# [SOLVED] fourier series of exponential function

• Oct 12th 2008, 10:09 PM
[SOLVED] fourier series of exponential function
Edit: sorry, figured it out: I should have been expecting the coefficients to have an imaginary part

Hi, I am doing some revision and am trying to find the fourier series for the exponential function between $\displaystyle -\pi$ and $\displaystyle \pi$. I am fairly sure all the coefficients should be real but I keep getting stuck with an imaginary part.

Please tell me where I am wrong:

$\displaystyle e^t = \sum_{n=-\infty}^{\infty}f_ne^{int}$where
$\displaystyle f_n = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^te^{-int}dt$
$\displaystyle f_n = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{t-int}dt$
$\displaystyle f_n = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{t(1-in)}dt$
$\displaystyle f_n = \frac{1}{2\pi}\frac{1}{1-in}[e^te^{-int}]_{-\pi}^{\pi}$
$\displaystyle f_n = \frac{1}{2\pi}\frac{1}{1-in}[e^{\pi}e^{-i\pi n}-e^{-\pi}e^{i\pi n}]$
$\displaystyle f_n = \frac{1}{2\pi}\frac{1}{1-in}[e^{\pi}(-1)^n-e^{-\pi}(-1)^n]$
$\displaystyle f_n = \frac{1}{2\pi}\frac{1}{1-in}(-1)^n2\sinh(\pi)$
$\displaystyle f_n = \frac{1}{\pi}\frac{1}{1+n^2}(1+in)(-1)^n\sinh(\pi)$

Thanks a lot