I have to find an even Fourier series (i.e. only cosine-terms) of $$f(x)=\cosh(x-1)$$ in the interval $0 \leq x \leq 1$ with periodicity 2.
(EDIT: With my calculated Fourier coefficients I end up with
$$f(x)=\frac{\sinh(2)}{2}+\sum_{n\geq 1} \frac{(-1)^n \sinh(2)}{1+\pi^2 n^2}\cos(\pi n x)$$
because the series is supposed to be even.)
(EDIT 2: For $f(x)=\text{cosh}(1-|x|)$, $|x| < 1$, we get
$$a_n = \frac{2e\pi n\sin(\pi n)+e^2-1}{e \pi^2 n^2 +e}=\frac{e^2-1}{e\pi^2 n^2 +e}$$
Therefore we get
$$f(x)=\frac{e^2 -1}{2e}+\sum_{n \geq1} \frac{e^2-1}{e\pi^2 n^2 +e}\cos(\pi n x)$$
I just quickly used WolframAlpha.)
With that I have to show, that
$$\sum_{n=1}^\infty \frac{1}{\pi^2 n^2 +1}=\frac{1}{e^2-1}$$
I've got trouble with the limits of integration because the length of the interval isn't equal to 2. We never had such an example in class.
I would be grateful for any kind of help! Especially with the second part.
Thank you!
Best Answer
To prove your sum note that $$\sum_{n\ge1}\frac{1}{n^2\pi^2+1}=\frac{i}{2}\sum_{n\ge1}\left(\frac{1}{n\pi+i}-\frac{1}{n\pi-i}\right)=\frac{i}{2}\sum_{n\in\Bbb Z\setminus\{0\}}\frac{1}{n\pi+i}.$$Taking the log-derivative of $$\sin s=s\prod_{n\in\Bbb Z\setminus\{0\}}\left(1-\frac{s}{n\pi}\right)$$gives $$\cot s=\frac{1}{s}-\sum_{n\in\Bbb Z\setminus\{0\}}\frac{1}{n\pi-s}.$$The case $s=-i$ gives $$\sum_{n\ge1}\frac{1}{n^2\pi^2+1}=\frac{i}{2}\left(\frac{1}{-i}-\cot(-i)\right)=\frac{\coth 1-1}{2}=\frac{\frac{e^2+1}{e^2-1}-1}{2}=\frac{1}{e^2-1}.$$