What is the followed integration process to get this result?

Question

I'd like to understand how $r[x]$ is gotten. I tried to apply Riemann, but I guess I haven't mastered it yet.

How to get this: $$ r[k]=\dfrac{1}{2\pi}\int_{-\pi}^\pi{s(w)e^{jkw}dw} $$

from this: $$ s(w)=\sum_{k=-\infty}^\infty {r[k]}e^{-jkw} $$

Answer 1

Under the assumption that the following holds true $$s(w)=\sum_{k=-\infty}^\infty {r[\color{red}{k}]}e^{-jkw}$$ the integral on the right hand side evaluates to (interchanging the integral and summation) $$\begin{align}\dfrac{1}{2\pi}\int_{-\pi}^\pi{s(w)e^{jkw}dw}&=\dfrac{1}{2\pi}\int_{-\pi}^\pi \sum_{m=-\infty}^\infty {r[m]}e^{-jmw}{e^{jkw}dw}\\&=\frac{1}{2\pi}\sum_{m=-\infty}^\infty r[m]\int_{-\pi}^{\pi}e^{j(k-m)w}\ dw\end{align}$$ Using the following result $$\int_{-\pi}^{\pi}e^{j(k-m)w}\ dw=2\pi\delta(k-m)$$ we have $$\frac{1}{2\pi}\sum_{m=-\infty}^\infty r[m]\int_{-\pi}^{\pi}e^{j(k-m)w}\ dw=\frac{1}{2\pi}\sum_{m=-\infty}^\infty r[m]\delta(k-m)=r[k]$$ so that $$\dfrac{1}{2\pi}\int_{-\pi}^\pi{s(w)e^{jkw}dw}=r[k]$$