find the complex Fourier series of the given periodic function $f(t) = e^{-t}$

Question

I am trying to find the complex Fourier series of the given periodic function $$f(t) = e^{-t}$$ over the interval $(-3,3)$ and $f(t+6) = f(t)$

My attempt:

The half period of the function is $3$, therefore the coefficient $c_0$ is

$$c_0 = \frac{1}{2L} \int_{-L}^{L} f(t)dt = \frac{1}{6} \int_{-3}^{3}e^{-t} dt = \frac{1}{6}\left[ -e^{-t}\big|_{-3}^{3}\right] = -\frac{e^{-3}}{6} + \frac{e^3}{6 } = \frac{e^3-e^{-3}}{6}$$

Then for $n \ne 0 $

$$c_n = \frac{1}{2L} \int_{-L}^{L} f(t)e^{-int}dt = \frac{1}{6}\int_{-3}^{3} e^{-t}e^{-int} dt = \frac{1}{6} \int_{-3}^{3} e^{(-1-in)t} = \frac{1}{6} \left[ \frac{1}{(-1-in)} e^{(-1-in)t} \bigg|^3_{-3} \right] \\ = \frac{1}{6} \left[ \frac{e^{-t} e^{-int}}{-1-in}\bigg| ^3_{-3} \right] = \frac{1}{6(-1-in)} (e^{-3}e^{-3int}-e^3e^{3int}) $$

So my question is how do I use $f(t+6) = f(t)$ and then put all these things together to find the complex Fourier series of the given function?

Answer 1

The complex Fourier expansion of a $T$-periodic function $t\mapsto ft)$ has the form $$f(t)\rightsquigarrow \sum_{k=-\infty}^\infty c_ke^{2k\pi i t/T}\ ,$$whereby the coefficients $c_k$ are given by the formula $$c_k={1\over T}\int_{\rm 1\ full\ period}f(t)e^{-2k\pi i t/T}\>dt\qquad(k\in{\mathbb Z}) .\tag{1}$$ In the case at hand the period length is $6$, and we use as "$1$ full period" the interval $[{-3},3]$, in which $f(t)$ is given by the simple expression $e^{-t}$. The formula $(1)$ then gives $$c_k={1\over 6}\int_{-3}^3\exp\left(-\left(1+{k\pi i \over 3}\right)t\right)\>dt\qquad(k\in{\mathbb Z})\ .\tag{2}$$ Since $1+{k\pi i\over3}\ne0$ for all $k\in{\mathbb Z}$ we can compute the integrals $(2)$ without any exception handling. The result is $$c_k={\sinh(3+ik\pi)\over 3+ik\pi}={(-1)^k\over 3+ik\pi}\sinh 3\qquad(k\in{\mathbb Z})\ .\tag{3}$$

Answer 2

$$A(n) = \frac{1}{6}\int_{-3}^3 e^{-t}e^{\frac{-2\pi int}{6}} dt = \frac{1}{6} \int_{-3}^3 e^{(-(\frac{3+n\pi i }{3}t)}dt = \frac{1}{6} \frac{-3}{3+n\pi i} e^{(-(\frac{3+n\pi i }{3 }t)} \bigg| _{-3}^3 \\ = \frac{-1}{2(3+n\pi i )} [e^{-(3+n\pi i ))} - e^{3+n\pi i }] = (\frac{1}{3+n\pi i })e^{n\pi i }\frac{(e^3-e^{-3})}{2} = (\frac{1}{3+n\pi i})(cos(n\pi)+isin(n\pi))*\frac{(e^3-e^{-3})}{2} = \frac{1}{3+n\pi i } ((-1)^n+0i)cosh3 = \frac{(-1)^ncosh3}{3+n\pi i }$$

Therefore,

$$f(t) \approx \sum_\infty^{-\infty} A(n) e^{\frac{2\pi int}{6}} = \sum_\infty^{-\infty} \left[ \frac{(-1)^ncosh(3)}{3+n\pi i }\right]e^{\frac{\pi i nt}{3}}$$