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Calculate the sum of a series of lorentzain function?

GROUPS:

Consider the following code:

IN: = Sum[1/((x + n)^2 + a^2), {n, -Infinity, Infinity}]


Out: = (\[Pi] Sinh[2 a \[Pi]])/(a (-Cos[2 \[Pi] x] + Cosh[2 a \[Pi]]))

But if I input

Sum[a/((x + n)^2 + a^2), {n, -Infinity, Infinity}]

just by multiplying a factor a, the result is totally different.

1/2 I (PolyGamma[0, 1 - I a - x] - PolyGamma[0, 1 + I a - x] + 
   PolyGamma[0, -I a + x] - PolyGamma[0, I a + x])

Apparently there is something wrong here... Could anybody help me?

POSTED BY: Yu Li
Answer
3 months ago

You can apply FullSimplify[] and the result will be more appropriate with your expectations:

In[10]:= Sum[1/((x + n)^2 + a^2), {n, -Infinity, Infinity}]
Out[10]= (\[Pi] Sinh[2 a \[Pi]])/(a (-Cos[2 \[Pi] x] + Cosh[2 a \[Pi]]))

In[11]:= Sum[a/((x + n)^2 + a^2), {n, -Infinity, Infinity}]
Out[11]= 1/2 (\[Pi] Coth[a \[Pi] - I \[Pi] x] + \[Pi] Coth[a \[Pi] + I \[Pi] x])

In[12]:= FullSimplify[1/2 I (PolyGamma[0, 1 - I a - x] - PolyGamma[0, 1 + I a - x] + PolyGamma[0, -I a + x] - PolyGamma[0, I a + x])]
Out[12]= 1/2 \[Pi] (Coth[\[Pi] (a - I x)] + Coth[\[Pi] (a + I x)])
POSTED BY: Valeriu Ungureanu
Answer
3 months ago

Thanks. FullSimplify did help.

and if we set

S1 = Sum[1/((x + n)^2 + a^2), {n, -Infinity, Infinity}];

S2 = Sum[a/((x + n)^2 + a^2), {n, -Infinity, Infinity}];

S1*a-S2 turn out to be zero.

POSTED BY: Yu Li
Answer
3 months ago

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