MHB An interesting two variables function....

[chisigma]

An interesting question has been posted in...

Infinite Series Conditional Converging

... about the following two variables function...

$\displaystyle \sigma(x,y)= \sum_{k=1}^{\infty} \frac{1}{(k+x)\ (k+y)}$ (1)

... and in particular it has been requested if the domain of $\sigma(*,*)$ must be restricted to the quarter of plane $x>0,y>0$ or may be that, with the exception of a discrete set of points, it could be also $x \le 0, y \le 0$. What is Your answer?...

Kind regards

$\chi$ $\sigma$

 

[CaptainBlack]

An interesting question has been posted in...

Infinite Series Conditional Converging

... about the following two variables function...

$\displaystyle \sigma(x,y)= \sum_{k=1}^{\infty} \frac{1}{(k+x)\ (k+y)}$ (1)

... and in particular it has been requested if the domain of $\sigma(*,*)$ must be restricted to the quarter of plane $x>0,y>0$ or may be that, with the exception of a discrete set of points, it could be also $x \le 0, y \le 0$. What is Your answer?...

Kind regards

$\chi$ $\sigma$

Since for sufficiently large \(k\) we have \(\frac{1}{(k+x)\ (k+y)}\sim k^{-2}\) this is convergent for all \(x,y \in \mathbb{R} \backslash \mathbb{Z}_{-}\) (and is obviously undefined for if either \(x\) or \(y\) are in \(\mathbb{Z}_{-}\) )

CB

 

[chisigma]

Of course what CB says is exact and that shows that also the most 'qualified' textbooks aren't free from mistakes. The two variable function...

$\displaystyle \sigma(x,y)= \sum_{k=1}^{\infty} \frac{1}{(k+x)\ (k+y)}$ (1)

... has been described in...

http://www.mathhelpboards.com/f15/difference-equation-tutorial-draft-part-i-426/#post2494

... and it was found to be...

$\displaystyle \sigma(x,y)= \frac{\phi(y)-\phi(x)}{y-x}$ (2)

... where...

$\displaystyle \phi(z)= \frac{d}{d z}\ \ln z!$ (3)

Now the function $z!$ is defined for all values of z with the exception of the negative integers, and that must be true also for $\sigma(x,y)$. Anyway the proposed question is very interesting because permits us to arrive to a nice result. If we consider the particular case $\displaystyle x=y=\frac{1}{2}$ and $\displaystyle x=y=-\frac{1}{2}$ we from (1) derive...

$\displaystyle \sigma(-\frac{1}{2}, -\frac{1}{2}) - \sigma(\frac{1}{2}, \frac{1}{2}) = 4$ (4)

In the particular case x=y=z the (2) becomes...

$\displaystyle \sigma(z,z)= \phi^{\ '}(z)$ (5)

... and because the McLaurin expansion of $\phi(z)$ [see the above indicated post...] is...

$\displaystyle \phi(z)= - \gamma + \sum_{k=2}^{\infty} (-1)^{k}\ \zeta(k)\ z^{k-1}$ (6)

... it is also...

$\displaystyle \phi^{\ '}(z)= \sum_{k=2}^{\infty} (-1)^{k}\ (k-1)\ \zeta (k)\ z^{k-2}$ (7)

... so that, combining (4) and (7) we arrive to write...

$\displaystyle \sum_{k=1}^{\infty} \frac{k}{2^{2k-1}}\ \zeta(2k+1) = 1$ (8)

It is well known that for the positive integers n odd with n>1 the function $\zeta(n)$ is a sort of 'unknown land'... may be that (8) can open some breach...

Kind regards

$\chi$ $\sigma$