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Question: Is there a closed form expression for $f(z)$ defined in formula $(1)$ below?

$$f(z)=\sum\limits_{k=1}^\infty \log\left(1+\frac{z^2}{k^2}\right),\quad\Re(z)>0\tag{1}$$

I believe the sum in formula $(1)$ above converges with no branches for $\Re(z)>0$. I tried simplifying this sum to a closed form representation using Mathematica but was unsuccessful.

Gary
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Steven Clark
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2 Answers2

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$$f(z)=\sum\limits_{k=1}^\infty \log\left(1+\frac{z^2}{k^2}\right)=\log \Bigg[\prod\limits_{k=1}^\infty \left(1+\frac{z^2}{k^2}\right)\Bigg]=\log \left(\frac{\sinh (\pi z)}{\pi z}\right)$$

Claude Leibovici
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  • Really simple. Nice. – K.defaoite May 11 '22 at 07:01
  • The formula works for $s>0$, but unfortunately not for $Re(s)>0\land\Im(s)\ne 0$ which I believe is related to the branch cut of the $\log(z)$ function. The imaginary part of the formula does not evaluate correctly for $Re(s)>0\land\Im(s)\ne 0$. – Steven Clark May 11 '22 at 14:35
  • @StevenClark The function $g(z)= \frac{{\sinh (\pi z)}}{{\pi z}}$ is holomorphic and non-zero if $\Re z>0$. Thus, since the open right half-plane is simply connected, $g(z)$ has a holomorphic logarithm there. Thus if $f(z)$ is analytic for $\Re z>0$ and $f(z)=\log g(z)$ for $z>0$, then $f(z)=\log g(z)$ for $\Re z>0$ as well. – Gary May 11 '22 at 22:52
  • @Gary I'm not sure I understand your point. The $\log\left(1+\frac{z^2}{k^2}\right)$ term of $f(z)$ has branch cuts along the positive and negative imaginary axis for $|\Im(z)|>1$ related to the branch cut of $\log(z)$ along the negative real axis. Therefore the function $f(z)$ has no branches for $\Re(z)>0$. – Steven Clark May 12 '22 at 02:45
  • @Gary The function $g(z)=\frac{{\sinh(\pi z)}}{{\pi z}}$ may be holomorphic, but $\log g(z)$ has many branches for $\Re(z)>0$. The real and imaginary parts of $g(z)$ both oscillate between positive and negative values when evaluated along the line $z=1+i t$ for example, and when the real part of $g(z)$ is negative and the imaginary part of $g(z)$ transitions from positive to negative, the imaginary part of $\log g(z)$ takes a branch. – Steven Clark May 12 '22 at 02:45
  • @StevenClark Do you know that if a function $g$ is analytic on a simply connected domain $D$ and $g(z)\neq 0$ for any $z\in D$, then $g$ has an analytic, single-valued logarithm on $D$? Use this with the $g$ specified in my comment and with $D$ being the right half-plane. Choose the branch of the $\log$ so that $\log g(0)=0$. – Gary May 12 '22 at 03:11
  • I'm evaluating the logarithm using Mathematica which I believe uses the principal branch. I believe you're saying $\log\left(\frac{\sinh(\pi z)}{\pi z}\right)$ can be analytically continued to eliminate the branches. Is it analogous to what I did in my question at https://math.stackexchange.com/q/4447013? The question here and my linked question are actually related as they're both related for formulas that I've derived for $\log\zeta(s)$. – Steven Clark May 12 '22 at 03:20
  • Yes, choose the branch of $\log$ so that $\log(\sinh(\pi z)/(\pi z))$ is real and positive if $z>0$ (note that the argument of the $\log$ is $>1$ if $z>0$). Now this function extends to the whole of $\Re z>0$ as a single-valued analytic function. Since according to your claim it agrees with your $f$ for $z>0$, it must agree with it for $\Re z>0$ as well. This is plain mathematics, what Mathematica does is another question. – Gary May 12 '22 at 03:23
  • @Gary Thanks for the clarification. I believe I can compensate for the branches of Mathematica's evaluation of $\log\left(\frac{\sinh(\pi z)}{\pi z}\right)$ the same way I did in my linked question. – Steven Clark May 12 '22 at 03:36
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We have the product representation of the Gamma function $$\Gamma(z)=\frac{e^{-\gamma z}}{z}\prod_{k\ge1}e^{z/k}\left(1+\frac{z}{k}\right)^{-1},$$ which lends itself to the series $$\log\Gamma(z)=-\gamma z-\log z+\sum_{k\ge1}\left[\frac{z}{k}-\log\left(1+\frac{z}{k}\right)\right].$$ Thus $$\begin{align} \log\Gamma(z)+\log\Gamma(-z)=&-\gamma z-\log z+\sum_{k\ge1}\left[\frac{z}{k}-\log\left(1+\frac{z}{k}\right)\right]\\ &+\gamma z-\log(-z)+\sum_{k\ge1}\left[-\frac{z}{k}-\log\left(1-\frac{z}{k}\right)\right]\\ =& -\log(z)-\log(-z)-\sum_{k\ge1}\left[\log\left(1+\frac{z}{k}\right)+\log\left(1-\frac{z}{k}\right)\right]\\ =&-\log(-z^2)-\sum_{k\ge1}\log\left(1-\frac{z^2}{k^2}\right). \end{align}$$ Hence $$\log\Gamma(iz)\Gamma(-iz)=-\log(z^2)-\sum_{k\ge1}\log\left(1+\frac{z^2}{k^2}\right),$$ so $$f(z)=-\log(z^2)-\log\Gamma(iz)\Gamma(-iz),$$ which can likely be simplified using the Gamma reflection formula.

clathratus
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  • Thanks for your answer! I was a bit skeptical that a closed form expression could be found. But shouldn't $\log\Gamma(iz)\Gamma(-iz)$ in your second to last formula be $\log\Gamma(iz)+\log\Gamma(-iz)$ and in your last formula be $\log\Gamma(iz)-\log\Gamma(-iz)$? – Steven Clark May 11 '22 at 03:59
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    @StevenClark Just use http://dlmf.nist.gov/4.36.E1 as shown by Claude. No need for the gamma function. – Gary May 11 '22 at 05:48
  • @Gary The formula posted by Claude works for $s>0$ but unfortunately not for $Re(s)>0\land\Im(s)\ne 0$ which I believe is related to the branch cut of the $\log(z)$ function, whereas the formula posted by clathratus works for $\Re(s)>0\land\Im(s)\ne 0$ as well as $s>0$. The imaginary part of the formula posted by Claude does not evaluate correctly for $Re(s)>0\land\Im(s)\ne 0$. – Steven Clark May 11 '22 at 14:32
  • @StevenClark I was simplifying $\log\Gamma(iz)+\log\Gamma(-iz)=\log\Gamma(iz)\Gamma(-iz)$. If this is not valid, we have $$f(z)=-\log(z^2)-\left[\log\Gamma(iz)+\log\Gamma(-iz)\right]$$ – clathratus May 11 '22 at 20:07
  • @StevenClark The function $g(z)= \frac{{\sinh (\pi z)}}{{\pi z}}$ is holomorphic and non-zero if $\Re z>0$. Thus, since the open right half-plane is simply connected, $g(z)$ has a holomorphic logarithm there. Thus if $f(z)$ is analytic for $\Re z>0$ and $f(z)=\log g(z)$ for $z>0$, then $f(z)=\log g(z)$ for $\Re z>0$ as well. – Gary May 11 '22 at 22:53
  • @clathratus Thanks for the clarification. I originally assumed $\log\Gamma(iz)\Gamma(-iz)$ was a typo, so I've just been using $\log\Gamma(iz)+\log\Gamma(-iz)$. But since you indicated it was a simplification instead of a typo I checked and the simplification doesn't appear to be valid, so I'll just continue to use $\log\Gamma(iz)+\log\Gamma(-iz)$ for now. – Steven Clark May 12 '22 at 03:06