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!!!
Best Answer
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!!!