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Old 2019-02-01, 17:28   #1
jrsousa2
 
Dec 2018
Miami

29 Posts
Default 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
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Old 2019-02-01, 18:21   #2
chalsall
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"Chris Halsall"
Sep 2002
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Sigh...

Interestingly, this forum has some seriously serious people.

And yet we regularly get twits posting nonsense.

We tend to deal with them diplomatically.
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Old 2019-02-01, 22:55   #3
jrsousa2
 
Dec 2018
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Haha
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Old 2019-02-02, 04:27   #4
jrsousa2
 
Dec 2018
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The question I had is, "Is it possible to obtain a complex harmonic progression formula using the \psi (derivative of the Gamma) function?
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Old 2019-02-05, 02:11   #5
jrsousa2
 
Dec 2018
Miami

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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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