Answer to Question #196987 in Math for Hanata yuji

Question

If f(t)= e^iwt at the interval (-π, π) determine fourier series of the f(t) ?

Answer 1

Here the half-period is $L=\pi.$ Therefore, the coefficient $c_0$ is

$$c_0=\dfrac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}f(t)dt=\dfrac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}e^{iwt}dt$$

$$=\dfrac{i}{2iw\pi}(e^{iw\pi}-e^{-iw\pi})=\dfrac{\sin{(w\pi)}}{w\pi}$$

For $n\not=0$

$$c_n=\dfrac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}f(t)e^{-{in\pi t \over \pi}}dt=\dfrac{1}{2\pi}\displaystyle\int_{-\pi}^{\pi}e^{iwt}e^{-int}dt$$

$$=\dfrac{1}{2\pi i(w-n)}(e^{i(w-n)\pi}-e^{-e^{i(w-n)\pi}})$$

$$=\dfrac{\sin((w-n)\pi)}{ (w-n)\pi}$$

$$f(t)=e^{iwt}=\dfrac{\sin{(w\pi)}}{w\pi}+\displaystyle\sum_{-\infin}^{\infin}\dfrac{\sin((w-n)\pi)}{ (w-n)\pi}e^{int}$$