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

Joanna Benson

Joanna Benson

Answered

2022-01-18

How to show the following equality?
n=01a2+n2=1+aπcothaπ2a2

Answer & Explanation

John Koga

John Koga

Expert

2022-01-18Added 33 answers

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)=12πf(x)exwdx
f(n)=2πF(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)=π21aea|w|
Applying Poisson formula, we have
n=01n2+a2=πan=0e2anπ=πan=0rn
n=01n2+a2=πa11e2aπ
Now, I leave it to you to manipulate the above expression to reach the form
n=01a2+n2=1+aπcothaπ2a2
You can use the identity
cothx=coshxsinhx=ex+exexex=e2x+1e12x
Alex Sheppard

Alex Sheppard

Expert

2022-01-19Added 36 answers

It is well known that
n=f(n)=j=1kResπcot(πz)f(z)
Assume a0
To find the residues of g(z)=πcot(πz)1a2+n2, we see
1a2+n2=1(n+ia)(nia)
so g has poles at z1=ia and z2=ia. Their respective residues, b1 and b2 can be found
b1=Resg(z)=limziaπcot(πz)zia}{(z+ia)(zia)}=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=11a2+k2

Thus
k=01a2+k2=πcoth(πa)2a+12a2=πcoth(πa)+12a2

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?