I've seen it done a few times, it's always roughly apparent that "oh yeah, okay, they've dropped this, pulled this there, yeah, okay, kind of convincing" – but I've never why or how to do it for myself.
I probably don't understand it enough to be searching in the right way, but I'm unable to find anything useful by Googling.
There's a really nice mathmo derivation that I barely understand a word of here.
And a thoroughly unhelpful Physicist "thus it is this" non-explanation here.
But how about a nice not-mathematically-proper-but-it-works method for Engineers?
To be clear, I want to know how to go from:
$$
\lim_{\Delta\omega\rightarrow0} \sum_{n=-\infty}^{\infty} \dfrac{G(n\cdot\Delta\omega)\cdot\Delta\omega}{2\pi} \cdot e^{jn\cdot\Delta\omega\cdot t}
$$
to:
$$
\dfrac{1}{2\pi} \int_{-\infty}^{\infty}G(\omega) e^{j\omega t} \cdot d\omega
$$
and vice versa, and know how to do this for other problems, since as I said, I can sort of see that we've kept the limits, integrated wrt the limit, and removed the summation variable \$n\$. But if I did that on another problem, I wouldn't know I was 'doing it right' – and going in the other direction, how do we know where to put the \$n\$ back in?
Best Answer
You just have to roll through the definitions. There are some small pitfalls but as long as all limits are defined, which they will be with appropriate adjectives on \$G\$ then we are ok. There are several definitions of the symbol \$\int\$. The one which you want to use is the Riemann integral. To simplify the notation I will set \$F(\omega) = \frac{1}{2\pi}G(\omega)e^{j \omega t}\$.
Now the Riemann integral \$\int_a^b F(\omega) d\omega\$ still has a pretty complicated definition involving sups and infs but when it does exist and \$a\$ and \$b\$ are integers then it is equal to $$\int_a^b F(\omega) d\omega =\lim_{j \rightarrow \infty} \sum_{n = aj}^{bj-1} F(n \frac{1}{j})\frac{1}{j}.$$
That is instead of taking \$\Delta \omega \rightarrow 0\$ it is good enough to take \$\Delta \omega = \frac{1}{j} \$ with \$j \rightarrow \infty \$. This is the definition you learned in Calculus class.
The improper integral is defined as $$\int_{-\infty}^{\infty}F(\omega) d\omega = \lim_{a \rightarrow -\infty , b \rightarrow \infty} \int_a^b F(\omega)d\omega.$$
Again, if this limit exists then this is just the limit (for \$k\$ an integer) $$\int_{-\infty}^{\infty}F(\omega)d\omega = \lim_{k \rightarrow \infty}\int_{-k}^k F(\omega)d\omega.$$
Combining all of this and not discussing when/how to changing the order of taking limits is valid we obtain $$\begin{eqnarray*} \int_{-\infty}^{\infty} F(\omega) d\omega &=& \lim_{k \rightarrow \infty} \int_{-k}^k F(\omega)d\omega\\ &=& \lim_{k \rightarrow \infty}\lim_{j \rightarrow \infty} \sum_{n = -kj}^{kj-1} F(n \frac{1}{j})\frac{1}{j}\\ &=& \lim_{j \rightarrow \infty}\lim_{k \rightarrow \infty} \sum_{n = -kj}^{kj-1} F(n \frac{1}{j})\frac{1}{j}\\ &=& \lim_{j \rightarrow \infty} \sum_{n = -\infty}^{\infty} F(n \frac{1}{j})\frac{1}{j}\\ &=& \lim_{\Delta \omega \rightarrow 0} \sum_{n = -\infty}^{\infty} F(n \Delta \omega)\Delta \omega \end{eqnarray*}$$