The Fourier Coefficients of $\cos(x)$

Question

To begin I simply went with the definition: $$f(t)=\sum_{n=-\infty}^{\infty} f_k\cdot e^{ikw_ot} \implies f_k=\frac{1}{T}\int_{0}^{T} f(t)\cdot e^{-ikw_ot} \,dt$$ And before plugging $\cos(x)$ I first expressed it in therm of complex exponentials: $$f_k=\frac{1}{T}\int_{0}^{T} cos(t)\cdot e^{-ikw_ot} \,dt=\frac{1}{T}\int_{0}^{T} \frac{e^{it}+e^{-it}}{2}\cdot e^{-ikw_ot} \,dt = \frac{1}{2T}\Bigg(\int_{0}^{T} e^{it}\cdot e^{-ikw_ot} \,dt+\int_{0}^{T} e^{-it}\cdot e^{-ikw_ot} \,dt\Bigg)$$ After going through both of the integrals I end up witht the following result: $$f_k = \frac{-i}{4\pi}\cdot\Bigg[\frac{1}{1-kw_o}(e^{i(1-kw_o)T}-1)+\frac{1}{1+kw_o}(e^{i(1+kw_o)T}-1) \Bigg]$$ However I could only find very little to verify this expression and the length of the result makes me wonder if I am missing any simplification.

Answer 1

First you should note that your very first assertion, "$f(t)=\sum_{n=-\infty}^{\infty} f_k\cdot e^{ikw_ot} \implies f_k=\frac{1}{T}\int_{0}^{T} f(t)\cdot e^{-ikw_ot} \,dt$" is simply wrong, unless you begin by specifying the value of $T$. (From what you write one gets the impression that that formula is valid for every $T$; that's obviously impossible, right?)

In any case there's no need to do any calculation. You have $$\frac12e^{it}+\frac12e^{-it}=\cos(t)=\sum_{n=-\infty}^{\infty} f_k\cdot e^{ikw_ot};$$ now since $\cos$ has period $2\pi$ we must have $\omega_0=1$, and then the uniqueness of the coefficients for Fourier series shows that $$f_k=\begin{cases}\frac12,&(k=\pm1), \\0,&(k\ne\pm1).\end{cases}$$(You already know two Fourier series for $\cos(t)$, and they must be the same.)