How to get real valued functions from a complex exponential Fourier Series?

Question

Imagine you want to represent the following function as a Fourier series: $$ f(x) = \cases{ 1 \; \; \; \; \text{if} \; \; 0 < x \leq 1 \\ 0 \; \; \; \; \text{if} \; \; 1 < x \leq 2 \\ } $$ Calculate the $c_n$ coefficients: $$ c_n = \frac{1}{2}\int_0^2{f(x) e^{-\pi i n t} \; dx} \\ c_n = \cases{ 0 \; \; \; \; \text{if} \; n \; \text{is even} \\ \frac{1}{\pi i n} \; \; \; \; \text{if} \; n \; \text{is odd} \\ } $$ Finally we arrive to: $$ f(x) \sim \sum_{n = 1, 3, 5...}^{\infty} \frac{1}{\pi i n} e^{\pi i n t} \\ f(x) \sim \frac{1}{\pi}\sum_{k = -\infty}^{\infty} \frac{1}{ i (2k + 1)} e^{\pi i (2k + 1) t} $$ The fourier series, using only real values looks like: $$ f(x) \sim \frac{1}{2} + \frac{2}{\pi}\sum_{k = 0}^{\infty} \frac{1}{ 2k + 1} \sin{(2k \pi t + \pi t)} $$ How do I arrive to this expession from the exponential? $$ \sin{(x)} = \frac{e^{ix} - e^{-ix}}{2i} $$ I'm almost sure, I have to use this identity, but I don't know how and where?

Answer 1

One can get an expansion in terms of real Fourier coefficients by extending the domain from $[0,2]$ to $[-2,2]$, and extending the function $f$ to be symmetric about $x=0$: $$ g(x) = \cases{ 1 \; \; \; \; \text{if} \; \; |x| \le 1 \\ 0 \; \; \; \; \text{if} \; \; 1 < |x| \le 2 }$$ The values of $f$ and $g$ are the same in the interval $[0,2]$, so the Fourier expansion of $g$ will also provide an expansion for $f$ in that domain.

Specifically, the Fourier expansion of $g$ is $$g(x) = \sum_{n\in\Bbb Z} \hat c(n) e^{\frac{i}{2}n\pi x},$$ where $$\hat c(n) = \frac{1}{4}\int_{-2}^{2} g(x) e^{-\frac{i}{2}n\pi x}dx.$$ Note that since $g$ is symmetric, $\hat c(n) = \hat c(-n)$ is real, and $$ g(x) = \hat c(0) + 2\sum_{n > 0}\hat c(n) \cos(n\pi x/2).$$ The rest is algebra.