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Energy signals, i.e., signals with finite energy $$\int_{-\infty}^{\infty}|f(t)|^2dt<\infty$$ have zero power and, consequently, a power spectrum that is equal to zero. They do have an energy density spectrum, which is the squared magnitude of their Fourier transform. You will only get a non-zero power spectrum, according to the definition in your ...
A more general expression states that for $M \geq N$: $$\sum_{n= N}^{n = M} c = (M-N+1) \cdot c$$ where the derivation simply relies on fact that the epxression has (M-N+1) terms : $$\sum_{n= N}^{n = M} c = \{ c + c + ... + c\} = (M-N+1) \cdot c$$ And when applied for your particular case (with $N = -M$) it becomes: $$\sum_{n= -M}^{n = M} c = (M-(-... 0 For instance, from -3 to 3, you have -3,\,-2\,-1,\,0\,1,\,2,\,3, hence 2\times 3+1 terms. More generally, the sum from -M to M is composed of 2M+1 terms: indices with m strictly negative (a total of M), those which m strictly positive (a total of M), plus one at zero (1). If all terms are the same constant c, the total is (2M+... 2 You are missing the effect of the integration interval on the integrand function... If you choose a period; T = 2 , and a periodic signal whose base period is s_0(t) = t in the interval [0,2]... Then the integrand in the shifted interval [-1,1] will be different as given by$$ s_1(t) = \begin{cases}{ t + 2 ~~~, -1 < t < 0 \\ t ~~~~~~~~~~ , ~ ...