How to get fourier transform from fourier series

[Ahmed Abdullah]

How you get Fourier transform from Fourier series? Do Fourier series becomes Fourier transform as L --> infinity?

http://mathworld.wolfram.com/FourierTransform.html

I don't understand where discrete A sub n becomes continuous F(k)dk ( where F(k) is exactly like A sub n in Fourier series)?
I also have a general question which is;
How to transform a discrete variable to a continuous variable in order to convert a summation to integral?

 

[Ahmed Abdullah]

I think I know the answer now.

" For a function periodic in [-L/2,L/2], Fourier series is
$$f(x) = \sum_{n=-\infty}^{\infty }A_{n}e^{i(2\pi nx{/}L)} \\A_{n} = 1{/}L \int_{-L{/}2}^{L{/}2}f(x)e^{-i(2\pi nx{/}L)}dx.$$"

The part that was bothering me was " The Fourier transform is a generalization of the complex Fourier series in the limit as L->infty. Replace the discrete $$A_{n}$$ with the continuous F(k)dk while letting n/L->k. Then change the sum to an integral, and the equations become
$$f(x) = \int_{-\infty}^{\infty} F(k)e^{2\pi ikx}dk \\ F(k) = \int_{-\infty}^{\infty} f(x) e^{-2\pi ikx}dx$$ "
My question was how $$A_{n}$$ becomes $$F(k) d(k)$$. Especially where the dk comes from?
$$k=n{/}L$$; then $$\Delta k = (n+1){/}L -n{/}L =1{/}L$$ , when $$L$$ goes to infinity $$\Delta k$$becomes dk.
So when $$L \rightarrow \infty; A_{n} =F(k)dk$$. Indeed! I am a happy man now :).