How to show the following equality? ∑n=0∞1a2+n2=1+aπcoth⁡aπ2a2

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John Koga · Accepted Answer

This problem is a direct application of Fourier transform and Poisson summation formula. Recalling the definition of Fourier transform and the Poisson summation formula respectively  F(w)=12π∫−∞∞f(x)e−xwdx  ∑−∞∞f(n)=2π∑−∞∞F(2nπ)  where F is the Fourier transform of f. Advancing with our problem, first, we compute the Fourier transform of f(x)=1x2+a2 wihich is equal to  F(w)=π21ae−a|w|  Applying Poisson formula, we have  ∫n=0∞1n2+a2=πa∑n=0∞e−2anπ=πa∑n=0∞rn  ⇒∑n=0∞1n2+a2=πa11−e−2aπ  Now, I leave it to you to manipulate the above expression to reach the form  ∑n=0∞1a2+n2=1+aπcoth⁡aπ2a2  You can use the identity  coth⁡x=coshxsinhx=ex+e−xex−e−x=e2x+1e12x

Alex Sheppard · Accepted Answer

It is well known that  ∑n=−∞∞f(n)=−∑j=1kResπcot⁡(πz)f(z)  Assume a≠0  To find the residues of g(z)=πcot⁡(πz)1a2+n2, we see  1a2+n2=1(n+ia)(n−ia)  so g has poles at z1=ia and z2=−ia. Their respective residues, b1 and b2 can be found  b1=Resg(z)=limz→iaπcot⁡(πz)z−ia}{(z+ia)(z−ia)}=πcot⁡(πia)12ia  And finally:  ∑k=−∞1a2+k2=−(b1+b2)=πcoth⁡(πa)a  To change the starting number from −∞ to 0, we divide the series, as it is symmetrical (i.e. g(n)=g(−n) ):  ∑k=−∞∞1a2+k2=πcoth⁡(πa)a=∑k=−∞−11a2+k2+1a2+∑k=1∞1a2+k2   Thus  ∑k=0∞1a2+k2=πcoth⁡(πa)2a+12a2=πcoth⁡(πa)+12a2

How to show the following equality?\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\po\coth a\pi}{2a^2}

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