if you should only integrate over one period in the $n$'s ? Like from $0$ to $N$ (possibly $2N$) in each $n_i$.
For the sum over $k$'s it looks like evaluations of the integrand. To simplify things, consider in 1D (without worrying about convergence, and for suitable $C$, and $\phi$ being the expression in the integral apart from the exponential at the end): $$ \hat{\phi}_p = \int_0^{N} \phi(n) e^{2\pi i p n/N} dn,\ p\in Z \ \ \mbox{and} \ \ \phi(m)=C\sum_{p\in Z} \hat{\phi}_p e^{-2\pi i p m/N} $$ Seems that you are only summing over terms with $p=Nk$ so you are calculating (apart from missing pre-factors): $$\sum_{k\in Z}\int_0^N \phi(n) e^{2\pi i k n} dn =\sum_{k\in Z} \hat{\phi}_{Nk}= \frac{1}{N}\sum_{m=0}^{N-1}\phi(m).$$ As I said, this suggest that you should only integrate over 1 period. The last also seems to have a reasonable limit as $N\rightarrow \infty$. Is this a reasonable direction for you problem (before attacking the evaluation of the last sum)?