The following infinite sum of exponential terms gives a Dirac comb:
$$ \sum_{n=-\infty}^\infty e^{i n x} = 2 \pi \sum_{n=-\infty}^\infty \delta(x - 2 \pi n) $$ Of course the sum doesn't strictly converge, but only in the same sense in which the Dirac delta-function is defined.

What is the result of a semi-infinite sum of such terms?
$$ \sum_{n=1}^\infty e^{i n x} =~? $$



(1) write the semi-infinite sum in the following way $$\sum_{n=1}^\infty e^{i n x} = \frac{1}{2} \sum_{n=-\infty}^\infty e^{i n x} + \frac{1}{2} \sum_{n=-\infty}^\infty \mathop{\rm sgn}(n) e^{i n x} - \frac{1}{2}.$$

The first summand is the Dirac-comb. So your question reduces to figuring out what the second term equals to...

(2) Lagrange's trigonometric identities might help there...

  • $\begingroup$ Got it - thanks! $\endgroup$
    – Joe
    Aug 5 '12 at 11:17

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