Complex form of Fourier series

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SUMMARY

The discussion focuses on the transformation of a Fourier series into its complex form. The original Fourier series is represented as $$\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos{\frac{2n\pi t}{b-a}}+b_n\sin{\frac{2n\pi t}{b-a}})$$ with coefficients defined by integrals over the interval [a, b]. The complex form is derived as $$f(t)=\sum_{-\infty}^{+\infty}Fne^{jnw0t}$$, where $$Fn$$ is calculated using the integral $$Fn=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-jnw0t}dt$$. A mistake was identified in the transformation process, where the coefficient was incorrectly derived as (2/T) instead of (1/T).

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Let function $f(t)$ is represented by Fourier series,
$$\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos{\frac{2n\pi t}{b-a}}+b_n\sin{\frac{2n\pi t}{b-a}}),$$
$$a_0=\frac{2}{b-a}\int_{a}^{b}f(t)dt,$$
$$a_n=\frac{2}{b-a}\int_{a}^{b}f(t)cos\frac{2n\pi t}{b-a}dt,$$
$$b_n=\frac{2}{b-a}\int_{a}^{b}f(t)sin\frac{2n\pi t}{b-a}dt,$$
where $$a$$ and $$b$$ are lower and upper boundary.

Here is how I transformed it in order to get complex form:

1.jpg


2.jpg


3.jpg


But here is what I find on web:
$$f(t)=\sum_{-\infty}^{+\infty}Fne^{jnw0t}$$, where $$Fn=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-jnw0t}dt$$

If I put in my solution a=-T/2, b=T/2, I will get $$Fn=\frac{2}{b-a}\int_{a}^{b}f(t)e^{-j\frac{2n\pi t}{b-a}}dt=\frac{2}{T/2-(-T/2))}\int_{-T/2}^{T/2}f(t)e^{-j\frac{2n\pi t}{T/2-(-T/2)}}dt=\frac{2}{T}\int_{-T/2}^{T/2}f(t)e^{-j\frac{2n\pi t}{T}}dt=
\frac{2}{T}\int_{-T/2}^{T/2}f(t)e^{-jnw0t}dt$$

You can see that I get (2/T)*integral but it should be (1/T)*integral. Whats wrong?
 
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Mistake found :)
 

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