Given are $N$ complex numbers $x_0,x_1,\cdots,n_{N-1}$. The discrete Fourier transformation is defined as:
$$k_m=\sum_{n=0}^{N-1} x_n e^{-i\frac{2\pi}{N}mn}~~~(1)$$
The inverse transformation is defined as:
$$x_n=\sum_{m=0}^{N-1}k_m e^{i\frac{2\pi}{N} n}~~~(2)$$
My goal is to perform the continuum limit. Then I should obtain the normal Fourier transformation formulas $f(x)=\int_{-\infty}^{\infty} dk e^{-ikx} f(k)$ and $f(k)=\int_{-\infty}^{\infty} dx f(x) e^{-ikx}$.
To do the proof I draw the following cartoons. Left: each $n$ is related to some $x$ value that will in the end be a continuous variable. The $x_n$'s will become the $f(x)$. Before performing the continuum limit the $x$ interval is finite. It reaches from $-L/2$ to $L/2$. The distance between two $x$ values before the continuum limit is $\Delta x=\frac{L}{N-1}$. The $x$ which will become the continuous variable after the continuum limit can be parameterized as $x=-\frac{L}{2}+n\Delta x$.
Analogous relations are also true in Fourier space. $m=0$ is related to a $k$ value of $-4\pi/L$ and $m=N-1$ is related to $4\pi/L$. the variable that will become continuous after taking the continuum limit is parameterized as $k=-\frac{4\pi}{L}+m\Delta k$ with $\Delta k=\frac{8\pi}{L(N-1)}$.
Now the continuum limit is $N\to \infty$ and $L\to \infty$ where $\Delta x=\frac{L}{N-1}\to 0$, i.e. $N$ diverges faster than $L$. One thing that is unclear right now is that in this case the $k$ range in Fourier space shrinks to zero since it goes from $-4\pi/L$ to $4\pi/L$?
Anyways, ignoring this for a moment my problem is more how I can rewrite $(1)$ as $f(k)=\int_{-\infty}^{\infty} dx f(x) e^{-ikx}$ after performing the continuum limit. It is trivial that $k_m$ becomes $f(k)$ and $x_n$ becomes $f(x)$. But what happens with the $m n$ factor in the exponential? Using the expressions for $n$ and $m$ I get
$mn=\frac{k+\frac{4\pi}{L}}{\Delta k}\frac{x+\frac{L}{2}}{\Delta x}=\left(x k + x\frac{4\pi}{L}+k\frac{L}{2}+2\pi\right)\frac{(N-1)^2}{8\pi}$
But this does not really make sense to me since first the result diverges in the continuum limit ($L\to \infty$) and second there are additional terms on top of the desired $xk$ term?
Also to convert the sum $\sum_{n=0}^{N-1}$ to an integral we need something like $\sum_{n=0}^{N-1}\Delta k\to \int dk$, but I am missing the $\Delta k$?