Fourier Series of $\frac{\sin(x)}{x}$

Question

Good afternoon! My teacher of signals and systems put in my test that calculate the Fourier coefficients for the function $f(x) = \frac{\sin x}{x}$. But ... How I can do?

I know that the function is even, hence the coefficient $B_n = 0$. For calculating $A_0$ would change the expression $\frac{\sin x}{x}$ for their respective series of Taylor? How I can calculate $A_n$?

Answer 1

The idea behind finding coefficients in Fourier series is "orthogonal functions". We can write $f(x)$ as an even function:

$$ f(x) = a_0 + \sum_{n=1}^{\infty} a_n \cos(nx) $$ Now if you multiply $f(x)$ by $\cos(mx)$ and take the integral from $-\pi$ to $\pi$:

$$ \begin{align} \int_{-\pi}^{\pi} f(x) \cos(mx) dx &= a_0 \int_{-\pi}^{\pi} \cos(mx) dx + \sum_{n=1}^{\infty} a_n \int_{-\pi}^{\pi} \cos(nx) \cos(mx) dx \\ &= a_0 (0) + \sum_{n=1}^{\infty} a_{n} \pi \delta_{mn} = \pi a_m \end{align} $$ therefore: $$ a_{m} = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(mx) dx = \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{\sin(x) \cos(mx)}{x} dx $$

Now you can write the integrand as a sum since:

$$ \frac{\sin(x) \cos(mx)}{x} = \frac{\sin[(1+m)x]+\sin[(1-m)x]}{2x} $$ then:

$$ \begin{align} a_{m} &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \left( \frac{\sin[(1+m)x]}{x} \right) dx + \frac{1}{2\pi} \int_{-\pi}^{\pi} \left( \frac{\sin[(1-m)x]}{x} \right) dx \\ &= Si(\pi +m \pi)+Si(\pi-m \pi) \end{align} $$

Answer 2

Assuming you want the Fourier series over the interval $[-\pi,\pi]$, we have this follow-up to Mahdi's answer:

$$\begin{eqnarray*}I_m=\int_{-\pi}^{\pi}\frac{\sin(x)\cos(mx)}{x}\,dx &=& \int_{0}^{\pi}\frac{\sin((1+m)x)}{x}\,dx-\int_{0}^{\pi}\frac{\sin((m-1)x)}{x}\,dx\\&=&\operatorname{Si}((m+1)\pi)-\operatorname{Si}((m-1)\pi)\end{eqnarray*}$$ where $\operatorname{Si}$ stands for the sine integral function. Since:

$$ I_m = \int_{(m-1)\pi}^{(m+1)\pi}\frac{\sin x}{x}\,dx =(-1)^m\int_{-\pi}^{\pi}\frac{\sin x}{x+m\pi}\,dx=(-1)^m\int_{0}^{\pi}\frac{x^2}{x^2-m^2\pi^2}\,\sin x\,dx$$ a quite good approximation for $|I_m|$ is provided by: $$|I_m|\approx\frac{1}{\pi^2 m^2}\int_{0}^{\pi}x^2\sin x\,dx = \frac{1}{m^2}\left(1-\frac{4}{\pi^2}\right).$$