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I have the function $f(x) = e^{2x}$ in the interval $(0,2\pi]$

Using the formula $\int_0^{2\pi } e^{2x} e^{-inx}\, dx $

I get that $ {\mathbf{\gamma}}^{}_{n}= \frac1{2\pi} \frac{(e^{4\pi}-1)}{(2-in)} $

To complete the fourier serie I have $f(x) =\sum_{-\infty}^{+\infty} {\mathbf{\gamma}}^{}_{n} e^{inx} $

Now my problem is that I have to calculate the sums of:

$$ \sum_{-\infty}^{+\infty} {\mathbf{\gamma}}^{}_{n}.$$ This is easy I think, because I use $x=0$ here $f(x) =\sum_{-\infty}^{+\infty} {\mathbf{\gamma}}^{}_{n} e^{inx} $ and I get the summatory by calculating the lateral limits in the function

$$ \sum_{-\infty}^{+\infty} (-1)^n{\mathbf{\gamma}}^{}_{n}$$ Here I dont know what to do, I was thinking on dividing the summatory in negative and positive terms, but I dont know how to go after that.

Jyrki Lahtonen
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German
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1 Answers1

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$$f(x)=e^{2x}\sim \sum_{n=-\infty}^{\infty}\gamma_ne^{inx}$$ \begin{align} \gamma_n &= \frac{1}{2\pi}\int_0^{2\pi}e^{2x}e^{-inx}\;dx = \frac{1}{2\pi}\frac{e^{(2-in)2\pi} - 1}{2-in} = \frac{1}{2\pi}\frac{e^{4\pi} - 1}{2-in} \cdot \frac{2+in}{2+in}\\ &=\frac{1}{\pi}e^{2\pi}\frac{e^{2\pi}-e^{-2\pi}}{2} \frac{2+in}{n^2+4} = \frac{1}{\pi}e^{2\pi}\sinh(2\pi)\frac{2+in}{n^2+4} \end{align}

We note that pairing positive and negative coefficients gives:

$$\gamma_n +\gamma_{-n} = \frac{1}{\pi}e^{2\pi}\sinh(2\pi)\frac{(2+in)+(2+i(-n))}{n^2+4} = \frac{4}{\pi}e^{2\pi}\sinh(2\pi)\frac{1}{n^2+4} $$

$$\sum_{n=-\infty}^{\infty}\gamma_n = \gamma_0 + \sum_{n=1}^{\infty} \big(\gamma_n+\gamma_{-n}\big) = \frac{1}{2\pi}e^{2\pi}\sinh(2\pi) + \frac{4}{\pi}e^{2\pi}\sinh(2\pi)\sum_{n=1}^\infty \frac{1}{n^2+4}$$

So the question now is what does the left hand side converge to? We know that the Fourier series converges on $(0,2\pi)$ and this shape repeats every $2\pi$. If the series converges between, then it converges to the average value, e.g., at $x=0$, $$\sum_{n=-\infty}^{\infty}\gamma_n = \frac{f(0^+)+f(2\pi^-)}{2} = \frac{e^0+e^{4\pi}}{2} = e^{2\pi}\frac{e^{2\pi}+e^{-2\pi}}{2} = e^{2\pi}\cosh(2\pi)$$

A bit of rearranging gives: $$\sum_{n=1}^\infty \frac{1}{n^2+4} = \frac{\pi}{4}\coth(2\pi) - \frac{1}{8}$$

adfriedman
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