The Fourier series
Any piece-wise regular periodic function (finite # of discontinuities, finite # of extreme values) can be written as a series of imaginary exponentials.
Assume f(x) is a periodic function of x with fundamental period L.
f(x) = ∑-∞+∞Cnexp(iknx).
Here kn = n2π/L, Δk = kn+1 - kn = 2π/L.  The coefficients Cm are given by
Cm = (1/L)∫0L f(x) exp(-ikmx)dx.
Using eikx = cos(kx) + i sin(kx) we can also write
f(x) = A0/2 + ∑n=1∞ Ancos(knx) + ∑n=1∞ Bnsin(knx),
with
A0 = (2/L)∫0Lf(x)dx,
Am = (2/L)∫0Lf(x)cos(kmx)dx,
Bm = (2/L)∫0Lf(x)sin(kmx)dx.
We have An = (Cn + C-n),  Bn = i(Cn - C-n),  A0 = 2C0,  n > 0.

Fourier's theorem states that any periodic function with spatial period (wavelength) L can be synthesized by a sum of harmonic functions whose spatial periods (wavelengths) are integral submultiples of L, (such as L/2, L/3, ...).  In the limit  L --> ∞ Fourier's theorem can be generalized to
f(x) = (1/(2π)1/2)∫-∞+∞ f(k)exp(ikx)dk 
f(k) = (1/(2π)1/2)∫-∞+∞ f(x)exp(-ikx)dx
Here f(x) and f(k) are Fourier transforms of each other.

Show:
Assume f(x) is a periodic function with period L.
f(x) = ∑-∞+∞ Cnexp(iknx)/
Cn = (1/L)∫0Lf(x)exp(-iknx)dx = (1/L)∫-L/2L/2f(x)exp(-iknx)dx.
Using Δk  = 2πf, Δk/(2π) = f = 1/L, we write
Cn = (Δk/(2π))∫-L/2L/2f(x)exp(-iknx)dx,
f(x) = (1/(2π))∑-∞+∞exp(iknx)Δk ∫-L/2L/2f(x')exp(-iknx')dx'.
As L --> ∞ , Δk --> 0, and this becomes
f(x) = (1/(2π))∫-∞+∞exp(iknx)dk ∫-∞+∞f(x')exp(-iknx')dx'.
Defining f(k) = (1/(2π)1/2)∫-∞+∞ f(x)exp(-ikx)dx
we have f(x) = (1/(2π)1/2)∫-∞+∞ f(k)exp(ikx)dk.

Note: A Fourier transform is a linear transform;
f(x) = c1f1(x) + c2f2(x)  implies  f(k) = c1f1(k) + c2f2(k) .