Fourier series of a function \(f(x)\) in interval \((-c,c)\) is given by
\begin{equation}
f(x) = \frac{a_{o}}{2}+\sum\limits_{n=1}^{\infty}a_{n}\cos \left(\frac{n\pi x}{c}\right)+b_{n}\sin \left(\frac{n\pi x}{c}\right)\tag{6.1.2}
\end{equation}
\begin{equation*}
= \frac{1}{2c}\int\limits_{-c}^{c}f(t)\,dt + \sum\limits_{n=1}^{\infty}\frac{1}{c}\int\limits_{-c}^{c}f(t)\cos \left(\frac{n\pi t}{c}\right)\cos \left(\frac{n\pi x}{c}\right)\,dt
\end{equation*}
\begin{equation*}
+ \sum\limits_{n=1}^{\infty}\frac{1}{c}\int\limits_{-c}^{c}f(t)\sin \left(\frac{n\pi t}{c}\right)\sin \left(\frac{n\pi x}{c}\right)\,dt
\end{equation*}
\begin{equation*}
= \frac{1}{2c}\int\limits_{-c}^{c}f(t)\,dt+ \frac{1}{c}\int\limits_{-c}^{c}f(t)\left[\sum\limits_{n=1}^{\infty}\cos \frac{n\pi (t-x)}{c}\right]\,dt
\end{equation*}
\begin{equation}
=\frac{1}{2c}\int\limits_{-c}^{c}f(t)\left[1+2\sum\limits_{n=1}^{\infty}\cos \frac{n\pi (t-x)}{c}\right]\,dt\tag{6.1.3}
\end{equation}
since cosine functions are even function, we have \(\cos (-\theta) = \cos \theta\text{.}\) Hence the expression
\begin{equation*}
1+2\sum\limits_{n=1}^{\infty}\cos \frac{n\pi (t-x)}{c} = 1+\sum\limits_{n=1}^{\infty}\left\{\cos \frac{n\pi (t-x)}{c}+\cos \frac{-n\pi (t-x)}{c}\right\}
\end{equation*}
\begin{equation*}
= 1+ \sum\limits_{n=-\infty}^{\infty}\left\{\cos \frac{n\pi (t-x)}{c}\right\}
\end{equation*}
\begin{equation*}
\therefore f(x) = \frac{1}{2c}\int\limits_{-c}^{c}f(t)\,dt + \frac{1}{2\pi}\int\limits_{-c}^{c}f(t)\left[\sum\limits_{n=-\infty}^{\infty}\left\{\frac{\pi}{c}\cos \frac{n\pi (t-x)}{c}\right\}\right]\,dt
\end{equation*}
Now assume that \(c\) increases indefinitely and making use of definition of integral as a limit of sum, we can write -
\begin{equation*}
\frac{n\pi}{c}=p \quad \text{and}\quad \frac{\pi}{c}=\frac{p}{n}= \,dp
\end{equation*}
\begin{equation*}
\therefore \quad \lim\limits_{c\to \infty}\left[\frac{\pi}{c} \sum\limits_{n=-\infty}^{\infty}\cos \frac{n\pi (t-x)}{c}\right] = \int\limits_{-\infty}^{\infty}\cos p(t-x)\,dp
\end{equation*}
and
\begin{equation}
f(x) = 0+ \frac{1}{2\pi} \int\limits_{-\infty}^{\infty} f(t)\,dt\int\limits_{-\infty}^{\infty}\cos p(t-x)\,dp\tag{6.1.4}
\end{equation}
here
\begin{equation*}
\frac{1}{2c}\int\limits_{-c}^{c}f(t)\,dt = 0
\end{equation*}
at \(c\rightarrow \infty\) and making use of the fact
\begin{equation*}
\int\limits_{-\infty}^{\infty} |f(x)|\,dx \quad \text{ is convergent. }
\end{equation*}
\begin{equation}
f(x) = \frac{1}{2\pi}\int\limits_{p=-\infty}^{\infty}\int\limits_{t=-\infty}^{\infty}f(t)\cos p(t-x)\,dp \,dt\tag{6.1.5}
\end{equation}
which is known as Fourier integral.
