Joanna Benson

2022-01-18

How to show the following equality?
$\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{a}^{2}+{n}^{2}}=\frac{1+a\pi \mathrm{coth}a\pi }{2{a}^{2}}$

John Koga

Expert

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
$F\left(w\right)=\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(x\right){e}^{-xw}dx$
$\sum _{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(n\right)=\sqrt{2\pi }\sum _{-\mathrm{\infty }}^{\mathrm{\infty }}F\left(2n\pi \right)$
where F is the Fourier transform of f. Advancing with our problem, first, we compute the Fourier transform of $f\left(x\right)=\frac{1}{{x}^{2}+{a}^{2}}$ wihich is equal to
$F\left(w\right)=\sqrt{\frac{\pi }{2}}\frac{1}{a}{e}^{-a|w|}$
Applying Poisson formula, we have
${\int }_{n=0}^{\mathrm{\infty }}\frac{1}{{n}^{2}+{a}^{2}}=\frac{\pi }{a}\sum _{n=0}^{\mathrm{\infty }}{e}^{-2an\pi }=\frac{\pi }{a}\sum _{n=0}^{\mathrm{\infty }}{r}^{n}$
$⇒\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{n}^{2}+{a}^{2}}=\frac{\pi }{a}\frac{1}{1-{e}^{-2a\pi }}$
Now, I leave it to you to manipulate the above expression to reach the form
$\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{a}^{2}+{n}^{2}}=\frac{1+a\pi \mathrm{coth}a\pi }{2{a}^{2}}$
You can use the identity
$\mathrm{coth}x=\frac{\text{cosh}x}{\text{sinh}x}=\frac{{e}^{x}+{e}^{-x}}{{e}^{x}-{e}^{-x}}=\frac{{e}^{2x}+1}{{e}_{1}^{2x}}$

Alex Sheppard

Expert

It is well known that
$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}f\left(n\right)=-\sum _{j=1}^{k}Res\pi \mathrm{cot}\left(\pi z\right)f\left(z\right)$
Assume $a\ne 0$
To find the residues of $g\left(z\right)\phantom{\rule{0.222em}{0ex}}=\pi \mathrm{cot}\left(\pi z\right)\frac{1}{{a}^{2}+{n}^{2}}$, we see
$\frac{1}{{a}^{2}+{n}^{2}}=\frac{1}{\left(n+ia\right)\left(n-ia\right)}$
so g has poles at ${z}_{1}=ia$ and ${z}_{2}=-ia$. Their respective residues, ${b}_{1}$ and ${b}_{2}$ can be found
${b}_{1}=Resg\left(z\right)=\underset{z\to ia}{lim}\pi \mathrm{cot}\left(\pi z\right)\frac{z-ia}{\right\}}\left\{\left(z+ia\right)\left(z-ia\right)\right\}=\pi \mathrm{cot}\left(\pi ia\right)\frac{1}{2ia}$
And finally:
$\sum _{k=-\mathrm{\infty }}\frac{1}{{a}^{2}+{k}^{2}}=-\left({b}_{1}+{b}_{2}\right)=\frac{\pi \mathrm{coth}\left(\pi a\right)}{a}$
To change the starting number from $-\mathrm{\infty }$ to 0, we divide the series, as it is symmetrical (i.e. $g\left(n\right)=g\left(-n\right)$ ):
$\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{{a}^{2}+{k}^{2}}=\frac{\pi \mathrm{coth}\left(\pi a\right)}{a}=\sum _{k=-\mathrm{\infty }}^{-1}\frac{1}{{a}^{2}+{k}^{2}}+\frac{1}{{a}^{2}}+\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{a}^{2}+{k}^{2}}$

Thus
$\sum _{k=0}^{\mathrm{\infty }}\frac{1}{{a}^{2}+{k}^{2}}=\frac{\pi \mathrm{coth}\left(\pi a\right)}{2a}+\frac{1}{2{a}^{2}}=\frac{\pi \mathrm{coth}\left(\pi a\right)+1}{2{a}^{2}}$

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