Fourier transform

From the Fourier Transform to the Inverse Fourier Transform

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

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 $\underset{T_{\zeta }\to \mathrm{\infty }}{\lim }\gamma (t)=\beta (t)$
The Fourier Series representation of $\gamma (t)$ is
$\gamma (t)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_k{e}^{j2\pi fkt}$ where $f=\frac{1}{T_{\zeta }}$
and ${\alpha }_k=\frac{1}{T_{\zeta }}\underset{-\frac{T_{\zeta }}{2}}{\overset{\frac{T_{\zeta }}{2}}{\int }}\gamma (t)e^{-j2\pi kt}dt$
${\alpha }_k$ can now be rewritten as ${\alpha }_k=\frac{1}{T_{\zeta }}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\beta (t)e^{-j2\pi kt}dt$
From our initial identity then, we can write ${\alpha }_k$ as ${\alpha }_k=\frac{1}{T_{\zeta }}\mathrm{B}(kf)$
and $\gamma (t)$ becomes $\gamma (t)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{T_{\zeta }}\mathrm{B}(kf)e^{j2\pi fkt}$
Now remember that $\beta (t)=\underset{T_{\zeta }\to \mathrm{\infty }}{\lim }\gamma (t)$ and $\frac{1}{T_{\zeta }}=f.$
Which means that $\beta (t)=\underset{f\to 0}{\lim }\gamma (t)=\underset{f\to 0}{\lim }\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}f\mathrm{B}(kf)e^{j2\pi fkt}$
Which is just to say that $\beta (t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\mathrm{B}(f)e^{j2\pi fkt}df$

So we have that ${ℱ}[\beta (t)]=\mathrm{B}(f)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\beta (t)e^{-j2\pi ft}dt$
Further ${{ℱ}}^{-1}[\mathrm{B}(f)]=\beta (t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\mathrm{B}(f)e^{j2\pi fkt}df$