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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.

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  • $\begingroup$ What are your definitions (constants, for example)? How is $C_n$ defined? What have you tried? $\endgroup$ – AlexR Jul 23 '14 at 20:56
  • $\begingroup$ I edited my post to show what I have tried so far. Please let me know what steps to take thank you. $\endgroup$ – Academic Jul 23 '14 at 21:23
  • $\begingroup$ See math notation guide. $\endgroup$ – user147263 Jul 23 '14 at 21:24
  • $\begingroup$ I'm having an incredibly difficult time using the math notation currently so I will edit later. $\endgroup$ – Academic Jul 23 '14 at 21:43
  • $\begingroup$ Why did you undo the work healthier did to make your post readable? That's self-destructive. $\endgroup$ – blue Jul 23 '14 at 22:20
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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.

$$C_n = \frac{(-1)^n}{\pi(\pi+2in)} \left( e^{\frac{\pi^2}{2}}-e^{-\frac{\pi^2}{2}}\right).$$

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