# PSD in MATLAB, problem with the coefficient $\frac{1}{2\pi}$

Using the FFT to mimic the Fourier Transform I have this question.

The definition that I use for the PSD $$S(f)$$ is:

$$S(f)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\exp\left(-i\tau2\pi f\right)r(\tau)d\tau$$

where $$r(t)$$ is the correlation function.

The FFT in MATLAB use this formula

$$X(k) = \sum_{n=1}^N x(n)\exp\left(-j2\pi\frac{(k-1)(n-1)}{N}\right), \qquad 1 \leq k \leq N.$$

So in order to replace the PSD I did:

S=abs(fft(r))*delta_t


where delta_t is the time step. In other words, translating the code in formula I did:

$$S(k)=\sum_{n=1}^{N}\exp\left(-i2\pi \frac{(k-1)(n-1)}{N}\right)r(n)\Delta T$$

After using this formula I applied this, checking the result $$\int_{-\infty}^{\infty}S(f)df=r.m.s. \big\{x(t)\big\}$$

where $$x(t)$$ is the original signal. This formula is perfectly verified.

Now the problem is that I noticed that in the MATLAB's fft there is no coefficient $$\frac{1}{2\pi}$$.

Can you give me an explanation of why this happens? I know that there are a lot of different coefficient that I can use for the Fourier transform pair, but I don't know if this is related to my problem.

Thanks