What is the closed form for ∑n=−∞∞?

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nick1337 · Accepted Answer

In my opinion, this is a rather nice example of application of the Poisson summation formula: ∑n∈Zf(n)=∑k∈Zf^(k), f^(v)=∫−∞∞f(x)e−2πivxdx Namely, setting f(x)=1(x−a)2+b2, we find f^(v)=∫−∞∞e−2πivxdx(x−a)2+b2=e−2πiva∫−∞∞e−2πivxdxx2+b2=πbe−2πiva−2π|v|b where we have assumed that b&amp;gt;0 and calculated the last integral using residues. Therefore, the sum we are trying to calculate reduces to geometric series: ∑n∈Z1(n−a)2+b2=πb∑k∈Ze−2πika−2π|k|b=πsinh⁡2πbb(cosh⁡2πb−cos⁡2πa)

star233 · Accepted Answer

This one can also be done with the standard technique of using the πcot⁡(πz) multiplier, integrating f(z)=πcot⁡(πz)(z−a)2+b2 along a circle using the same technique as here. We integrate f(z) along a circle of radius R with R going to infinity and the integral disappears in the limit so that the residues sum to zero (actually a square with vertices (±(N+1/2),±(N+1/2)) N a positive integer is easier to handle computationally). The poles of f(z) other than at the integers are at z0,1=a±ib and the residues are Res(f(z);z=z0)=πcot⁡(πz)2(z−a)|z=z0=πcot⁡(π(a+bi))2bi=π2biieiπ(a+bi)+e−iπ(a+bi)eiπ(a+bi)−e−iπ(a+bi) and Res(f(z);z=z1)=πcot⁡(πz)2(z−a)|z=z1=πcot⁡(π(a−bi))−2bi=πcot⁡(π(a−bi))−2bi=π−2biieiπ(a−bi)+e−iπ(a−bi)eiπ(a−bi)−e−iπ(a−bi) Now put x=eiπa and y=e−πb. The first residue becomes Res(f(z);z=z0)=π2bxy+1/x/yxy−1/x/y and the second residue is Res(fz);z=z1)=−π2bx/y+y/xx/y−y/x Adding the two contributions and simplifying with Eulers

RizerMix · Accepted Answer

∑n=−∞∞1(n−a)2+b2 =∑n=−∞−11(n−a)2+b2+1a2+b2+∑n=1∞1(n−a)2+b2 =∑n=1∞1(n+a)2+b2+1a2+b2+∑n=1∞1(n−a)2+b2 =−1a2+b2+[∑n=0∞1(n−a)2+b2+(a→−a)] ∑n=0∞1(n−a)2+b2=∑n=0∞1[n−(a+bi)][b−(a−bi)] =ψ(a+bi)−ψ(a−bi)(a+bi)−(a−bi)=2iψ(a+bi)2bi=ψ(a+bi)b ∑n=−∞∞1(n−a)2+b2=−1a2+b2+ψ(a+bi)+ψ(−a+bi)b

What is the closed form for \sum_{n=-\infty}^{\infty}?

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