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Let $a=\exp (2\pi i u) $ where $0<u<1$. Then it can be proved using Liouville's theorem that $$f(z, a) =\frac{1}{z}+\sum_{n=1}^{\infty}\left(\frac{a^{n}} {z+n} +\frac{a^{-n}} {z-n}\right)=2\pi i\cdot\frac{a^{-z}} {1-\exp(-2\pi iz)}, z\neq 0$$

Is there any other way to prove this preferably using some general techniques for summation of a series?

For $a=1$ it is easy to see that $f(z, a) =\pi\cot\pi z$ (based on infinite product of $\sin\pi z$) and I am expecting some similar approach which works for $a\neq 1$.

Note: For the uninitiated the above identity is crucial to the proof of Kronecker's second limit formula. Apart from this step the formula involves nothing but simple algebraic manipulation and is thus at the same level as the above identity. My interest in Kronecker's formula ignited due to this question.

Paramanand Singh
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  • @reuns: I thought that the series for cotangent was a consequence of infinite product of sine. Let's not worry about this. If there is any other approach, that's fine. – Paramanand Singh Oct 22 '17 at 05:08
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    What proof of $\pi \cot \pi z =\frac1z \sum_{n\ge 1} \frac{1}{z+n}+\frac{1}{z-n} $ would you like to extend ? And there is nothing simpler than starting from $2\pi i\frac{a^{-z}} {1-\exp(-2\pi iz)}-\frac{1}{z}-\sum_{n=1}^{\infty}\frac{a^{n}} {z+n} +\frac{a^{-n}} {z-n}$ is an entire function – reuns Oct 22 '17 at 05:09
  • @reuns: entire function approach is the one I have mentioned (Liouville's theorem). But this sort of seems indirect. – Paramanand Singh Oct 22 '17 at 05:11
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    Indirect, or one of the most powerful tool we know in maths (without it we wouldn't know elliptic functions, modular forms, Riemann surfaces, $\zeta(s)$, class field theory..) – reuns Oct 22 '17 at 05:13
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    @reuns: i personally like the proof of $1+2+\dots +n=n(n+1)/2$ by reversing the sum as $n+\dots+2+1$ and adding with original. This beats the proof by induction by a wide margin. Modern authors have been so biased by complex analytic techniques that other simpler direct methods have become obsolete. And it has added to a lot of mathematics which has become unnecessarily complicated. The fields which you have mentioned were handled by Ramanujan using much simpler methods to achieve much more than the modern guys. – Paramanand Singh Oct 22 '17 at 05:20

1 Answers1

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This series is one of the simplest cases of a Fourier series:

$\displaystyle e^{-i2\pi uz} = \frac{1}{2\pi}\sum\limits_{k=-\infty}^{+\infty} e^{i2\pi uk} \int\limits_0^{2\pi} e^{-i(z+k)t}dt = \frac{1}{2\pi} \sum\limits_{k=-\infty}^{+\infty} e^{i2\pi uk} \frac{1-e^{-i2\pi(z+k)}}{i(z+k)} $

$\hspace{1.5cm}\displaystyle = \frac{1- e^{-i2\pi z} }{i2\pi} \sum\limits_{k=-\infty}^{+\infty} \frac{ e^{i2\pi uk}}{z+k} $

(with which the formula in the question follows)

Paramanand Singh
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user90369
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