Editing Poisson summation formula (section)

==Derivations==
We prove that,<ref name="Pinsky"/> if <math>s\in L^1(\mathbb R)</math>, then the (possibly divergent) Fourier series of <math>s_P(x)</math> is
<math display="block">s_{_P}(x)\sim \sum_{k=-\infty}^\infty \frac{1}{P}S\left(\frac{k}{P}\right)e^{2\pi i k/P}.</math>
When <math>s(x)</math> is a Schwartz function, this establishes equality in {{EquationNote|Eq.2}} of the previous section.

First, the periodization <math>s_P(x)</math> converges in <math>L^1</math> norm to an <math>L^1([0,P])</math> function which is periodic on <math>\mathbb R</math>, and therefore integrable on any interval of length <math>P.</math> We must therefore show that the Fourier series coefficients of <math>s_{_P}(x)</math> are <math display="inline"> \frac{1}{P} S\left(\frac{k}{P}\right)  </math> where <math display="inline"> S\left( f \right)   </math> is the [[Fourier transform#Definition|Fourier transform]] of <math display="inline"> s\left( x \right)   </math>. (Not <math display="inline"> S\left[ k \right]   </math>, which is the Fourier coefficient of <math>s_{_P}(x)</math>.)

Proceeding from [[Fourier series#Definition|the definition of the Fourier coefficients]] we have''':'''

<math display="block">\begin{align}
S[k]\ &\triangleq \ \frac{1}{P}\int_0^{P} s_{_P}(x)\cdot e^{-i 2\pi \frac{k}{P} x}\, dx\\
&=\ \frac{1}{P}\int_0^{P}
     \left(\sum_{n=-\infty}^{\infty} s(x + nP)\right)
     \cdot e^{-i 2\pi\frac{k}{P} x}\, dx\\
&=\ \frac{1}{P}
     \sum_{n=-\infty}^{\infty}
        \int_0^{P} s(x + nP)\cdot e^{-i 2\pi\frac{k}{P} x}\, dx,
\end{align}</math>

where the interchange of summation with integration is justified by [[dominated convergence]]. With a [[Integration by substitution|change of variables]] (<math>\tau = x + nP</math>), this becomes the following, completing the proof of {{EquationNote|Eq.2}}:

<math display="block">\begin{align}
S[k] =
\frac{1}{P} \sum_{n=-\infty}^{\infty} \int_{nP}^{(n+1)P} s(\tau) \ e^{-i 2\pi \frac{k}{P} \tau} \ \underbrace{e^{i 2\pi k n}}_{1}\,d\tau
\ =\ \frac{1}{P} \int_{-\infty}^{\infty} s(\tau) \ e^{-i 2\pi \frac{k}{P} \tau} d\tau \triangleq \frac{1}{P}\cdot S\left(\frac{k}{P}\right)
\end{align}.</math>

This proves {{EquationNote|Eq.2}} for <math>L^1</math> functions, in the sense that the right-hand side is the (possibly divergent) Fourier series of the left-hand side.  Similarly, if <math>S(f)</math> is in <math>L^1(\mathbb R)</math>, a similar proof shows the corresponding version of {{EquationNote|Eq.3}}.<!-- The proof of Eq.3 may involve the Fourier transform of S(f), so the Fourier transform of the Fourier transform of s(x), and the Dirac delta function. -->

Finally, if <math>s_{_P}(x)</math> has an [[absolutely convergent]] Fourier series, then {{EquationNote|Eq.2}} holds as an equality almost everywhere. This is the case, in particular, when <math>s(x)</math> is a Schwartz function. Similarly, {{EquationNote|Eq.3}} holds when <math>S(f)</math> is a Schwartz function.