Fourier Series

The trigonometric notation

Let \(n \in \mathbb{N}^*\) be a natural number greater than or equal to 1.

As this proof is quite long, we have split it into several successive steps.

1. In the ideal world: on \( \bigl[-\pi; \pi\bigr]\)

   Let \(f(x)\) be a \(2\pi\)-periodic function.

   In this world, the angular frequency is \(\omega = \frac{2\pi}{2\pi} = 1\). The sines and cosines are therefore simply written as \(\sin(nx)\) and \(\cos(nx)\).

   Fourier's initial hypothesis is that this function can be reconstructed as a sum of cosines and sines:

   $$f(x) = a_0 + \sum_{n = 1}^{+ \infty} a_n \cos(nx) + \sum_{n = 1}^{+ \infty} b_n \sin(nx) \qquad(H)$$

   The goal is then to determine the coefficients \((a_0)\) and \((a_n, b_n)\) for all \(n\) as well.

   1. Isolating \( a_0 \)

      By simply integrating our starting function \((H)\), we get:

      $$\int_{-\pi}^{\pi} f(x) \hspace{0.3em} dx = \int_{-\pi}^{\pi} a_0 \hspace{0.3em} dx + \int_{-\pi}^{\pi} \sum_{n = 1}^{+ \infty} a_n \cos(nx) \hspace{0.3em} dx + \int_{-\pi}^{\pi} \sum_{n = 1}^{+ \infty} b_n \sin(nx) \hspace{0.3em} dx $$

      Since the \(cos\) and \(sin\) functions are \(2\pi\)-periodic, they cancel out over the interval from \(-\pi\) to \(\pi\):

      $$\int_{-\pi}^{\pi} f(x) \hspace{0.3em} dx = \int_{-\pi}^{\pi} a_0 \hspace{0.3em} dx + \underbrace{ \int_{-\pi}^{\pi} \sum_{n = 1}^{+ \infty} a_n \cos(nx) \hspace{0.3em} dx } _{=0} + \underbrace{ \int_{-\pi}^{\pi} \sum_{n = 1}^{+ \infty} b_n \sin(nx) \hspace{0.3em} dx } _{=0} $$

      Which, after cleaning up, yields:

      $$\int_{-\pi}^{\pi} f(x) \hspace{0.3em} dx = \int_{-\pi}^{\pi} a_0 \hspace{0.3em} dx$$
      $$\int_{-\pi}^{\pi} f(x) \hspace{0.3em} dx = 2\pi a_0 $$
      $$a_0 = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x) \hspace{0.3em} dx $$

   2. Calculating the harmonics \( a_n \)

      Let \(m \in \mathbb{N}^*\) be a natural number greater than or equal to 1.

      To isolate the coefficient \(a_n\), we multiply equation \((H)\) by \(\cos(mx)\):

      $$f(x) \cos(mx) = a_0 \cos(mx) + \sum_{n = 1}^{+ \infty} a_n \cos(nx) \cos(mx) + \sum_{n = 1}^{+ \infty} b_n \sin(nx) \cos(mx) $$

      Then we integrate it from \(-\pi\) to \(\pi\):

      $$\int_{-\pi}^{\pi} f(x) \cos(mx) \hspace{0.3em} dx = \underbrace{ \int_{-\pi}^{\pi} a_0 \cos(mx) \hspace{0.3em} dx } _{(A)} + \underbrace{ \int_{-\pi}^{\pi} \sum_{n = 1}^{+ \infty} a_n \cos(nx) \cos(mx) \hspace{0.3em} dx } _{(B)} + \underbrace{ \int_{-\pi}^{\pi} \sum_{n = 1}^{+ \infty} b_n \sin(nx) \cos(mx) \hspace{0.3em} dx } _{(C)} \qquad (1) $$

      From this point forward, many terms cancel out:

      • Part \((A)\) cancels out in all cases:

        $$ \forall m \in \mathbb{N}^*, \ A = \int_{-\pi}^{\pi} a_0 \cos(mx) \hspace{0.3em} dx \qquad (A)$$
        $$ A = a_0\left[ \frac{\sin(mx)}{m}\right]_{-\pi}^{\pi} $$
        $$ A = a_0 \left(\frac{\sin(m\pi)}{m} - \frac{\sin(-m\pi)}{m} \right) $$
        $$ A = a_0 \left( \frac{\sin(m\pi)}{m} + \frac{\sin(m\pi)}{m} \right)$$
        $$ A = a_0 \left( \frac{2\sin(m\pi)}{m} \right) $$

        All sine functions that are multiples of \(\pi\) are equal to \(0\), therefore:

        $$ A = 0 $$

      • Part \((B)\) cancels out only if \((n \neq m)\):

        $$ \forall m \in \mathbb{N}^*, \ B = \int_{-\pi}^{\pi} \sum_{n = 1}^{+ \infty} a_n \cos(nx) \cos(mx) \hspace{0.3em} dx \qquad (B)$$

        Now, we know from trigonometric addition formulas that:

        $$ \forall (\alpha, \beta) \in \hspace{0.04em} \mathbb{R}^2, $$

        $$ \cos(\alpha + \beta) + \cos(\alpha - \beta) = 2 \cos(\alpha) \cos(\beta) $$

        So, in our case:

        $$ B = \int_{-\pi}^{\pi} \sum_{n = 1}^{+ \infty} a_n \frac{ \cos\Bigl((n + m)x\Bigr) + \cos\Bigl((n - m)x\Bigr) }{2} \hspace{0.3em} dx$$

        By linearity of the integral , we can switch both symbols \(\int\) and \(\sum\):

        $$ B = \sum_{n = 1}^{+ \infty} \int_{-\pi}^{\pi} a_n \frac{ \cos\Bigl((n + m)x\Bigr) + \cos\Bigl((n - m)x\Bigr) }{2} \hspace{0.3em} dx $$
        $$ B = \sum_{n = 1}^{+ \infty} \frac{1}{2} a_n \left( \int_{-\pi}^{\pi} \cos\Bigl((n + m)x\Bigr) + \int_{-\pi}^{\pi} \cos\Bigl((n - m)x\Bigr) \hspace{0.3em} dx \right) $$
        
        
        $$ \underline{\text{Case 1: When } (n \neq m)} $$
        $$ B = \sum_{\substack{n = 1 \\ n \neq m}}^{+ \infty} \frac{1}{2} a_n \left(\left[ \frac{\sin(n + m)x}{n + m}\right]_{-\pi}^{\pi} + \left[ \frac{\sin(n - m)x}{n - m}\right]_{-\pi}^{\pi} \right) $$
        $$ B = \sum_{\substack{n = 1 \\ n \neq m}}^{+ \infty} \frac{1}{2} a_n \left(\frac{\sin\Bigl((n + m)\pi\Bigr)}{n + m} - \frac{\sin\Bigl((n + m)(-\pi)\Bigr)}{n + m} + \frac{\sin\Bigl((n - m)\pi\Bigr)}{n - m} - \frac{sin\Bigl((n - m)(-\pi)\Bigr)}{n - m} \right) $$

        All sine functions that are multiples of \(\pi\) are equal to \(0\), therefore:

        $$ B = \sum_{\substack{n = 1 \\ n \neq m}}^{+ \infty} \frac{1}{2} a_n \left(\underbrace{\frac{2\sin\Bigl((n + m)\pi\Bigr)}{n + m}} _{=0} + \underbrace{\frac{2\sin\Bigl((n - m)\pi\Bigr)}{n - m} } _{=0} \right) $$
        $$ (n \neq m) \implies B = 0 $$
        
        
        $$ \underline{\text{Case 2: When } (n = m)} $$

        All terms cancel out except one when \((n = m)\):

        $$ \begin{cases} (n + m) = 2m \implies \cos\Bigl((n + m)x\Bigr) = \cos(2mx) \\ (n - m) = 0 \implies \cos\Bigl((n - m)x\Bigr) = 1 \end{cases} $$
        $$ B = \frac{1}{2} a_m \left( \int_{-\pi}^{\pi} \cos(2mx) \, dx + \int_{-\pi}^{\pi} dx \right) $$
        $$ B = \frac{1}{2} a_m \left( \left[ \frac{\sin(2mx)}{2}\right]_{-\pi}^{\pi} + \bigl[ x \bigr]_{-\pi}^{\pi} \right) $$
        $$ B = \frac{1}{2} a_m \left( \frac{\sin(2m \pi)}{2} - \frac{\sin\Bigl(2m (-\pi) \Bigr)}{2} + \pi -(- \pi) \right) $$
        $$ B = \frac{1}{2} a_m \Bigl( \underbrace{\sin(2m \pi)} _{=0} + 2\pi \Bigr) $$
        $$ (n = m) \implies B = a_m \pi $$

        Comment:

        Conveniently enough, the case \((n = m)\) must be treated separately anyway, because \(B\) would not have been defined due to its denominator.

      • Part \((C)\) cancels out in all cases:

        $$ \forall m \in \mathbb{N}^*, \ C = \int_{-\pi}^{\pi} \sum_{n = 1}^{+ \infty} b_n \sin(nx) \cos(mx) \hspace{0.3em} dx \qquad (C)$$

        Now, we know from trigonometric addition formulas that:

        $$ \forall (\alpha, \beta) \in \hspace{0.04em} \mathbb{R}^2, $$

        $$ \sin(\alpha + \beta) + \sin(\alpha - \beta) = 2 \sin(\alpha) \cos(\beta) $$

        So, in our case:

        $$ C = \int_{-\pi}^{\pi} \sum_{n = 1}^{+ \infty} b_n \frac{ \sin\Bigl((n + m)x\Bigr) + \sin\Bigl((n - m)x\Bigr) }{2} \hspace{0.3em} dx$$

        $$ C = \sum_{n = 1}^{+ \infty} \int_{-\pi}^{\pi} b_n \frac{ \sin\Bigl((n + m)x\Bigr) + \sin\Bigl((n - m)x\Bigr) }{2} \hspace{0.3em} dx $$

        $$ C = \sum_{n = 1}^{+ \infty} \frac{1}{2} b_n \left( \int_{-\pi}^{\pi} \sin\Bigl((n + m)x\Bigr) + \int_{-\pi}^{\pi} \sin\Bigl((n - m)x\Bigr) \hspace{0.3em} dx \right) $$

        $$ C = \sum_{n = 1}^{+ \infty} \frac{1}{2} b_n \left(\left[ \frac{\cos(n + m)x}{n + m}\right]_{-\pi}^{\pi} + \left[ \frac{\cos(n - m)x}{n - m}\right]_{-\pi}^{\pi} \right) $$
        $$ C = \sum_{n = 1}^{+ \infty} \frac{1}{2} b_n \left(\frac{\cos\Bigl((n + m)\pi\Bigr)}{n + m} - \frac{\cos\Bigl((n + m)(-\pi)\Bigr)}{n + m} + \left( \frac{\cos\Bigl((n - m)\pi\Bigr)}{n - m} - \frac{\cos\Bigl((n - m)(-\pi)\Bigr)}{n - m} \right) \right) $$

        $$ C = \sum_{n = 1}^{+ \infty} \frac{1}{2} b_n \left( \underbrace{ \frac{\cos\Bigl((n + m)\pi\Bigr)}{n + m} - \frac{\cos\Bigl((n + m)\pi\Bigr)}{n + m} } _{=0} + \left( \underbrace{ \frac{\cos\Bigl((n - m)\pi\Bigr)}{n - m} - \frac{\cos\Bigl((n - m)\pi\Bigr)}{n - m} } _{=0} \right) \right) $$

        $$ C = 0 $$

      • Conclusion:

        Now, if we take our formula \((1)\) and evaluate it when \((n = m)\):

        $$ (n = m) \implies \int_{-\pi}^{\pi} f(x) \cos(nx) \hspace{0.3em} dx = a_n \pi $$

        Which gives,

        $$ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \hspace{0.3em} dx $$

   3. Calculating the harmonics \( b_n \)

      This section is carried out in the same manner as before, but this time multiplying by \(sin(mx)\):

      $$\int_{-\pi}^{\pi} f(x) \sin(mx) \hspace{0.3em} dx = \underbrace{ \int_{-\pi}^{\pi} a_0 \sin(mx) \hspace{0.3em} dx } _{(A')} + \underbrace{ \int_{-\pi}^{\pi} \sum_{n = 1}^{+ \infty} a_n \cos(nx) \sin(mx) \hspace{0.3em} dx } _{(B')} + \underbrace{ \int_{-\pi}^{\pi} \sum_{n = 1}^{+ \infty} b_n \sin(nx) \sin(mx) \hspace{0.3em} dx } _{(C')} \qquad (1') $$

      • Part \((A')\) cancels out in all cases:

        $$ \forall m \in \mathbb{N}^*, \ A' = \int_{-\pi}^{\pi} a_0 \sin(mx) \hspace{0.3em} dx \qquad (A')$$
        $$ A' = a_0 \left[ -\frac{\cos(mx)}{m}\right]_{-\pi}^{\pi} $$
        $$ A'= a_0 \left(-\frac{\cos(m\pi)}{m} + \frac{\cos(-m\pi)}{m} \right)$$
        $$ A'= a_0 \left(-\frac{\cos(m\pi)}{m} + \frac{\cos(m\pi)}{m} \right)$$
        $$ A'= 0 $$

      • Part \((B')\) cancels out in all cases:

        $$ \forall m \in \mathbb{N}^*, \ B' = \int_{-\pi}^{\pi} \sum_{n = 1}^{+ \infty} a_n \cos(nx) \sin(mx) \hspace{0.3em} dx \qquad (B')$$

        Now, we know from trigonometric addition formulas that:

        $$ \forall (\alpha, \beta) \in \hspace{0.04em} \mathbb{R}^2, $$

        $$ \sin(\alpha + \beta) - \sin(\alpha - \beta) = 2 \cos(\alpha) \sin(\beta) $$

        This is almost the same as part \((C)\) previously calculated, but with a minus sign \((-)\) instead of a plus sign \((+)\):

        $$ B' = \int_{-\pi}^{\pi} \sum_{n = 1}^{+ \infty} a_n \frac{ \sin\Bigl((n + m)x\Bigr) - \sin\Bigl((n - m)x\Bigr) }{2} \hspace{0.3em} dx$$

        The steps are therefore identical, and so is the result:

        $$ B' = \sum_{n = 1}^{+ \infty} \frac{1}{2} a_n \left( \underbrace{ \frac{\cos\Bigl((n + m)\pi\Bigr)}{n + m} - \frac{\cos\Bigl((n + m)\pi\Bigr)}{n + m} } _{=0} - \left( \underbrace{ \frac{\cos\Bigl((n - m)\pi\Bigr)}{n - m} - \frac{\cos\Bigl((n - m)\pi\Bigr)}{n - m} } _{=0} \right) \right) $$

        $$ B' = 0 $$

      • Part \((C')\) cancels out only if \((n \neq m)\):

        $$ \forall m \in \mathbb{N}^*, \ C' = \int_{-\pi}^{\pi} \sum_{n = 1}^{+ \infty} b_n \sin(nx) \sin(mx) \hspace{0.3em} dx \qquad (C')$$

        Now, we know from trigonometric addition formulas that:

        $$ \forall (\alpha, \beta) \in \hspace{0.04em} \mathbb{R}^2, $$

        $$ cos(\alpha - \beta) - \cos(\alpha + \beta) = 2 \sin(\alpha) \sin(\beta) $$

        This is almost the same as part \((B)\) previously calculated:

        $$ C' = \int_{-\pi}^{\pi} \sum_{n = 1}^{+ \infty} b_n \frac{ \cos\Bigl((n - m)x\Bigr) - \cos\Bigl((n + m)x\Bigr) }{2} \hspace{0.3em} dx$$
        
        
        $$ \underline{\text{Case 1: When } (n \neq m)} $$
        $$ C' = \sum_{\substack{n = 1 \\ n \neq m}}^{+ \infty} \frac{1}{2} b_n \left(\left[ \frac{\sin(n - m)x}{n - m}\right]_{-\pi}^{\pi} - \left[ \frac{\sin(n + m)x}{n + m}\right]_{-\pi}^{\pi} \right) $$
        $$ C' = \sum_{\substack{n = 1 \\ n \neq m}}^{+ \infty} \frac{1}{2} b_n \left(\frac{\sin\Bigl((n - m)\pi\Bigr)}{n - m} - \frac{\sin\Bigl((n - m)(-\pi)\Bigr)}{n - m} - \left( \frac{\sin\Bigl((n + m)\pi\Bigr)}{n + m} - \frac{sin\Bigl((n + m)(-\pi)\Bigr)}{n + m} \right) \right) $$

        All sine functions that are multiples of \(\pi\) are equal to \(0\), therefore:

        $$ C' = \sum_{\substack{n = 1 \\ n \neq m}}^{+ \infty} \frac{1}{2} b_n \left(\underbrace{\frac{2\sin\Bigl((n + m)\pi\Bigr)}{n + m}} _{=0} - \underbrace{\frac{2\sin\Bigl((n - m)\pi\Bigr)}{n - m} } _{=0} \right) $$
        $$ (n \neq m) \implies C' = 0 $$
        
        
        $$ \underline{\text{Case 2: When } (n = m)} $$

        Then:

        $$ \begin{cases} (n - m) = 0 \implies \cos\Bigl((n - m)x\Bigr) = 1 \\ (n + m) = 2m \implies \cos\Bigl((n + m)x\Bigr) = \cos(2mx) \end{cases} $$
        $$ C' = \frac{1}{2} b_n \left( \int_{-\pi}^{\pi} dx - \int_{-\pi}^{\pi} \cos(2mx) \right) $$
        $$ C' = \frac{1}{2} b_n \left( \bigl[ x \bigr]_{-\pi}^{\pi} - \left[ \frac{sin(2mx)}{2}\right]_{-\pi}^{\pi} \right) $$
        $$ C' = \frac{1}{2} b_n \left( + \pi -(- \pi) - \left( \frac{sin(2m \pi)}{2} - \frac{sin\Bigl(2m (-\pi) \Bigr)}{2} \right) \right) $$
        $$ C' = \frac{1}{2} b_n \Bigl( 2\pi + \underbrace{sin(2m \pi)} _{=0} \Bigr) $$
        $$ C' = \pi b_n $$
        $$ (n = m) \implies C' = b_n \pi $$

      • Conclusion:

        In the same manner as before, evaluating \((1')\) when \((n = m)\):

        $$ (n = m) \implies \int_{-\pi}^{\pi} f(x) \sin(nx) \hspace{0.3em} dx = b_n \pi $$

        Which gives,

        $$ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \hspace{0.3em} dx $$

2. Change of variables: moving to the physical world \( \bigl[0; T\bigr]\)

   Now, let's leave the ideal world and move toward any physical signal \(f(t)\) that is \(T\)-periodic and starts at \((t = 0\text{ s})\). Its angular frequency is then \(\omega = \frac{2\pi}{T}\).

   To transfer our previous formulas, we proceed using two consecutive steps.

   1. Shifting the window

      Since the function \(f(x)\) is \(2\pi\)-periodic, calculating its area between \(-\pi\) and \(\pi\) is equivalent to calculating it between \(0\) and \(2\pi\).

   2. Change of variables

      We now want to transition from \(x\) (the angle) to the variable \(t\) (time). The relationship between the two is:

      $$ x = \omega t$$
      $$ \left(\text{with } \omega = \frac{2\pi}{T}\right) $$

      We must also transform our differential \(dx\):

      $$ dx = \omega \, dt$$
      
      

      Fourier's starting hypothesis is now written as:

      $$f(t) = a_0 + \sum_{n = 1}^{+ \infty} a_n \cos(n \omega t) + \sum_{n = 1}^{+ \infty} b_n \sin(n \omega t) $$
      
      

      Thus, the three previous relations transform and become:

      $$a_0 = \frac{1}{2\pi}\int_{0}^{T} f(t) \hspace{0.3em} \omega \, dt $$
      $$a_0 = \frac{\omega}{2\pi}\int_{0}^{T} f(t) \, dt $$
      $$a_0 = \frac{1}{T}\int_{0}^{T} f(t) \, dt $$
      $$ \forall n \in \mathbb{N}^*,$$
      $$ a_n = \frac{1}{\pi} \int_{0}^{T} f(t) \cos(n \omega t) \hspace{0.3em} \omega \, dt $$

      $$ a_n = \frac{\omega}{\pi} \int_{0}^{T} f(t) \cos(n \omega t) \, dt $$

      $$ a_n = \frac{2}{T} \int_{0}^{T} f(t) \cos(n \omega t) \, dt $$

      $$ b_n = \frac{1}{\pi} \int_{0}^{T} f(t) \sin(n \omega t) \hspace{0.3em} \omega \, dt $$

      $$ b_n = \frac{\omega}{\pi} \int_{0}^{T} f(t) \sin(n \omega t) \, dt $$

      $$ b_n = \frac{2}{T} \int_{0}^{T} f(t) \sin(n \omega t) \, dt $$
      $$ \text{with } \begin{cases} T : \text{Period of the function } f(t) \\ \omega = \frac{2\pi}{T} : \text{Pulsation } \end{cases} $$

      Why isolate \(a_0\)?

      Why isolate this famous coefficient \(a_0\), when the formula could simply have been:

      $$f(t) = \sum_{n = 0}^{+ \infty} a_n \cos(n \omega t) + \sum_{n = 0}^{+ \infty} b_n \sin(n \omega t) $$

      Because the exception for \((n = 0)\) is the one and only moment in the entire series where the function loses its wave nature to "freeze" into a constant.

      • The fair sharing rule (for \(n \geqslant 1\))

        The fundamental relation of trigonometry states that for any angle \(X\):

        $$ \cos^2(X) + \sin^2(X) = 1 $$

        If we integrate this relation over a full time period \(\bigl[0; T\bigr]\) (by setting \(X = n\omega t\)), we obtain the following equality:

        $$ \int_{0}^{T} \cos^2(n\omega t) \, dt + \int_{0}^{T} \sin^2(n\omega t) \, dt = \int_{0}^{T} 1 \, dt $$

        The total integral (on the right) is exactly equal to \(T\). For any oscillating frequency (\(n \geqslant 1\)), the cosine and the sine have strictly the same shape, simply shifted in time. Therefore, they share the result of the integral equally:

        • The cosine takes half of the area: \(\frac{T}{2}\)
        • The sine takes the other half: \(\frac{T}{2}\)

        It is this perfect 50/50 split that gives rise to the famous factor \(\frac{2}{T}\) used to calculate the harmonics \(a_n\) and \(b_n\).

      • The collapse of the pair (for \(n = 0\))

        Now let's look at what happens for the zero frequency (\(n=0\)) when applying this same sharing logic.

        The base equation remains true:

        $$ \cos^2(0) + \sin^2(0) = 1 $$

        The total integral must therefore still equal \(T\). However, the nature of the sine breaks the balance:

        $$ \sin(0) = 0 $$

        The sine collapses completely and its integral is \(0\). It no longer does its part of the work. The sharing of the total integral then becomes:

        $$ \int_{0}^{T} \cos^2(0) \, dt + 0 = T $$

      • Conclusion

        Because the sine cancels out at \((n=0)\), it no longer "draws" its half of the integral. The cosine is left alone to bear 100% of the result.

        The integral of the cosine therefore suddenly jumps from \(\frac{T}{2}\) (when the sine was helping it) to \(T\) (when the sine abandons it).

        This is what makes the mathematical result twice as large, and what forces us to divide the term \(a_0\) by 2 to restore the true physical value of the signal!

The exponential notation

If we take the previous function:

This series becomes:

However, \(\frac{1}{i} = -i\). Therefore,

Then, by grouping the terms:

We thus have, by replacing the previous expression:

We can now invert the indices in the second sum so that the sums become symmetrical:

By observing that for \(n = 0\), the exponential cancels out and equals \(1\) (\(e^{i \cdot 0 \cdot \omega t} = 1\)), we can define the constant term as the coefficient \(c_0\) of our sequence:

By injecting \(c_0\) and reorganizing the limits of the second sum using the change of index \(k = -n\), the three blocks of the equation connect perfectly:

With all the pieces of the puzzle aligned from \(-\infty\) to \(+\infty\), we can condense everything into a single, unique summation:

1. Calculation of the coefficient \( c_n \)

   If we use the same method as before, but this time with the exponential, we start from the general function:

   $$f(t) = \sum_{-\infty}^{+\infty} c_n \ e^{in\omega t} $$

   For an \((m \geqslant 1) \), we will this time use \(e^{-i m\omega t}\):

   $$f(t) \ e^{-i m\omega t} = \sum_{-\infty}^{+\infty} c_n \ e^{in\omega t} \ e^{-i m\omega t} $$
   $$f(t) \ e^{-i m\omega t} = \sum_{-\infty}^{+\infty} c_n \ e^{i (n - m)\omega t} $$

   Then we integrate:

   $$ \int_0^T f(t) \ e^{-i m\omega t} \, dt = \int_0^T \sum_{-\infty}^{+\infty} c_n \ e^{i (n - m)\omega t} \, dt $$
   
   
   $$ \underline{\text{Case 1: When } (n \neq m)} $$
   $$ \int_0^T f(t) \ e^{-i m\omega t} \, dt = \int_0^T \sum_{-\infty}^{+\infty} c_n \ e^{i (n - m)\omega t} \, dt $$

   By linearity , we can swap the symbols \(\int\) and \(\sum\):

   $$ \int_0^T f(t) \ e^{-i m\omega t} \, dt = \sum_{-\infty}^{+\infty} \int_0^T c_n \ e^{i (n - m)\omega t} \, dt $$
   $$ \int_0^T f(t) \ e^{-i m\omega t} \, dt = \sum_{-\infty}^{+\infty} c_n \left[ \frac{e^{i (n - m)\omega t}}{i (n - m)\omega} \right]_0^T $$
   $$ \int_0^T f(t) \ e^{-i m\omega t} \, dt = \sum_{-\infty}^{+\infty} c_n \left( \frac{e^{i (n - m)\omega T}}{i (n - m)\omega} - \frac{e^{i (n - m)\omega \times 0}}{i (n - m)\omega} \right) $$

   However, \(\omega T = 2\pi\), therefore:

   $$ \int_0^T f(t) \ e^{-i m\omega t} \, dt = \sum_{-\infty}^{+\infty} c_n \left( \frac{e^{i (n - m) 2\pi }}{i (n - m)\omega} - \frac{1}{i (n - m)\omega} \right) $$
   $$ \int_0^T f(t) \ e^{-i m\omega t} \, dt = \sum_{-\infty}^{+\infty} c_n \left( \underbrace{\frac{1}{i (n - m)\omega} - \frac{1}{i (n - m)\omega}} _{=0} \right) $$
   $$ \int_0^T f(t) \ e^{-i m\omega t} \, dt = 0 $$
   
   
   $$ \underline{\text{Case 2: When } (n = m)} $$
   $$ \int_0^T f(t) \ e^{-i m\omega t} \, dt = \int_0^T c_n \ e^{i (n - m)\omega t} \, dt $$
   $$ \int_0^T f(t) \ e^{-i m\omega t} \, dt = c_n \int_0^T \ e^{i \times 0 \times \omega t} \, dt $$
   $$ \int_0^T f(t) \ e^{-i m\omega t} \, dt = c_n \int_0^T \, dt $$
   $$ \int_0^T f(t) \ e^{-i m\omega t} \, dt = c_n \ T $$

   And with this notation, we have a unique coefficient \(c_n\):

   $$ c_n = \frac{1}{T} \int_{0}^{T} f(t) e^{-in\omega t} \, dt $$