Let:
$$h(t)=\frac{\sin(\pi t(2N+1))}{\sin(\pi t)}$$
$$I=\int_\frac{-1}{2}^\frac{1}{2} h(t) dt$$
when $N\rightarrow\infty$ , obviously (with a change of variable $v=\pi t(2N+1)$ ):
$$I=2\int_0^\frac{1}{2} \frac{\sin(\pi t(2N+1))}{\sin(\pi t)} dt=2\int_0^\frac{\pi (2N+1)}{2} \frac{\sin(v)}{\sin(\frac{v}{2N+1})\pi (2N+1)} dv$$
and since $N\rightarrow \infty$ we have $I=\frac{2}{\pi}\int_0^\infty \frac{sin(v)}{v} dv=1$
It seems to me that we can use the same reasoning for different integral bounds, e.g for
$I=\int_\frac{-3}{2}^\frac{3}{2} h(t) dt\hspace{0.4cm}$ or even $\hspace{0.4cm} I=\int_{-\infty}^{\infty} h(t) dt$
and still obtain $I=1$. But the problem is that $h(t$) is a periodic function (with period 1), since it can easily be shown that,
$$\sum_{k=-N}^N \hspace{0.4cm} e^{2\pi ikt}=h(t)$$
I therefore would expect $\int_\frac{-3}{2}^\frac{3}{2} h(t) dt$ to be equal to $3$, and $\int_{-\infty}^{\infty} h(t) dt$ to be $\infty$ , and certainly not $1$. just to clarify the context, i'm trying to show that $\sum_{k=-N}^N \hspace{0.4cm} e^{2\pi ikt}=h(t)$ is equivalent to a dirac comb, and that's what led me to the integral of $h(t)$...
So... What am i missing? Any help will be immensely appreciated.