3 - The Game Simplified

Lets look at what happens if our signals are not periodic. We can achieve this by setting our period T to infinity such that $\lim _{T\to \infty }\sum _{n=-\infty }^{\infty }(1/T\int _{-T/2}^{T/2}x(t^{'})e^{-j2\pi nt^{'}/T}dt^{'})e^{j2\pi nt/T},\!$
where
$\ 1/T\int _{-T/2}^{T/2}x(t^{'})e^{-j2\pi nt^{'}/T}dt^{'}=\alpha _{n}\!$
first we need to remove the restiction x(t) = x(t + T) by following these steps.
1/T $\Longrightarrow df\!$
n/T $\Longrightarrow df\!$
$\sum _{n=-\infty }^{\infty }1/T\Longrightarrow \int _{-\infty }^{\infty }()df\!$
$\alpha _{n}\Longrightarrow X(f)\!$
this leads us to the equation
$x(t)=\lim _{T\to \infty }\sum _{n=-\infty }^{\infty }(1/T\int _{-T/2}^{T/2}x(t^{'})e^{-j2\pi nt^{'}/T}dt^{'})e^{j2\pi nt/T},\!$
and if we replace n/T with f and take the integral with respect to f we get
$x(t)=\int _{-\infty }^{\infty }(\int _{-\infty }^{\infty }x(t^{'})e^{-j2\pi ft^{'}}dt^{'})e^{j2\pi ft}df,\!$
where $\int _{-\infty }^{\infty }x(t^{'})e^{-j2\pi ft^{'}}dt^{'}=X(f)\!$
simplifying the equation to
$x(t)=\int _{-\infty }^{\infty }X(f)e^{j2\pi ft}df=<X(f)|e^{j2\pi ft}>\!$ = Inverse Fourier Transform
and
$X(f)=\int _{-\infty }^{\infty }x(t)e^{-j2\pi ft}dt=<x(t)|e^{j2\pi ft}>=\!$Fourier Transform
Now using the x(t) equation and rearranging it gives us $x(t)=\int _{-\infty }^{\infty }(\int _{-\infty }^{\infty }x(t^{'})e^{-j2\pi ft^{'}}dt^{'})e^{j2\pi ft}df=\int _{-\infty }^{\infty }x(t^{'})(\int _{-\infty }^{\infty }e^{j2\pi f(t-t^{'})}df)dt^{'}\!$
where
$e^{j2\pi f(t-t^{'})}=\delta (t-t^{'})\!$
Similarly for X(f)
$X(f)=\int _{-\infty }^{\infty }(\int _{-\infty }^{\infty }X(f^{'})e^{j2\pi f^{'}t}df^{'})e^{-j2\pi ft}dt=\int _{-\infty }^{\infty }X(f^{'})(\int _{-\infty }^{\infty }e^{j2\pi t(f^{'}-f)}dt)df^{'}\!$
where
$e^{j2\pi t(f^{'}-f)}=\delta (f^{'}-f)=\delta (f-f^{'})\!$

This works out nicely for us in both the time and frequency domain because this give us the inpulse function for both where they are non-zero only when t = t' or f = f' depending on which equation you use