evaluating a sum using fourrier analysis

[fallen frost]

Rotor 😑
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[fallen frost]

Rotor 😑

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[tardy schooner]

Can you make a document and share it? Maybe the bot can't handle it.

[quasi plume]

let a>0, we can evaluate $\sum_{n=1}^{+\infty} \frac{1}{n^{2}+a^{2}}$ using complex analysis, but i wanted to share a different approach using fourrier analysis that i found while testing different fourrier series expansions lets consider $\forall x \in ]-\pi, \pi[, f(x)=e^{ax}$ which is extended by a $2\pi$ period, in that case let's find $\mathcal{F}(f)$, we can write $ \ f(x)=\frac{a_0}{2}+\sum_{k=1}^{+\infty} a_k cos(kx)+\sum_{k=1}^{+\infty} b_k sin(kx)$ im not going to do the whole work here, but calculating each coefficient using the integral formula we find that $a_0=\frac{2sinh(a\pi)}{a\pi}$ $ \ \forall n \geq 1, a_n=\frac{2a(-1)^{n}sinh(a\pi)}{\pi(n^{2}+a^{2})}$ and $ \ b_n=\frac{2n(-1)^{n+1} sinh(a\pi)}{\pi(n^{2}+a^{2})}$, thus we have that $ \ \forall x \in ]-\pi, \pi[, e^{ax}=\frac{2sinh(a\pi)}{\pi} \left(\frac{1}{2a}+\sum_{n=1}^{+\infty} \frac{(-1)^{n}}{n^{2}+a^{2}} \left(acos(nx)-nsin(nx) \right)\right)$ the idea is now to evaluate $f$ at $\pi$ but the function is undefined at that point, but using jordan's theorem, the fourrier series converges to the the average of the limits on the left and right side ie $\mathcal{F}(f)(\pi)=cosh(a\pi)$ (the limit from the right side is $e^{a\pi}$ and left side is $e^{-a\pi}$ by $2\pi$ periodicity) so we have that $cosh(a\pi)=\frac{2sinh(a\pi)}{\pi} \left( \frac{1}{2a} +a\sum_{n=1}^{\infty} \frac{1}{n^{2}+a^{2}} \right)$ and so isolating the sum we finally get $\sum_{n=1}^{+\infty} \frac{1}{n^{2}+a^{2}}=\frac{1}{2a} \left(\pi coth(a\pi)-\frac{1}{2a} \right)$

[fallen frost]

Rotor 😑

[quasi plume]

i found my mistake

what do you think? @tardy schooner