If $a$ is integer and $\textbf{i} b$ is not integer then:

$\sum_{k=1}^{n}\frac{1}{a i k+b}=-\frac{1}{2b}+\frac{1}{2(a i n+b)}+\frac{2\pi}{e^{2\pi b}-1}\int_{0}^{1}e^{\pi(a i n+2b)u}\sin{(\pi a n u)}\cot{(\pi a u)}\,du$

(If $a$ is not integer we can make it integer by putting it in evidence.)

As some examples of applications, we can use the above to produce the curious formula below:

\begin{multline} \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 \end{multline}

Or the below:

$\sum_{k=1}^{n}\frac{1}{k^2+1}=-\frac{1}{2}+\frac{1}{2(n^2+1)}+\frac{4\pi}{e^{4\pi}-1}\int_{0}^{1} e^{4\pi u}\cos{[2\pi n(1-u)]}\sin{(2\pi n u)}\cot{(2\pi u)}\,du$

I have two questions:

a) How do you get the same complex harmonic progression using $\psi$?

b) If we know all the roots of $x^{2k}+1=0$, and a linear combination such that $\sum_{j}c_j/(x-x_j)=1/(x^{2k}+1)$, we can produce a formula for $\sum_{j}1/(j^{2k}+1)$. Is such a linear combination known?


Not an answer but a remark about b) :

Let us give the name $f(x):=x^{2k}+1$

There exists an identity having the form


and its coefficients are unique. They are :


(by computation of the residue at a simple pole, as usually done in complex function theory).

An example : using formula above for $k=2$, with $w:=e^{i \pi/4}$, one gets


I have no idea if the formulas you deduce are classical or not.

Edit : an idea concerning the first formula (may be a deadend)

As the initial part of your formula is of the form :

$$\sum _{i=m}^{n}f(i)={\frac {f(n)+f(m)}{2}}+\int _{m}^{n} something$$

I have seen a connection with the powerful Euler-MacLaurin formula. I copy-paste it from the corresponding Wikipedia article : https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula

$$\displaystyle \sum _{i=m}^{n}f(i)=\int _{m}^{n}f(x)\,dx+{\frac {f(n)+f(m)}{2}}+\sum _{k=1}^{\lfloor p/2\rfloor }{\frac {B_{2k}}{(2k)!}}\underbrace{(f^{(2k-1)}(n)-f^{(2k-1)}(m))}_{A}+R_{p}$$

where the $B_{2k}$ are Bernoulli numbers.

Of course, there would be some transformation work, but I think it is valuable to try it. Ideally if all the derivatives'values in part denoted $A$ are zero, it is perfect because we don't have to cope with Bernoulli numbers...

  • $\begingroup$ That's great, I didn't know that. When you say classical formula, which one do you mean? $\endgroup$ – JR Sousa Feb 2 at 19:24
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    $\begingroup$ I mean the one you deduce from the decomposition. Besides, I have provided an example for the formula with residues. $\endgroup$ – Jean Marie Feb 2 at 19:42
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    $\begingroup$ I have different remarks. As it is more confortable to writ them in an answer, I do it in an Edit. $\endgroup$ – Jean Marie Feb 2 at 19:51
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    $\begingroup$ A double pole can be treated by considering that it results from the "coalescence" of two simple poles (a single pole in $x_0$, fixed, and a mobile pole at $x_0+\varepsilon$ with $\varepsilon \to 0$). $\endgroup$ – Jean Marie Apr 7 at 13:12
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    $\begingroup$ There is a formula for the computation of the residue in a pole $z_0$ with ordre $n>1$ which is : $\text{Res}(f,z_0) = \lim_{z\to z_0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}[(z - z_0)^{n}f(z)]$ $\endgroup$ – Jean Marie Apr 7 at 17:32

Answer to part a)

The sum in question is

$$s(n) = \sum_{k=1}^n \frac{1}{i \;a\; k+b}$$

1. Integral formula for the harmonic number

The harmonic number is defined as

$$H_n = \sum_{k=1}^n \frac{1}{k}$$


$$\frac{1}{k} = \int_0^1 x^{k-1} \, dx$$

and interchanging integral and sum observing

$$\sum_{k=1}^n x^{k-1} =\frac{1-x^n}{1-x}$$

we find

$$H_n =\int_0^1 \frac{1-x^n}{1-x} \, dx\tag{1}$$

2. Integral formula for $s(n)$

Applying the similar procedure to $s(n)$ we arrive under the x-integral at the sum

$$\sum _{k=1}^n x^{i a k+b-1} = \frac{x^{i a} x^{b-1} \left(1-x^{i a n}\right)}{1-x^{i a}}$$

Changing variables $x^{i a}\to y$ and rotating the integral path back to the real $y$-axis we have

$$s(n) =\frac{1}{i\; a} \int_{0}^1 \frac{\left(1-y^n\right) y^{-\frac{i b}{a}}}{1-y} \,dy$$

In the numerator of the integrand multiplying out and adding and subtracting $1$ this can be written as the sum of two terms

$$s(n) = \frac{1}{i\;a}\left( H_{n-\frac{i b}{a}} - H_{-\frac{i b}{a}}\right)\tag{2}$$

Here we have used the definition (1).

Now using the relation between the harmonic number and the polygamma function

$$H_z = -\gamma + \psi (z+1)\tag{3}$$

where $\gamma$ is Euler's gamma we obtain finally

$$s(n) =\frac{1}{i \;a}\left(\psi (1+n-\frac{i b}{a}) - \psi (1-\frac{i b}{a})\right)\tag{4}$$

  • $\begingroup$ I find this $\psi$ function so artificial, unwieldy and cumbersome, which is why I wanted to find a better one. $\endgroup$ – JR Sousa Feb 4 at 4:38
  • $\begingroup$ @ JR S Just take the Harmonic number $H_x$ instead, see (2). $\endgroup$ – Dr. Wolfgang Hintze Feb 4 at 6:42
  • $\begingroup$ Wow, I don't understand why this post go no votes, whereas my other formula for the real harmonic progression got so many. Are people scared of complex numbers? Perhaps they should see all the general formula, not just the simple case. $\endgroup$ – JR Sousa Feb 5 at 0:56
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    $\begingroup$ @ JR S It is my experience that the voting behaviour here can't be consistently explained by logic. So don't worry - be happy. $\endgroup$ – Dr. Wolfgang Hintze Feb 8 at 10:43
  • $\begingroup$ Wolfgang Haha, thank you. They are really misleading. $\endgroup$ – JR Sousa Feb 8 at 20:37

This is not an answer to items a) or b), but if you're 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.

I hope the mods don't pick on this answer. My goal was to know how much simpler my formulae are than $\psi$, because I intend to keep digging for more such formulae, and it's getting quite challenging.


Besides, this new formula complements the previous one for integer parameters, posted here.


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