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I find proofs of Abel-Plana formula $\sum_{n=0}^{\infty} f(n)-\int_{0}^{\infty} f(x)\text{d}x=\frac{1}{2}f(0)+\text{i}\int_{0}^{\infty}\frac{f(\text{i}t)-f(-\text{i}t)}{e^{2\pi t}-1}$ where $f$ is a function analytic everywhere in $\mathbb{C}$ simply saying that it follows from the argument principle for an appropriate choice of $g$ and $\gamma$.

The argument principle states that for any function $f$ that is meromorphic inside the piecewise regular simple closed curve $\gamma$ and analytic and non-null on $\gamma$, and for any function $g$ analytic on and inside $\gamma$, if $f$'s zeros inside $\gamma$ are $a_1,...,a_m$, of multiplicity $\alpha_1,...,\alpha_m$, and its poles inside $\gamma$ are $b_1,...,b_n$, of multiplicity $\beta_1,...,\beta_n$, we have

$\int_{\gamma}g(z)\frac{f'(z)}{f(z)}\text{d}z=2\pi\text{i}\Big(\sum_{k=1}^{m}\alpha_kg(a_k)-\sum_{k=1}^{n}\beta_kg(b_k) \Big)$

but I can't see what steps we would have to do to reach Abel-Plana formula...

Thank you all for any help!!!

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    $\begingroup$ Given $f(z)$, what is $\int_{\gamma} f(z) \pi\cot(\pi z) dz$ where $\gamma$ is the contour from $\infty + i\epsilon \to i\epsilon \to $ semi-circle at $0$ $ \to -i\epsilon \to \infty - i\epsilon$? $\endgroup$ – achille hui Jun 24 '14 at 16:27
  • $\begingroup$ I heartily thank you, Achille!!! I put $f$ as $g$ and $\sin(\pi z)$ as $f(z)$ in the formula above and I see that $\frac{d\sin(\pi z)}{dz}=\pi\cos(\pi z)$ has simple zeros in every integer, therefore I have $\int_{\pi/2}^{3\pi/2}f(\varepsilon e^{it})\pi\cot(\pi\varepsilon e^{it})i\varepsilon e^{it}dt+\int_{0}^{\infty}f(i\varepsilon+t)\pi\cot(\pi(i\varepsilon+t))dt+\int_{0}^{\infty}f(-i\varepsilon+t)\pi\cot(\pi(-i\varepsilon+t))dt=2\pi i\sum_{n=0}^{\infty}f(n)$ $\endgroup$ – Self-teaching worker Jun 25 '14 at 9:27
  • $\begingroup$ where, if I'm not wrong, according to a theorem about indented path integrals, the first integral should be $\int_{\pi/2}^{3\pi/2}f(\varepsilon e^{it})\pi\cot(\pi\varepsilon e^{it})i\varepsilon e^{it}dt=\pi i\text{Res}_{z=0}f(z)\pi\cot(\pi z)=i\pi^2 f(0)$, but I can't go further... :-( Thank you so much again!!! $\endgroup$ – Self-teaching worker Jun 25 '14 at 9:32
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    $\begingroup$ To complete the proof, rewrite $\pi\cot(\pi z)$ as $$\pi\cot(\pi z) = -2\pi i\begin{cases} {\displaystyle\;\frac{1}{e^{-2\pi i z} - 1} + \frac12},&z \in (i\epsilon,\infty+i\epsilon)\\ {\displaystyle\;\frac{1}{e^{2\pi i z} - 1} - \frac12},&z \in (-i\epsilon,\infty-i\epsilon) \end{cases}$$ For both the upper and lower branch of $\gamma$, $\pi\cot(\pi z)$ now composes of two pieces. In each case, the first piece is something that can be evaluated by deforming the contour from the positive real axis to appropriate part of imaginary axis. $\endgroup$ – achille hui Jun 25 '14 at 9:47
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    $\begingroup$ BTW, $\text{Res}_{z=0} f(z)\pi\cot(\pi z) = f(0)$, not $\pi f(0)$. $\endgroup$ – achille hui Jun 25 '14 at 9:49
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EDIT: this "answer" contains errors, as explained by Achille Hui. I have tried to amend it in the next "answer".

You are very kind: thank you a lot! So we have $\int_{\pi/2}^{3\pi/2}f(\varepsilon e^{it})\pi\cot(\pi\varepsilon e^{it})i\varepsilon e^{it}dt+\int_{0}^{\infty}f(i\varepsilon+t)\pi\cot(\pi(i\varepsilon+t))dt-\int_{0}^{\infty}f(-i\varepsilon+t)\pi\cot(\pi(-i\varepsilon+t))dt=2\pi i\sum_{n=0}^{\infty}f(n)$ [EDIT: I wrote wrong sings!] and therefore, as $\varepsilon\to 0$, $\int_{0}^{\infty}f(i\varepsilon+t)\pi\cot(\pi(i\varepsilon+t))dt-\int_{0}^{\infty}f(-i\varepsilon+t)\pi\cot(\pi(-i\varepsilon+t))dt=2\pi i\sum_{n=0}^{\infty}f(n)-\pi i\text{Res}_{z=0}f(z)\pi\cot(\pi z)\to2\pi i\sum_{n=0}^{\infty}f(n)-\pi i f(0)$.

Rewriting $\pi\cot(\pi z)$ and dividing by $2\pi i$ we get $\int_{0}^{\infty}\frac{f(i\varepsilon+t)}{e^{2\pi i(i\varepsilon+t)}-1}+\frac{1}{2}f(i\varepsilon+t)dt+\int_{0}^{\infty}\frac{f(-i\varepsilon+t)}{e^{-2\pi i(-i\varepsilon+t)}-1}+\frac{1}{2}f(-i\varepsilon+t)dt\to\sum_{n=0}^{\infty}f(n)-\frac{1}{2}f(0)$

If we knew that $\int_{0}^{\infty}f(\pm i\varepsilon+t)$ converges we could be allowed to write $\int_{0}^{\infty}\frac{f(i\varepsilon+t)}{e^{2\pi i(i\varepsilon+t)}-1}dt+\int_{0}^{\infty}\frac{f(-i\varepsilon+t)}{e^{-2\pi i(-i\varepsilon+t)}-1}dt+\int_{0}^{\infty}\frac{1}{2}f(i\varepsilon+t)+\frac{1}{2}f(-i\varepsilon+t)dt\to\sum_{n=0}^{\infty}f(n)-\frac{1}{2}f(0)$ but can we be sure that such a step is allowed?

Then it would be easy to change variables [EDIT: I had convinced myself that it is possible to change variable as in real integration, which is not true in general] to write $\int_{0}^{\infty}\frac{f(-i\varepsilon+s)}{e^{-2\pi i(-i\varepsilon+s)}-1}ds+\int_{0}^{\infty}\frac{f(i\varepsilon+s)}{e^{2\pi i(i\varepsilon+s)}-1}ds+\int_{0}^{\infty}\frac{1}{2}f(i\varepsilon+t)+\frac{1}{2}f(-i\varepsilon+t)dt$ $=i\int_{0}^{\infty}\frac{f(-i\varepsilon+it)}{e^{2\pi(\varepsilon+t)}-1}dt-i\int_{0}^{\infty}\frac{f(i\varepsilon-it)}{e^{2\pi(t-\varepsilon)}-1}dt+\int_{0}^{\infty}\frac{1}{2}f(i\varepsilon+t)+\frac{1}{2}f(-i\varepsilon+t)dt$ which would be the searched result if we could substitute $\varepsilon$ with 0, but I am not sure how we can pass to the limit under the integral sign: can we and, if we can, why can we? I heartily thank you again!!!

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    $\begingroup$ If I remember correctly, the typical assumption for Abel-Plana formula is $f(z)$ is analytic in the left half-plane and falls off at least as fast as $\frac{1}{z^{1+\delta}}$. All the integrals you write down converges absolutely away from the origin. It won't cause you any problem by taking the limit $\epsilon \to 0$ there. The only place you need to be a little bit more careful is what happens near the origin. In any event, all these integrals are contour integrals, you only need to worry about what happens at the end points. $\endgroup$ – achille hui Jun 25 '14 at 14:49
  • $\begingroup$ Wow, real time answer: I $\infty$-ly thank you!!! $\endgroup$ – Self-teaching worker Jun 25 '14 at 14:52
  • $\begingroup$ I'm noticing that I deformed the contour $\infty+i\varepsilon\to i\varepsilon\to$ semi-circle at $0\to-i\varepsilon\to\infty-i\varepsilon$ changing $\infty+i\varepsilon$ into $-i\infty+i\varepsilon$ and $\infty-i\varepsilon$ into $-i\infty+i\varepsilon$, by changing $\int_{0}^{\infty}\frac{f(\pm i\varepsilon+s)}{e^{\pm 2\pi i(\pm i\varepsilon+s)}-1}ds$ into $\mp i\int_{0}^{\infty}\frac{f(\pm i\varepsilon\mp it)}{e^{2\pi(\mp\varepsilon+t)}-1}dt$, respectively: haven't I? I'm not sure it is correct... $\endgroup$ – Self-teaching worker Jun 27 '14 at 11:00
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    $\begingroup$ Doesn't look right. Should be something like $$ \int_{i\epsilon+\infty}^{i\epsilon} \frac{f(s)}{e^{\color{red}{\bf -}2\pi i s} - 1} ds = -\int_0^\infty \frac{f(i\epsilon + s)}{e^{\color{red}{\bf -}2\pi i (i\epsilon + s)}-1} ds \quad\to\quad -i\int_\epsilon^\infty \frac{f(it)}{e^{2\pi t}-1} dt $$ You need to rotate the upper branch of contour $\;i\epsilon + \infty \to i\epsilon\;$ to $\;i\epsilon \to i\infty\;$. i.e to the positive imaginary axis. Remember, during the deformation, you need to make sure the contour avoid the origin. $\endgroup$ – achille hui Jun 27 '14 at 12:49
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    $\begingroup$ Another thing is on the upper branch, it should be $e^{\color{red}{\bf -}2\pi i s}$ that appear in the denominator. When you deform the upper branch of contour, you need that minus sign to suppress the contribution from the end point near infinity. $\endgroup$ – achille hui Jun 27 '14 at 12:49
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I deeply thank you. I made a mess, confusing the orientation. Trying to correct what I wrote: let $\gamma_{\varepsilon}$ be the semicircle at 0 $\gamma_{\varepsilon}(t)=\varepsilon e^{it},t\in[\pi/2,3\pi/2]$ and $\gamma$ our contour around the real positive semiaxis approaching infinity. A theorem about indented path let us infer that $\lim_{\varepsilon\to 0}\int_{\gamma_{\varepsilon}}f(z)\pi\cot(\pi z)dz=\text{Res}_{z=0}f(z)\pi\cot(\pi z)=\pi if(0)$.

We have

$\int_{\gamma}f(z)\pi\cot(\pi z)dz=\int_{i\varepsilon +\infty}^{i\varepsilon}f(z)\pi\cot(\pi z)dz+\int_{\gamma_{\varepsilon}}f(z)\pi\cot(\pi z)dz+\int_{-i\varepsilon}^{-i\varepsilon+\infty}f(z)\pi\cot(\pi z)dz$ $=-2\pi i\int_{i\varepsilon +\infty}^{i\varepsilon}\frac{f(z)}{e^{-2\pi iz}-1}+\frac{1}{2}f(z)dz+\int_{\gamma_{\varepsilon}}f(z)\pi\cot(\pi z)dz+2\pi i\int_{-i\varepsilon}^{-i\varepsilon+\infty}\frac{f(z)}{e^{2\pi iz}-1}+\frac{1}{2}f(z)dz$ $=2\pi i\sum_{n=0}^{\infty}f(n)$ and, rearranging the integrals and dividing by $2\pi i$,

$\sum_{n=0}^{\infty}f(n)-\int_{0}^{\infty}\frac{f(i\varepsilon +t)+f(-i\varepsilon +t)}{2}dt=\frac{1}{2\pi i}\int_{\gamma_{\varepsilon}}f(z)\pi\cot(\pi z)dz-\int_{i\varepsilon +\infty}^{i\varepsilon}\frac{f(z)}{e^{-2\pi iz}-1}dz+\int_{-i\varepsilon}^{-i\varepsilon+\infty}\frac{f(z)}{e^{2\pi iz}-1}dz$

The paths $\pm i\varepsilon\to\pm i\varepsilon+\infty$ of the last two integrals can respectively be deformed into $\pm i\varepsilon\to\pm i\varepsilon\pm i\infty$, as you have told me, saving me from a terrible confusion, to get

$-\int_{i\varepsilon +\infty}^{i\varepsilon}\frac{f(z)}{e^{-2\pi iz}-1}dz=\int_{0}^{\infty} \frac{f(i\varepsilon+t)}{e^{-2\pi(i\varepsilon+t)}-1}dt\quad\to\quad-\int_{i\varepsilon +i\infty}^{i\varepsilon}\frac{f(z)}{e^{-2\pi iz}-1}dz=i\int_{\varepsilon}^{\infty} \frac{f(it)}{e^{2\pi t}-1}dt$

$\int_{-i\varepsilon}^{-i\varepsilon+\infty}\frac{f(z)}{e^{2\pi iz}-1}dz=\int_{0}^{\infty}\frac{f(-i\varepsilon+t)}{e^{2\pi(-i\varepsilon+t)}-1}dt\quad\to\quad\int_{-i\varepsilon}^{-i\varepsilon-i\infty}\frac{f(z)}{e^{2\pi iz}-1}dz=-i\int_{\varepsilon}^{\infty} \frac{f(-it)}{e^{2\pi t}-1}dt$

which, taking both the limits to $\infty$, already written, and $\varepsilon\to 0$, proves the formula.

Correct?

Is the reason behind such a deformation the fact that the path $\pm i\varepsilon\to\pm i\varepsilon+R$ is, respectively with + and -, homotopic and with identical ends, on an open set where the integrand functions are analytic, to the union of the path $\pm i\varepsilon\to\pm i\varepsilon\pm iR$ with some path -say $\lambda_{R_{\pm}}$- linking the point $\pm i\varepsilon\pm R$ with the point $\pm i\varepsilon+R$ and the integral of the integrand functions approaches 0 on $\lambda_{R_{\pm}}$ as $R$ approaches $\infty$: am I right? If I am right, how can we know that the integrand functions $-\frac{f(z)}{e^{-2\pi iz}-1}=\frac{f(z)\cot(\pi z)}{2i}+\frac{f(z)}{2}$ and $\frac{f(z)}{e^{2\pi iz}-1}=\frac{f(z)\cot(\pi z)}{2i}-\frac{f(z)}{2}$ vanish on $\lambda_{R_{\pm}}$ as $R$ approaches $\infty$? I thought about Darboux's inequality and something similar to Jordan's lemma, and used some $\lambda_{R_{\pm}}$ like $\lambda_{R_{+}}(\theta)=i\varepsilon+Re^{i\theta},\theta\in[0,\pi/2]$ and $\lambda_{R_{-}}(\theta)=-i\varepsilon+Re^{i\theta},\theta\in[-\pi/2,0]$, but I am not sure I can apply them to prove that $\int_{\lambda_{R_{+}}}-\frac{f(z)}{e^{-2\pi iz}-1}dz\xrightarrow{R\to\infty} 0$ and $\int_{\lambda_{R_{-}}}\frac{f(z)}{e^{2\pi iz}-1}dz\xrightarrow{R\to\infty} 0$ or, as it is enough, that $\int_{\lambda_{R_{-}}}\frac{f(z)}{e^{2\pi iz}-1}dz+\int_{\lambda_{R_{+}}}-\frac{f(z)}{e^{-2\pi iz}-1}dz\xrightarrow{R\to\infty}0$...

I $\aleph_0$-ly thank you!

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