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The Wikipedia page on group velocity uses the superposition of two cosines (of slightly different wavenumbers and angular frequencies) to show that group velocity $v_g = \frac{d\omega}{dk}.$

The sum yields to the product of two cosines $ 2 cos(kx-\omega t)cos(\Delta k x - \Delta \omega t)).$

The right-hand cosine plays the role of modulator and it's phase velocity is $ \frac{\Delta \omega}{\Delta k}$

So that all makes sense and hence, we have $v_g = \frac{d\omega}{dk}.$

What I quite haven't understand is the continuous superposition.

For the continuous superposition the formula is $$ \int dk \Phi (k) e^{i(kx-\omega t)} .$$

$\Phi (k)$ being the amplitude (we make the assumption that it narrowly peaks around a certain value $k_0$), and the $exp$ preferably used because matter waves can't be made of $cos$ ...

The (continuous) summation is made between different $exp$ at slightly different wavenumbers (around $k_0$) BUT I can't see how the summation is made at slighly different angular frequencies since $\omega$ is now a function of k whose shape is absolutely unknown (if $k_1$ and $k_2$ are close we don't have the guanrantee that $\omega (k_1)$ will be close to $\omega (k_2)$).

So why wouldn't, similarly to the discrete case, be a summation over slightly different $\omega$'s?

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The continuity of the function $\omega(k)$ guarantees that $\omega(k_1)$ and $\omega(k_2)$ will be close if $k_1$ and $k_2$ are sufficiently close. Formally, the continuity of $\omega(k)$ at $k_1$ means that you can make $|\omega(k_2) - \omega(k_1)|$ arbitrarily small by choosing $k_2$ such that $|k_2 - k_1|$ is sufficiently small.

If $\omega$ is a monotonic function of $k$ (at least in an interval where $\Phi(k)$ is non-zero), it is invertible, so you can express $k$ as a function of $\omega$ and use frequency as the integration variable via a change of variables. Namely, given the function $k(\omega)$ with $k_1 = k(\omega_1)$ and $k_2 = k(\omega_2)$,

$$ \int\limits_{k_1}^{k_2}dk\ \Phi(k)e^{i[kx-\omega(k)t]} = \int\limits_{\omega_1}^{\omega_2}d\omega\ \frac{dk}{d\omega}(\omega)\ \Phi\left(k(\omega)\right)e^{i[k(\omega)x-\omega t]} $$

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