mersenneforum.org Formula for complex harmonic progression
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 2019-02-01, 17:28 #1 jrsousa2   Dec 2018 Miami 29 Posts Formula for complex harmonic progression If $a$ is integer and $i b$ is not integer: $\sum_{k=1}^{n}\frac{1}{a i k+b}=-\frac{1}{2b}+\frac{1}{2(a i n+b)}+\frac{2\pi}{e^{2\pi b}-1}\int_{0}^{1}e^{\pi(a i n+2b)u}\sin{(\pi a n u)}\cot{(\pi a u)}\,du$ Particularly, we can use the above to produce the below formulae: $\sum_{k=1}^{n}\frac{1}{k^2+1}=-\frac{1}{2}+\frac{1}{2(n^2+1)}+\frac{4\pi}{e^{4\pi}-1}\int_{0}^{1} e^{4\pi u}\cos{[2\pi n(1-u)]}\sin{(2\pi n u)}\cot{(2\pi u)}\,du$ $\sum_{k=1}^{n}\frac{1}{k^2+2k+2}=-\frac{1}{4}+\frac{1}{2(n^2+2n+2)}+\frac{2\pi}{e^{4\pi}-1}\int_{0}^{1}\left[e^{4\pi(1-u)}+e^{4\pi u}\right]\cos{2\pi(n+2)u}\sin{2\pi n(1-u)}\cot{2\pi(1-u)}\,du$ That answers the question I had on my other post. Last fiddled with by jrsousa2 on 2019-02-01 at 17:29
 2019-02-01, 18:21 #2 chalsall If I May     "Chris Halsall" Sep 2002 Barbados 23·5·251 Posts Sigh... Interestingly, this forum has some seriously serious people. And yet we regularly get twits posting nonsense. We tend to deal with them diplomatically.
 2019-02-01, 22:55 #3 jrsousa2   Dec 2018 Miami 1D16 Posts Haha
 2019-02-02, 04:27 #4 jrsousa2   Dec 2018 Miami 29 Posts The question I had is, "Is it possible to obtain a complex harmonic progression formula using the $\psi$ (derivative of the Gamma) function?
 2019-02-05, 02:11 #5 jrsousa2   Dec 2018 Miami 29 Posts For who's interested in knowing more about how the above formula was derived and how it can be generalized for higher powers, please check my 3rd paper. Since getting new formulas published in journals is not very easy, I'm using sites like this to bring attention to them. My next formula should tackle the Riemann Zeta function. https://arxiv.org/abs/1902.01008

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