Fourier transform

An initially identity that is useful: $x(t)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt$

Suppose that we have some function, say $\beta (t)$, that is nonperiodic and finite in duration.
This means that $\beta(t)=0$ for some $T_\alpha < \left | t \right |$

Now let's make a periodic function $\gamma(t)$ by repeating $\beta(t)$ with a fundamental period $T_\zeta$. Note that $\lim_{T_\zeta \to \infty}\gamma(t)=\beta(t)$
The Fourier Series representation of $\gamma(t)$ is
$\gamma(t)=\sum_{k=-\infty}^\infty \alpha_k e^{j2\pi fkt}$ where $f={1\over T_\zeta}$
and $\alpha_k={1\over T_\zeta}\int_{-{T_\zeta\over 2}}^{{T_\zeta\over 2}} \gamma(t) e^{-j2\pi kt}\,dt$
$\alpha_k$ can now be rewritten as $\alpha_k={1\over T_\zeta}\int_{-\infty}^{\infty} \beta(t) e^{-j2\pi kt}\,dt$
From our initial identity then, we can write $\alpha_k$ as $\alpha_k={1\over T_\zeta}\Beta(kf)$
and $\gamma(t)$ becomes $\gamma(t)=\sum_{k=-\infty}^\infty {1\over T_\zeta}\Beta(kf) e^{j2\pi fkt}$