Complex Fourier Series and using the square norm Find the complex Fourier series of $f(x)=e^{(-πx/2)}$ on $-π < x < π$ Discuss the significance of $|C_n|$ in the solution.
I've tried so far Using the Complex Fourier Series:
$$
%% ![](http://www.math24.net/images/4frs5.gif)
            f(x) = \sum_{n=-\infty}^{\infty} C_{n}e^{in\pi x/L},
$$ 
where $C_n$ is 
$$
%% ![](http://www.math24.net/images/4frs6.gif)
              C_{n} = \frac{1}{2L}\int_{-L}^{L}f(x)e^{-in\pi x/L}\,dx,
    \;\;\; n = 0,\pm 1,\pm 2,\cdots\;.
$$
So I set it up like so
$$
%% Cn = 1/2π ∫ from -π to π e^(-πx/2)*[e^(-inπx/π)]
C_{n} = \frac{1}{2\pi}\int_{-\pi}^{\pi} e^{-\pi x/2} e^{-in\pi x/\pi}\,dx
$$
After a few steps I did integration by parts where:
$$
%% dv = e^(-inx)dx and v= (i/n)(e^(-inx)) and du = -πe^(-πx/2) and u =e^(-πx/2)
   dv = e^{-inx}dx\;\;\; \mbox{ and }\;\;\; v = (i/n)e^{-inx},\\
   du = -\pi e^{-\pi x/2}\;\;\; \mbox{ and }\;\;\; u = 2e^{-\pi x/2},
$$
but after I used integration by parts I can't seem to simplify it in anyway where I can use any complex identity. Therefore, I haven't gotten to the square norm part of the problem. I was trying to incorporate where $e^{inπ}=(-1)^n$ or anything that gets me to a simpler solution. I also attempted to use $e^{ix}=\cos x + i\sin x$ but no cigar.
 A: The function $f(x)=e^{-\frac{\pi x}{2}}$ on $-\pi < x < \pi$ admits a periodic extension $\hat{f}$ on whole $\mathbb R$ with period $2L=2\pi$. For example, all points $x_n=2n\pi$, $n\in\mathbb Z$ are mapped into $\hat f(x_n)= 1$. Note that $\hat f(0):=f(0)=1$.
We want to find the Fourier coefficients $C_n$ of the periodic extension $\hat f$ s.t. 
$$
\hat f(x) = \sum_{n=-\infty}^{\infty} C_{n}e^{\frac{in\pi x}{\pi}},
$$ 
where $C_n$ is 
$$
C_{n} = \frac{1}{2\pi}\int_{-\pi}^{\pi}\hat f(x)e^{-\frac{in\pi x}{2\pi} }\,dx = \text{(def. of periodic extension)}=
\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)e^{-\frac{in\pi x}{2\pi} }\,dx,$$
i.e.
$$
2\pi C_{n} = \int_{-\pi}^{\pi} e^{-\frac{(\pi+2in)x}{2} }\,dx
= -\frac{2}{(\pi+2in)} \left( e^{-\frac{(\pi+2in)\pi}{2}}-e^{\frac{(\pi+2in)\pi}{2}}  \right)=\\
\frac{2}{(\pi+2in)} \left( e^{\frac{\pi^2}{2}+in\pi}-e^{-\frac{\pi^2}{2}-in\pi}\right);
$$
using $e^{in\pi}=e^{-in\pi}=(-1)^n$ one arrives at the final result, i.e.
$$C_n = \frac{(-1)^n}{\pi(\pi+2in)} \left( e^{\frac{\pi^2}{2}}-e^{-\frac{\pi^2}{2}}\right).$$
