Let $f(t)=e^{iwt}$ on $(-\pi,\pi).$Expand $f(t)$ in a complex exponential Fourier series of period $2\pi$.($w$ is not an integer.)

Fourier series of the function in complex form is $f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos nx +b_n\sin nx$

$\implies f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\frac{e^{inx}+e^{-inx}}{2}+\sum_{n=1}^{\infty}b_n\frac{e^{inx}-e^{-inx}}{2i} $

$\implies \frac{a_0}{2}+\sum_{n=1}^{\infty}(\frac{a_n+ib_n}{2})e^{-inx}+\sum_{n=1}^{\infty}(\frac{a_n-ib_n}{2})e^{inx}$

$\implies f(x)=\sum_{-\infty}^{\infty}C_n e^{inx}$,


The coefficients $C_n$ are called Complex Fourier coefficients.They are defined by the formulas


$C_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{iwt}dt=\frac{1}{\pi w}\sin(w\pi)$

$C_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{iwt}e^{-int}dt=\frac{1}{\pi (w-n)}\sin(w-n)\pi$

I'm not getting how to reduce $f(x)$ to Fourier series.

I've invested a great amount of time on solving this...Please give any hint

  • $\begingroup$ I don't understand the question. You wrote what the $C_n$-s are (according to you, at least: I did not check). $\endgroup$ – Gae. S. Nov 25 '19 at 10:07
  • $\begingroup$ @Gae.S.:See the edit... $\endgroup$ – Styles Nov 25 '19 at 10:20

There is a small typo in your coefficient, it should be $c_n = \frac{1}{2\pi}\int_{-\pi}^\pi \mathrm{e}^{\mathrm{i}wt}\mathrm{e}^{-\mathrm{i}nt}\mathrm{d}t$ for all $n\in\mathbb{Z}$ (there is a minus sign in the argument of the second exponential since you defined the serie with $\sum_n c_n\mathrm{e}^{+\mathrm{i}nt}$).

Your expression is correct. It is useless to compute $c_0$ separately here : $c_n = \frac{1}{(w-n)\pi}\sin (w-n)\pi$ for all $n\in\mathbb{Z}$.

Introducing the (normalized) sinc function you can write : $c_n = \operatorname{sinc}(w-n)$ for all $n\in\mathbb{Z}$. Thus, the notation is more compact and you can now extend the result to the general case where $w$ can be an integer (indeed $\operatorname{sinc}(0)=1$).

Thus, $f(t) = \sum_{n=-\infty}^\infty \operatorname{sinc}(w-n)\mathrm{e}^{+\mathrm{i}nt}$ for all $t\in(-\pi,\pi)$.

Now, if we introduce $g$ the "periodised"-$f$ function (a $2\pi$ periodic function such that $g(t) = f(t)$ for all $t\in(-\pi,\pi)$), we can see that $g$ can jump if $w$ is not an integer. Then, according to Dirichlet theorem

$$\tilde{g}(t) = \sum_{n=-\infty}^\infty \operatorname{sinc}(w-n)\mathrm{e}^{+\mathrm{i}nt}$$ for all $t\in\mathbb{R}$ where $g(t) = \frac{g(t^+)+g(t^-)}{2}$.

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