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Given that the Fourier cosine transform of $f(x)=e^{-x}$ is: $$f(x)=e^{-x}=\frac{1-e^{-π}}{π}+\frac{2}{π}\sum_{n=1}^\infty\frac{1+e^{-π}(-1)^{n+1}}{n^2+1}\cos({nx}),0\leq x \leq π$$ calculate the following summation: $$\sum_{n=0}^\infty\frac{(-1)^n}{4n^2+1}, 0<x<π$$

Here's my attempt:

We start by noticing that we need to find a way to create a similar looking summation in the Fourier transform. For that reason, I'm going to start by replacing where $x=\frac{π}{2}$.

That way we get: $$e^\frac{-π}{2}=\frac{1-e^{-π}}{π}+\frac{2}{π}\sum_{n=1}^\infty\frac{1-e^{-π}}{n^2+1}\cos(\frac{nπ}{2})$$

We understand that for $\frac{nπ}{2}=\frac{(2n+1)π}{2}$ we get $\cos{\frac{nπ}{2}}=0$, therefore we limit the answer to where $n$ is an even number, so we write $n=2m$ and since $\frac{(2n+1)π}{2}=\frac{π}{2}$ for $m=0$ we can write the summation starting from $m=0$.

$$\frac{πe^{\frac{-π}{2}}+e^{-π}-1}{2}=\sum_{m=0}^\infty\frac{1-e^{-π}(-1)^{2m+1}}{(2m)^2+1}\cos(mπ)$$ $$=>$$ $$\frac{πe^{\frac{-π}{2}}+e^{-π}-1}{2}=\sum_{m=0}^\infty\frac{1+e^{-π}}{4m^2+1}(-1)^m$$ $$=>$$ $$\sum_{m=0}^\infty\frac{(-1)^m}{4m^2+1}=\frac{πe^{\frac{-π}{2}}+e^{-π}-1}{2({1+e^{-π}})}$$

Let me know if you have any other ideas, or if I made a mistake!

Edit: I actually think I made a mistake. I said that I'd be taking the sum for m=0, when I know I can't do that. So, instead, I'll leave the summation as it is with m=1. Now after the end of the solution I provided here, we are going to take the given sum that we're asked to calculate, and say we put 0 were we have n and it's going to give us the first term. Therefore:

$$\frac{(-1)^0}{40^2+1}=1$$

So, we get the result:

$$\sum_{m=0}^\infty\frac{(-1)^m}{4m^2+1}=\frac{πe^{\frac{-π}{2}}+e^{-π}-1}{2({1+e^{-π}})}+1$$

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