Suppose we have the function $f(n) = \sqrt{n^{2}+a^{2}}$, where $a$ is real. I need to calculate the finite expression $$ \sum_{n = 0}^{\infty}f(n) - \int \limits_{0}^{\infty}f(x)dx $$ By using the Abel-Plana formula, I obtain $$ \sum_{n = 0}^{\infty}f(n) = \int \limits_{0}^{\infty}f(x)dx + \frac{a}{2} + i\int \limits_{0}^{\infty}\frac{f(ix) - f(-ix)}{e^{2\pi x}-1}dx $$ How to show that the last integral is $$ i\int\limits_{0}^{\infty}\frac{f(ix) - f(-ix)}{e^{2\pi x}-1}dx = -2\int\limits_{a}^{\infty} \frac{\sqrt{x^{2}-a^{2}}dx}{e^{2\pi x}-1}? $$ The problem is with the branching point $x = \pm ia$, which leads to appearance of the additional phase for $f(ix)$ for $x>a$, and for $f(ix)$ it is $+i$, while for $f(-ix)$ it is $-\pi$.
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Similarly: https://math.stackexchange.com/questions/1387950/how-to-evaluate-infinite-series-sum-limits-n-0-infty-sqrtb2n2-e-an – Simply Beautiful Art Jun 07 '17 at 17:28
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the sum in question diverge if we interpret it in a classical manner.. so what do have in mind exactly? – tired Jun 07 '17 at 18:10
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@tired : instead of the sum, we can consider its finite part: $$ \sum_{n = 0}^{\infty}f(n) - \int \limits_{0}^{\infty}f(x)dx $$ The only remained terms in the Abel-Plana formula are finite. – John Taylor Jun 07 '17 at 18:18
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ok fine... next question: where do you get the $e^{2\pi f(x)}$ in the denominator from? usually we have just $e^{2\pi x}$ (see wiki for instance)? – tired Jun 07 '17 at 18:33
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@tired : sorry, this is a misprint. – John Taylor Jun 07 '17 at 18:35
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then you should remove it... – tired Jun 07 '17 at 18:40
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@tired : it is already done. – John Taylor Jun 07 '17 at 18:45
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ok, next problem what can we do to handle the singulatiry at $x=0$ of your target integral? – tired Jun 07 '17 at 18:56
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@tired : I'm sorry for such mistakes, this is again a misprint. The correct lower limit should be $a$ instead of $0$... – John Taylor Jun 07 '17 at 19:07
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Are you also aware that your target function $f$ doesn't fit the usual criteria to apply abel-plana? we therefor have to be really careful to make sense out of the outlined calculations – tired Jun 07 '17 at 19:13
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shouldne be $$ \sqrt{a^{2}-x^{2}} $$ – Jose Garcia Jun 07 '17 at 19:14
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@tired : it turns out that the extra contributions in the Abel-Plana formula coming from the bypassing of the branching points tend to zero. – John Taylor Jun 07 '17 at 19:18
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@JoseGarcia : it seems that it shouldn't. – John Taylor Jun 07 '17 at 19:19
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I would rephrase and ask for a closed form (if existing) of $\int_{0}^{+\infty}\left( f(x)-f(\lfloor x\rfloor)\right),dx$ with $f(x)=\sqrt{x^2+a^2}$, instead of asking for the value of a clearly divergent series. – Jack D'Aurizio Jun 10 '17 at 22:10