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Here wikipedia it is said that the Dirac delta could be thought of as $$ \delta(x) = \left\{ \begin{array}{ll} \infty &, x = 0 \\ 0 &, x \ne 0 \end{array}\right. $$ and here that the Fourier-Transform of $\cos(2\pi k_0 x)$ is $$ \frac{1}{2} \left( \delta(k - k_0) + \delta(k + k_0) \right). $$ From computer programs I know that the spectrum gives the intensity of every cos/sin wave in the waveform, but where is the intensive (i.e. one) of the simple cosine waveform represented when i thought of $\delta(k-k_0)$ as being "stretched till infinity" at the point $k_0$?

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You are right in a sense. In a Fourier series, you have a single coefficient that tells you the "intensity" of each frequency: $$f = c +\sum_{n=1}^\infty a_n sin(nx) + b_n cos(nx)$$ you can just read off the $a_n$ and $b_n$ to find the contribution of any sinusoidal component.

The Fourier transform on the other hand, may yield a continuous frequency. If $\hat{f}$ is the Fourier transform of $f$, then $\int_a^b \hat{f}(\xi)d\xi$ gives you the contribution of the "range" of frequencies from $a$ to $b$. Here, the contributing frequencies are $k_0$ and $-k_0$, and since $\hat{f}$ is 0 elsewhere, you know these are the only values that matter. The integral over neighborhoods of these points will tell you the contributions. Don't think of the function $\delta(x)$ has having a value at 0, think of it only as having a nonzero integral over an infinitesimal domain.

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