Re: Strange integral

On 7 Сер, 23:26, "David C. Ullrich" <dullr...@xxxxxxxxxxx> wrote:
In article
<bc760f69-cca4-4b00-93d2-637167ddc...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,



 leox <leonid...@xxxxxxxxx> wrote:
On 7 óÅÒ, 18:48, Ray Vickson <RGVick...@xxxxxxx> wrote:
On Aug 7, 7:10 am, leox <leonid...@xxxxxxxxx> wrote:

 Do you mean to write 1/[(1-p*exp(it))*(1-q*exp(-it))] = 1 +
pq + (pq)^2 + ... + sum_{k,m:k < > m} p^k q^m exp((k-m)*t)? (writing m
instead of 'l', to avoid confusing 'l' = ell with '1' = one). Or did
you mean something else?

yes, I mean exact it

1/[(1-p*exp(it))*(1-q*exp(-it))] = 1 +
 pq + (pq)^2 + ... + sum_{k,m:k < > m} p^k q^m exp((k-m)*t)?

Anyway, don't you need to have |pq| < 1 to
 get a convergent series?


Ah, _there's_ your error.

No, if |p| < 1 and |q| < 1 then one pole is inside the circle
and one pole is outside the circle. (Where _are_ the poles,
anyway? Not at p and q.)


Yes, thanks...you are right:) really - the poles are p and 1/q .
the condition |pq|<1 implies that only one of them is inside the
circle |z|=1. Then by using residues or Cauchy's integral formula one
may calculate that the integral is equal exact to 1/(1-pq)

Very strange final. The integral is came from algebra and I was
interested only in formal manipulation with series, without any
convergence etc..
Thanks

p.s. Maple and Mathematica made mistake, not me only :)
.

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