New formula for complex harmonic progression

Question

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?

Answer 1

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

$$\sum_{j}\dfrac{c_j}{(x-x_j)}=\dfrac{1}{(x^{2k}+1)}=\dfrac{1}{f(x)}$$

and its coefficients are unique. They are :

$$c_j=\dfrac{1}{2k(x_j)^{2k-1}}=\dfrac{1}{f^{\prime}(x_j)}$$

(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

$$\dfrac{1}{(x^{4}+1)}=\dfrac14\left(\dfrac{w^{-3}}{(x-w)}+\dfrac{w^{-1}}{(x-w^3)}+\dfrac{w}{(x-w^5)}+\dfrac{w^{3}}{(x-w^7)}\right)$$

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...

Answer 2

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

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

Writing

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

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

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

Answer 3

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.

https://arxiv.org/abs/1902.01008

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