# Discrete fourier transform in different forms

For the discrete Fourier transform, it is defined by

$$f(k)=\sum_{s_i}\exp(-iks)\phi(s).~~~~~~~~~~~~~~~~~(ds-1)$$ here $$s_i=-(N-1)/2,-(N-2)/2,…..(N-1)/2$$.

For convenience, we also add the continuous Fourier transform

$$f(k)=\int_{-\infty}^{\infty}\exp(-iks)\phi(s)ds~~~~~~~~~~~~~~~~~(cn-1)$$

It can be seen that for the discrete case, the integral of the right-hand side of Eq.~(cn-1) is chosen at some special point, i.e., $$s_i=-(N-1)/2,-(N-2)/2,…..(N-1)/2$$. If you take $$N\to \infty$$, the above two formulas should consistent with each other.

While for the inverse discrete Fourier transform, it reads

$$\phi(s)=\frac{1}{N}\sum_{k_i}\exp(iks)f(k),~~~~~~~~~~~~~~~~~(ds-2)$$ where $$k_i=-\frac{2\pi}{N}\frac{N-1}{2},……\frac{2\pi}{N}\frac{N-1}{2}$$. Let $$N\to \infty$$, Eq.~(ds-2) can be shown by $$\phi(s)=?\frac{1}{N}\int_{-\pi}^{\pi}\exp(iks)f(k)dk~~~~~~~~~~~~~~~~~(ds-3)$$ I believe Eq.~(ds-3) is wrong if you let $$N\to \infty$$.

I see that in some books, they use

$$\phi(s)=\frac{1}{2\pi}\int_{-\pi}^{\pi}\exp(iks)f(k)dk~~~~~~~~~~~~~~~~~(ds-4)$$

Question 1): how to understand the equation (ds-3) and (ds-4). Why $$N$$ should be replaced by $$2\pi$$?

Question 2): If N is finite, we can use Eq.~(ds-1) for the discrete Fourier transform, and the inverse is given by (ds-2). While, for the discrete case, if $$N\to \infty$$, how can I get the inverse discrete Fourier transform? Can we use Eq.~(ds-2)? But If $$N\to \infty$$, it seems that Eq.~(ds-2) is not correct.

Question 3) If $$N\to \infty$$, can we use the continuous Fourier transform, i.e., $$\phi(s)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\exp(iks)f(k)dk~~~~~~~~(cn-2)$$ to estimate Eq.~($$ds-2$$)? It seems that Eq.($$ds-4$$) is different from Eq.~(cn-2). Is there some relation between Eq.~(ds-4) and Eq.(cn-2).

Question 4, whih formula is the discrete inverse Fourier tansfom in the limit of $$N\to \infty$$?

Any suggestions or related URL or books are welcome! Thanks!