Fourier transform

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

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 $\alpha _{k}={1 \over T_{\zeta }}\mathrm {B} (kf)$
and $\gamma (t)$ becomes $\gamma (t)=\sum _{k=-\infty }^{\infty }{1 \over T_{\zeta }}\mathrm {B} (kf)e^{j2\pi fkt}$
Now remember that $\beta (t)=\lim _{T_{\zeta }\to \infty }\gamma (t)$ and ${1 \over {T_{\zeta }}}=f.$
Which means that $\beta (t)=\lim _{f\to 0}\gamma (t)=\lim _{f\to 0}\sum _{k=-\infty }^{\infty }f\mathrm {B} (kf)e^{j2\pi fkt}$
Which is just to say that $\beta (t)=\int _{-\infty }^{\infty }f\mathrm {B} (f)e^{j2\pi fkt}\,df$

So we have that the Fourier Transform of $\beta (t)$ is $X(f)=\int _{-\infty }^{\infty }x(t)e^{-j2\pi ft}\,dt$