I would need some help to solve the next complex equation for $y\in\mathbb {R}$, which I already know to be real-valued:

$$ \frac {1} {2i}\left ((2\pi)^{\text {iy}}\text {}\text {Sin}\left (\frac {\pi} {2}\left(\frac {1} {2} + \text {iy} \right) \right)\text {}\Gamma\left (\frac {1} {2} -\text {iy} \right) - (2\pi)^{-\text {iy}}\text {}\text {Sin}\left (\frac {\pi} {2}\left (\frac {1} {2} - \text {iy} \right) \right)\Gamma\left (\frac {1} {2} + \text {iy} \right)\text {} \right) = 0 $$

I went on Wolfram Alpha, but nothing appears about the roots: Equation on WolframAlpha Does someone has any idea about? Thanks.

  • $\begingroup$ It's reduced to solve ( I assume $\displaystyle y \in {\mathbb R}$ ): $$ \Im\left(\,{\left[\, 2\pi\,\right]^{{\rm i}y} \over \Gamma\left(\,1/2 + {\rm i}y\,\right)}\,\right) = 0 $$ I guess you should use some numerical method. $\endgroup$ – Felix Marin Dec 15 '14 at 18:36
  • $\begingroup$ I don't know if is the right way, but it's similar to Riemann Zeta function functional equation in $s=1/2+iy$. $\endgroup$ – Marco Cantarini Dec 15 '14 at 18:40
  • $\begingroup$ You're right, @The_Cam. This equation is also the roots of $\arg\left (\zeta\left (\frac {1} {2} + iy \right) \right)$ for $\theta = \left\{\frac {\pi} {2}, 0, -\frac {\pi} {2}, ... \right\}$ $\endgroup$ – Eddy Khemiri Dec 15 '14 at 19:18

Let's first write your equation as an imaginary part : $$\tag{1}\Im\left((2\pi)^{iy}\,\sin\left(\frac {\pi}2\left(\frac 12+iy\right)\right)\Gamma\left(\frac 12-iy\right)\right)=0$$ For real values of $y$ the power of $2\pi$ and the $\sin$ won't provide zeros except the trivial $y=0$ (the $\sin$ function behaves nearly like $\sinh$ here) so that we will study only the zeros of $\;\Im\left(\Gamma\left(\frac 12+iy\right)\right)$ (which are also the zeros of $\;\Im\left(\Gamma\left(\frac 12-iy\right)\right)$ of course).

Let's start with an illustration of $\;\Im\left(\Gamma\left(x+iy\right)\right)\,$ to show our case $x=\frac 12$ in context : Im Gamma in complex plane

For other illustrations (including the real and absolute value parts) and interesting relations for $\Gamma$ in the complex plane see the thread "Real and imaginary part of Gamma function".

The relation $\left|\Gamma\left(\frac 12+iy\right)\right|=\sqrt{\dfrac{\pi}{\cosh(\pi\,y)}}\;$ found there allows us to 'rescale' the function so that the zeros become visible (else the oscillations would disappear quickly from the exponential decay far from $y=0$ as you may see in the previous picture).

Representation of $\;\Im\left(\Gamma\left(\dfrac 12+i\,y\right)\sqrt{\dfrac{\cosh\left(\pi\,y\right)}{\pi}}\right)$ : zeros of Im Gamma

with the first non negative zeros approximately given by $y_k$ in : \begin{array} {lll} k&y_k&a_k\\ \hline 0&0&0\\ 1&2.702691117402403870165565853&2.718281828\\ 2&5.053344767849197367797351047&5.058444527\\ 3&6.821889695106635313202928274&6.825074237\\ 4&8.373032938914556281390088770&8.375376509\\ 5&9.797707517468851913880784837&9.799572037\\ 6&11.13617463421067206562439664&11.13772771\\ 7&12.41062737183439804026853637&12.41196089\\ 8&13.63490243932245059948493022&13.63607244\\ \end{array}

(of course the negative solutions are symmetrical and... you would have obtained the same solutions using Wolfram alpha !)

At this point let's notice that the imaginary part of $\Gamma\left(\dfrac 12+i\,y\right)$ will be zero iff $\Im\left(\log\,\Gamma\left(\dfrac 12+i\,y\right)\right)=k\pi,\quad \text{with}\;k\in\mathbb{Z}\;$ so let's use for example following asymptotic expansion of $\,\log\Gamma$ :

$$\tag{2}\log\,\Gamma\left(z\right)\sim \left(z-\frac 12\right)\log\left(z\right)-z+\frac 12\log(2\pi)+\sum_{n=1}^\infty \frac{B_{2n}}{2n\,(2n-1)z^{2n-1}}$$

$$\tag{3}\log\,\Gamma\left(\frac 12+i\,y\right)\sim \left(iy\right)\log\left(\frac 12+i\,y\right)-\frac 12-i\,y+\frac 12\log(2\pi)+\frac 1{12\left(\frac 12+iy\right)}$$ The imaginary part of this will be (after expansion for $y\gg 1$ and simplification) : $$\tag{4}\Im\left(\log\,\Gamma\left(\frac 12+i\,y\right)\right)\sim y\,(\log(y)-1)+\frac 1{24\,y}$$ We want these values to be multiple of $\pi$ so let's try to revert the function of $y$ at the right (neglecting the $O(1/y)$ term) :

We may rewrite $\;f(y):=y\,(\log(y)-1)\;$ as $\;\dfrac{f(y)}e=\dfrac ye\log \dfrac ye\;$ and use the Lambert $W$ function's identity $\,W(x\log x)=\log x\;$ to deduce $$\tag{5}\log \dfrac ye=W\left(\frac ye\log \frac ye\right)=W\left(\frac {f(y)}e\right)$$ that we will rewrite as $\;\displaystyle y=e^{\large{1+W(f(y)/e)}}\;$ so that the $k$-th zero $y_k$ (defined for $k>0$ by $\,f(y_k)=(k-1)\pi\,$ from the choice of the asymptotic formula for $\log\,\Gamma$) will be approximatively given by : $$\tag{6}\boxed{\displaystyle y_k\approx a_k=\large{e^{\,1+W\left({(k-1)\,\pi}/e\right)}}},\quad\text{for}\;k\in\mathbb{N}^*$$

Concerning the zeros of $\zeta$ and their relations with the argument of $\zeta$ you may be interested by this thread.

ADDITION: To compare the zeros we obtained with these of the Riemann-Siegel $\theta$ function let's apply $(2)$ to $\;\displaystyle z=\frac 14+i\,\frac y2\,$ : $$\tag{7}\log\,\Gamma\left(\frac 14+i\,\frac y2\right)\sim \left(\frac {iy}2-\frac 14\right)\log\left(\frac 14+\frac {iy}2\right)-\frac 14-\frac {iy}2+\frac {\log(2\pi)}2+\frac 1{3+6\,iy}$$ getting thus for $y\gg 1$ : $$\tag{8}\Im\left(\log\,\Gamma\left(\frac 14+i\,\frac y2\right)\right)\sim \frac {y}2\,\left(\log\frac y2-1\right)-\frac {\pi}8+\frac 1{48\,y}$$ Subtracting $\;\log(\pi)\dfrac y2\;$ from the definition of $\theta$ we get $$\tag{9}\theta(y)\sim \frac {y}2\,\left(\log\frac y{2\,\pi}-1\right)-\frac {\pi}8+\frac 1{48\,y}$$ so that, neglecting the $O(1/y)$ terms and solving $\theta(y)=(k-1)\,\pi$ (we may replace $k$ by $\frac k2$ as needed), we get $$k-1=\frac {\theta(y)}{\pi}= \frac {y}{2\,\pi}\,\log\frac y{2\,\pi\,e}-\frac 18$$ and $\;\displaystyle \frac{k-\frac 78}e=\frac {y}{2\,\pi\,e}\,\log\frac y{2\,\pi\,e}\;$ that we will resolve as : $$\log \dfrac y{2\,\pi\,e}=W\left(\frac y{2\,\pi\,e}\log \frac y{2\,\pi\,e}\right)=W\left(\frac{k-\frac 78}e\right)$$ obtaining thus this approximation of the Gram points : $$\tag{10}\boxed{\displaystyle y_k\approx 2\,\pi\;e^{\large{1+W\left(\left(k-\frac 78\right)/e\right)}}},\quad\text{for}\;k\in\mathbb{N}^*$$

(a table of values is given in the link with a shift of one in the indice).

  • 1
    $\begingroup$ First, thanks a lot, @Raymond! This is great. I've upvoted your answer, and certainly will defined it as the final answer as it seems difficult to do better. I've learned a lot going through this post and others you wrote. But I've found a simpler formula with the same roots, so yet I'm thinking about if it's possible to get a more direct answer. I will post it soon: for the moment, I need yet to understand everything in your posts! $\endgroup$ – Eddy Khemiri Dec 17 '14 at 14:59
  • $\begingroup$ Glad you liked it @EddyKhemiri ! It would be very pleasant to see a simpler formula for the roots (I only provided an approximation...) since reverting $\Gamma$ is not that easy... Wishing you fun with all that in the meantime, $\endgroup$ – Raymond Manzoni Dec 17 '14 at 15:14
  • $\begingroup$ I've noticed that the roots of this equation are solutions of $\arg\left(\zeta\left(\frac {1}{2} + iy \right)\right) = \left\{\frac {\pi}{2}, 0, -\frac {\pi} {2}, ... \right\}$ as also part of the roots of $\Re\left(\zeta\left(\frac {1}{2}+iy\right)\right)$ and $\Im\left(\zeta\left(\frac {1}{2}+iy\right) \right)$. My question is: may it be possible to define the roots of the Zeta function relying only on the Gamma (or Factorial) function? or such relation already exists? $\endgroup$ – Eddy Khemiri Dec 17 '14 at 17:25
  • $\begingroup$ @EddyKhemiri: Well, from the link at the end of my answer we see that the asymptotic distribution of zeros is the same as the distribution of zeros of the Riemann-Siegel $\theta$ function but this seems to be all we can tell at this point : the actual positions are regular only for $\Gamma$ and not for $\zeta$! B.t.w. could you verify that the values I provided verify $\;\arg\left(\zeta\left(\frac 12+iy\right)\right)=k\pi/2$. I think that there is a relation between the zeros here and those of R-S $\theta$ but I don't know it yet... $\endgroup$ – Raymond Manzoni Dec 17 '14 at 18:23
  • 1
    $\begingroup$ the 1st result you found does not match, but the 2nd one works just perfectly with $2\pi e^{1 + W\left (\frac {k - \frac {7} {8}} {e} \right)} \text { for } k = m\frac {1} {2}, m\in\mathbb {N}$. So, half-values of k are the solutions such that $\arg\left(\zeta\left(\frac{1}{2}+iy\right)\right)=\pm\frac{\pi}{2}$, but it's not so surprising, finally. $\endgroup$ – Eddy Khemiri Dec 18 '14 at 17:31

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