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In one paper I saw this equality:

$$\sum_{\eta=-\infty}^{\infty}\frac{z}{(z+\eta)}=\pi z\cot(\pi z)$$ which is the same as $$\sum_{\eta=-\infty}^{\infty}\frac{1}{(z+\eta)}=\pi \cot(\pi z)$$ where summation is understood in the sense of a principal value. What does it mean?

In another paper I found the next expression:

$$\frac{\exp(2\pi iaz)}{\exp(2\pi iz)-1}=\frac{1}{2\pi i}\sum_{n=-\infty}^{\infty}\frac{\exp(2\pi ina)}{z-n}$$ for $a=0$ it is equivalent to $$\frac{1}{\exp(2\pi iz)-1}=\frac{1}{2\pi i}\sum_{n=-\infty}^{\infty}\frac{1}{z+n}$$ which is not exactly the same expression like in the first case. $$\sum_{n=-\infty}^{\infty}\frac{1}{z+n}=\pi Cot[\pi z]-i\pi$$

Where is my mistake?

If the second formula is wrong, what is the correct formula for the second case? $$\sum_{n=-\infty}^{\infty}\frac{\exp(2\pi ina)}{z+n}=?$$

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You probably mean $z + \eta$, not $z + \nu$, in the denominator of the expression in the first summation. –  L Spice Feb 22 '10 at 16:25
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By the way, maybe you know this already, but the first formula is a standard exercise in contour integration. Maybe that will shed some light on it for you. See also mathoverflow.net/questions/10948/what-is-sum-x-mathbbz-2/10975 –  j.c. Feb 23 '10 at 1:49
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2 Answers

up vote 2 down vote accepted

A principal-value sum (or integral) is usually one in which unconditional summation (or integration) does not converge, so one needs to sum in a particular way to achieve convergence. I suspect that, in this case, the necessary summation is symmetric, so that we consider $\lim_{N \to \infty} \sum_{n = -N}^{n = N} f(n)$ instead of $\sum_{n = 1}^\infty f(-n) + \sum_{n = 0}^\infty f(n)$.

It's not quite clear to me what your issue is with the two formulæ you mention. Since you are summing different functions ($1/(z + n)$ versus $z/(z + n)$), it is no surprise that the answers are different. What am I missing? (Sorry, I did not notice that you had already factored out the $z$.)

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the sum is over n, not z –  vilvarin Feb 22 '10 at 16:32
    
Certainly, that was silly of me. Thanks! –  L Spice Feb 22 '10 at 17:11
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See the answer of L Spice for the principal value bit.

For the second bit, the formula from the second paper is rather suspect. For example, $a=1/2$ produces the divergent sum (even in the principal value sense) $$\sum_{n=-\infty}^\infty \frac{(-1)^n}{z-n}.$$ And for your case $a=0$, $z=1/2$ yields $\sum(z-n)^{-1}=0$ by symmetry, so the formula cannot be right then either.

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thanks for the answer –  vilvarin Feb 22 '10 at 19:21
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