I want to compute the following product

$$\frac{1}{N}\sum_{t=1}^{N}\left(\sum_{s=-\infty}^{\infty}a_{s}\exp(2\pi is\frac{t}{N}\right)\left(\sum_{z=-\infty}^{\infty}a_{z}\exp(-2\pi iz\frac{t}{N})\right)$$ If $N\rightarrow\infty$ we can approximate it with integral: $$\int_{0}^{1}\left(\sum_{s=-\infty}^{\infty}a_{s}\exp(2\pi isx\right)\left(\sum_{z=-\infty}^{\infty}a_{z}\exp(2\pi izx)\right)dx=\sum_{s=-\infty}^{\infty}a_{s}\sum_{z=-\infty}^{\infty}a_{z}\int_{0}^{1}\exp(2\pi i(s-z))dx$$ Hence $s=z$ and we have $$\sum_{s=-\infty}^{\infty}a_{s}^{2}$$

On the other hand this product is equal to: $$\frac{1}{N}\left(\sum_{s=-\infty}^{\infty}a_{s}\right)\left(\sum_{z=-\infty}^{\infty}a_{z}\right)\left(\sum_{t=1}^{N}\exp(2\pi i(s-z)\frac{t}{N})\right)=\sum_{w=-\infty}^{\infty}\sum_{s=-\infty}^{\infty}a_{s}a_{s+Nw}$$

We have $$\sum_{s=-\infty}^{\infty}a_{s}^{2}+\sum_{w=-\infty,w\neq0}^{\infty}\sum_{s=-\infty}^{\infty}a_{s}a_{s+Nw}$$

Now I don't see why the second term will disappear.

Which is correct?