# Are all zeros of $\Gamma(s) \pm \Gamma(1-s)$ on a line with real part = $\frac12$ ?

The function $\Gamma(s)$ does not have zeros, but $\Gamma(s)\pm \Gamma(1-s)$ does.

Ignoring the real solutions for now and assuming $s \in \mathbb{C}$ then:

$\Gamma(s)-\Gamma(1-s)$ yields zeros at:

$\frac12 \pm 2.70269111740240387016556585336 i$ $\frac12 \pm 5.05334476784919736779735104686 i$ $\frac12 \pm 6.82188969510663531320292827393 i$ $\frac12 \pm 8.37303293891455628139008877004 i$ $\frac12 \pm 9.79770751746885191388078483695 i$ $\frac12 \pm 11.1361746342106720656243966380 i$ $\frac12 \pm 12.4106273718343980402685363665 i$

$\dots$

and

$\Gamma(s)+\Gamma(1-s)$ gives zeros at:

$\frac12 \pm 4.01094805906156869492043027819 i$ $\frac12 \pm 5.97476992595365858561703252235 i$ $\frac12 \pm 7.61704024553573658642606787126 i$ $\frac12 \pm 9.09805003388841581320246381948 i$ $\frac12 \pm 10.4760650707765536619292369200 i$ $\frac12 \pm 11.7804020877663106830617193188 i$ $\frac12 \pm 13.0283749883477570386353012761 i$

$\dots$

By multiplication, both functions can be combined into: $\Gamma(s)^2 - \Gamma(1-s)^2$

After playing with the domain of $s$ and inspecting the associated 3D output charts, I now dare to conjecture that all 'complex' zeros of this function must have a real part of $\frac12$.

Has this been proven? If not, appreciate any thoughts on possible approaches.

Thanks!

• Should be an easy consequence of Euler's reflection formula, I'd guess. $\Gamma(z) \Gamma(1-z) = \pi / \sin(\pi z)$. Feb 23, 2012 at 20:59
• @Marty: It does not seem so easy. Feb 23, 2012 at 21:27
• I guess something like this works: apply complex Stirling approximation to $|\Gamma(s)| = |\Gamma(1-s)|$ to show no non-real zeros with real part outside $[0,1]$, then contour integrate to count zeros of $\Gamma(s) \pm \Gamma(1-s)$ in a rectangle $[0,1]+i[-T,T]$, and compare with the number of roots of real part $1/2$ that can again be estimated by Stirling. Feb 23, 2012 at 23:52
• humbee (mathoverflow.net/users/21930/humbee) posted in an answer (now moved to this comment): A. Does this mean that Gamma(s) +- Gamma(1-s) has an Euler product? B. How does the existence of an Euler product affects the location of the on the critical strip? Mar 7, 2012 at 19:48
• How are my approximations of zeros with 5000 digits precision explained? Checked with precision 100 in sage, pari and maple.
– joro
Apr 9, 2012 at 9:07

In the first part, we show that there are no zeros for $z = s + i t$ with $|t| \ge 4$ .

Let $\psi(z):= \Gamma'(z)/\Gamma(z)$ be the digamma function. If $z = s + i t$, then $$\frac{d}{ds} |\Gamma(z)|^2 = \frac{d}{ds} \Gamma(z) \Gamma(\overline{z}) = |\Gamma(z)|^2 \left(\psi(z) + \psi(\overline{z})\right).$$ (Both $\Gamma(z)$ and $\psi(z)$ are real for real $z$, and so satisfy the Schwartz reflection principle.) The product formula for the Gamma function implies that there is an identity $$\psi(z) = - \ \gamma + \sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{z + n} \right) = 1 - \gamma + \sum_{n=1}^{\infty} \left(\frac{1}{n + 1} - \frac{1}{z + n} \right),$$ and hence $$\psi(z) + \psi(\overline{z}) = 2(1 - \gamma) + \sum_{n=1}^{\infty} \left(\frac{2}{n + 1} - \frac{1}{z + n} - \frac{1}{\overline{z} + n} \right).$$ Suppose that $z = s + i t$, and that $s \in [0,1]$. Then $$\frac{2}{n + 1} - \frac{1}{s + i t + n} - \frac{1}{s - i t + n} = \frac{2(s^2 + t^2 + n s - s - n)}{(1+n)(n^2 + 2 n s + s^2 + t^2)} \ge \frac{-2}{(n^2 + t^2)}.$$ (The last inequality comes from ignoring all the positive terms in the numerator, and then setting $s = 0$ in the denominator.) It follows that $$\psi(z) + \psi(\overline{z}) \ge 2(1 - \gamma) - \sum_{n=1}^{\infty} \frac{2}{n^2 + t^2},$$ which is positive for $t$ big enough, e.g. $|t| \ge 4$. On the other hand, $$\psi(z + 1) + \psi(\overline{z} + 1) = \psi(z) + \psi(\overline{z}) + \frac{1}{z} + \frac{1}{\overline{z}} = \psi(z) + \psi(\overline{z}) + \frac{2s}{|z|^2}.$$ In particular, if $\psi(z) + \psi(\overline{z})$ is positive for $s \in [0,1]$ for some particular $t$, it is positive for all $s$ and that particular $t$. It follows that, if $|t| > 4$, that $|\Gamma(s + it)|^2$ is increasing as a function of $s$. In particular, if $|t| > 4$, then any equality $$|\Gamma(s + i t)| = |\Gamma(1 - (s + i t))| = |\Gamma(1 - s + i t)|$$ implies that $s = 1/2$.

The second part is a continuation of the argument above, which completes the argument. (merged from a different answer.)

Let $C_n$ denote the square with vertices $[n \pm 1/2, \pm 4 I]$ for a positive integer $n$. We have the following inequalities for $z \in C_n$ and $n \ge 15$: $$|\sin(\pi z)| \ge 1, \quad z \in C_n.$$ $$|\Gamma(z)| \ge \frac{1}{2} \Gamma(n - 1/2),$$ $$|\Gamma(1-z)| \le \frac{\pi}{\Gamma(n - 1/2)} \le 1,$$ $$|\psi(1-z)|, |\psi(z)| \le 2 \log(n),$$

The first is easy, the second follows from Stirling's formula (this requires $n$ to be big enough, and also requires $z$ to have imaginary part at most $4$), the third follows from the previous two by the reflection formula for $\Gamma(z)$, the last follows by induction and by the formula $\psi(z+1) = \psi(z) + 1/z$. It follows that $$\left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\Gamma'(z)}{\Gamma(z)} - \frac{d/dz (\Gamma(z) + \theta \cdot \Gamma(1-z))}{\Gamma(z) + \theta\cdot \Gamma(1-z)} \right|$$ $$= \left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\theta \Gamma(1-z) (\psi(1-z) + \psi(z))} {\Gamma(z) + \theta \cdot \Gamma(1-z)} \right|$$ $$\le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \oint_{C_n} \frac{1} {|\Gamma(z) + \theta \cdot \Gamma(1-z)|}$$ $$\le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \cdot \frac{1}{1/2 \Gamma(n - 1/2) + 1} \ll 1,$$ where $\theta = \pm 1$ (or anything small) and $n \ge 15$, where the final inequality holds by a huuuge margin. It follows that $\Gamma(z) + \theta \cdot\Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the number of poles in $C_n$. Since $\Gamma(z)$ has no zeros and poles in $C_n$, it follows that $\Gamma(z) + \theta\cdot\Gamma(1-z)$ has the same number of zeros and poles. It has exactly one pole, and thus exactly one zero. If $\theta = \pm 1$ (and so in particular is real), by the Schwarz reflection principle, this zero is forced to be real. By symmetry, the same argument applies in the region $z = s + i t$ with $|t| \le 4$ and $s \le -15$. Combined with the above argument, this reduces the claim to $z = s + i t$ with $|s| \le 15$ and $|t| \le 4$ where the claim can be checked directly.

Hence all the zeros outside the box $z = s + it$ with $|t| \le 4$ and $|s| \le 15$ are either in $\mathbf{R}$, or lie on the line $1/2 + i \mathbf{R}$.

EDIT To clarify, I didn't actually check that there were no exceptional'' zeros in the box $\pm 15 \pm 4 I$, since I presumed that the original poster had done so. If $F(z) = \Gamma(z) - \Gamma(1-z)$, then computing the integral $$\frac{1}{2 \pi i} \oint \frac{F'(z)}{F(z)} dz$$ around that box, one obtains (numerically, and thus exactly) $1$. There are (assuming the OP at least computed the critical line zeros correctly) $2$ zeros in that range on the critical line. Along the real line in that range, there are $30$ poles and $25$ zeros. This means that there must be $1 + 30 - 25 = 6$ unaccounted for zeros. For such a zero $\rho$ off the line, by symmetry one also has $\overline{\rho}$, $1 - \rho$ and $1 - \overline{\rho}$ as zeros. Hence there must be either $1$ or $3$ pairs of zeros on the critical line, and either $1$ or $0$ quadruples of roots off the line. Varying the parameters of the integral, one can confirm there is a zero with $\rho \sim 2.7 + 0.3 i$, which is one of the four conjugates of the root found by joro. A similar argument applies for $\Gamma(z)+\Gamma(1-z)$. Hence:

Any zero of $\Gamma(z) - \Gamma(1-z)$ is either in $\mathbf{R}$, on the line $1/2 + i \mathbf{R}$, or is one of the four exceptional zeros $\{\rho,1-\rho,\overline{\rho},1-\overline{\rho}\}$. A similar calculation implies the same for $\Gamma(z) + \Gamma(1-z)$, except now with an exceptional set $\{\mu,1-\mu,\overline{\mu},1-\overline{\mu}\}$.

• There are some results in the literature that prove this monotonicity property for fairly small t but I don't recall offhand how small. Feb 24, 2012 at 3:44
• In fact the monotonicity property holds for $|t|>5/4$ but fails for $|t|\leq 1$. See math.ca/10.4153/CMB-2010-107-8 Feb 25, 2012 at 0:59

This is a continuation of the argument above, which completes the argument.

Let $C_n$ denote the square with vertices $[n \pm 1/2, \pm 4 I]$ for a positive integer $n$. We have the following inequalities for $z \in C_n$ and $n \ge 15$: $$|\sin(\pi z)| \ge 1, \quad z \in C_n.$$ $$|\Gamma(z)| \ge \frac{1}{2} \Gamma(n - 1/2),$$ $$|\Gamma(1-z)| \le \frac{\pi}{\Gamma(n - 1/2)} \le 1,$$ $$|\psi(1-z)|, |\psi(z)| \le 2 \log(n),$$

The first is easy, the second follows from Stirling's formula (this requires $n$ to be big enough, and also requires $z$ to have imaginary part at most $4$), the third follows from the previous two by the reflection formula for $\Gamma(z)$, the last follows by induction and by the formula $\psi(z+1) = \psi(z) + 1/z$. It follows that $$\left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\Gamma'(z)}{\Gamma(z)} - \frac{d/dz (\Gamma(z) + \theta \cdot \Gamma(1-z))}{\Gamma(z) + \theta\cdot \Gamma(1-z)} \right|$$ $$= \left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\theta \Gamma(1-z) (\psi(1-z) + \psi(z))} {\Gamma(z) + \theta \cdot \Gamma(1-z)} \right|$$ $$\le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \oint_{C_n} \frac{1} {|\Gamma(z) + \theta \cdot \Gamma(1-z)|}$$ $$\le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \cdot \frac{1}{1/2 \Gamma(n - 1/2) + 1} \ll 1,$$ where $\theta = \pm 1$ (or anything small) and $n \ge 15$, where the final inequality holds by a huuuge margin. It follows that $\Gamma(z) + \theta \cdot\Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the number of poles in $C_n$. Since $\Gamma(z)$ has no zeros and poles in $C_n$, it follows that $\Gamma(z) + \theta\cdot\Gamma(1-z)$ has the same number of zeros and poles. It has exactly one pole, and thus exactly one zero. If $\theta = \pm 1$ (and so in particular is real), by the Schwarz reflection principle, this zero is forced to be real. By symmetry, the same argument applies in the region $z = s + i t$ with $|t| \le 4$ and $s \le -15$. Combined with the above argument, this reduces the claim to $z = s + i t$ with $|s| \le 15$ and $|t| \le 4$ where the claim can be checked directly.

Hence all the zeros are either in $\mathbf{R}$, or lie on the line $1/2 + i \mathbf{R}$.

EDIT To clarify, I didn't actually check that there were no exceptional'' zeros in the box $\pm 15 \pm 4 I$, since I presumed that the original poster had done so. If $F(z) = \Gamma(z) - \Gamma(1-z)$, then computing the integral $$\frac{1}{2 \pi i} \oint \frac{F'(z)}{F(z)} dz$$ around that box, one obtains (numerically, and thus exactly) $1$. There are (assuming the OP at least computed the critical line zeros correctly) $2$ zeros in that range on the critical line. Along the real line in that range, there are $30$ poles and $25$ zeros. This means that there must be $1 + 30 - 25 = 6$ unaccounted for zeros. For such a zero $\rho$ off the line, by symmetry one also has $\overline{\rho}$, $1 - \rho$ and $1 - \overline{\rho}$ as zeros. Hence there must be either $1$ or $3$ pairs of zeros on the critical line, and either $1$ or $0$ quadruples of roots off the line. Varying the parameters of the integral, one can confirm there is a zero with $\rho \sim 2.7 + 0.3 i$, which is one of the four conjugates of the root found by joro. A similar argument applies for $\Gamma(z)+\Gamma(1-z)$. Hence:

Any zero of $\Gamma(z) - \Gamma(1-z)$ is either in $\mathbf{R}$, on the line $1/2 + i \mathbf{R}$, or is one of the four exceptional zeros $\{\rho,1-\rho,\overline{\rho},1-\overline{\rho}\}$. A similar calculation implies the same for $\Gamma(z) + \Gamma(1-z)$, except now with an exceptional set $\{\mu,1-\mu,\overline{\mu},1-\overline{\mu}\}$.

• Wonderful. You can simplify and strengthen the proof by using a generalized Rouché's theorem. This tells us that $\Gamma(z)+\theta\cdot\Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the number of poles in $C_n$ when $|\Gamma(1-z)|<|\Gamma(z)|$ holds on the boundary. This is equivalent to $\pi/|\sin(\pi z)|<|\Gamma(z)|^2$, hence it suffices to have $\pi<|\Gamma(z)|^2$ on $\partial C_n$. It seems that the last inequality holds for $n\geq 5$. Feb 25, 2012 at 0:15
• Very impressive, although I honestly have to say that fully understanding the proof is beyond my math skills. Still got the goosebumps from reading it though :-) The proof does induce two follow up questions: 1) could the function $\Gamma(s)^2 - \Gamma(1-s)^2$ be uniquely represented by an infinite product involving its 'complex' zeros (via Weierstrass factorization)? 2) is there a function for locating the zeros (similar to $Z(t)$ for the Riemann non trivial zeros)? Thanks.
– Agno
Feb 25, 2012 at 0:33
• @Agno: Rouché's theorem is contained in basic textbooks, and this is all you need (actually a slight generalization of it). Using this you can shorten the above proof to a few lines (e.g. no integrals), see my comment above. Feb 25, 2012 at 0:39
• @GH: Rouché? Touché!
– user631
Feb 25, 2012 at 3:19
• @Agno: the logarithmic derivative is the tool to count zeros and it is always available. Mar 6, 2012 at 18:30

I would like to expand on Guild of Pepperers's answer by noting that the zeros are essentially uniformly spaced and may easily be approximated to a high degree of accuracy. Using Stirling approximation, I obtained the formula $$\Gamma\left(\frac12+it\right) = \sqrt{\frac{2\pi}{1+e^{-2\pi|t|}}}\exp\left(-\frac\pi2|t|+i(t\log|t|-t+\varepsilon(t))\right),$$ valid for real $t$, where the error $\varepsilon(t)$ is an odd, bounded, real-valued function asymptotically equal to $\frac{1}{24t}$. (Indeed, $\varepsilon(t)$ has asymptotic and convergent expansions coming from the asymptotic and convergent versions of Stirling approximation, respectively.) We then have, for $s = \frac12+it$ on the critical line, $$\Gamma(s)+\Gamma(1-s) = 2\sqrt{\frac{2\pi}{1+e^{-2\pi|t|}}}e^{-\frac\pi2|t|}\cos\left(t\log|t|-t+\varepsilon(t)\right),$$ $$\Gamma(s)-\Gamma(1-s) = 2\sqrt{\frac{2\pi}{1+e^{-2\pi|t|}}}e^{-\frac\pi2|t|}\sin\left(t\log|t|-t+\varepsilon(t)\right).$$ One may show by means fair or foul that $t\log|t|-t+\varepsilon(t)$ is monotonically increasing for $|t|\geq1.05$, is bounded between $-0.96$ and $0.96$ for $|t|<1.05$, and is only zero when t = 0. Therefore, the zeros of $\Gamma(s)+\Gamma(1-s)$ on the critical line occur, with multiplicity one, very near those $t$ for which $t\log|t|-t$ is an odd integer multiple of $\frac{\pi}{2}$, and similarly for $\Gamma(s)-\Gamma(1-s)$ and the even integer multiples of $\frac{\pi}{2}$.

It's interesting that the number of zeros up to a given height $T$ is of the same order of magnitude, $T \log(T)$, as for the Riemann zeta function, but that these zeros have (essentially) uniform spacings rather than GUE spacings.

Just to point out there are very good approximation to complex zeros off your line of $$\Gamma(s)-\Gamma(1-s) \qquad(1)$$

At $\rho \approx -1.69711183621729718874218687438 - 0.305228379993226071272967719419 i$ (1) appears to vanish while $\Gamma(\rho) \approx 1.4648039 + 0.3642699441 i$

Root finding with better precision converges to $\rho$ while (1) still appear to vanish in both sage and gp/pari (modulo bugs).

Checked to precision $5000$ digits and (1) still appears to vanish.

Here is $\rho$ with $100$ digits of precision:

-1.697111836217297188742186874382163077146364585981726518217373889827452772242797069678994954785699956 - 0.3052283799932260712729677194188512919331197338088909477524842921187943642970297308885952936796125572*I


... for $\Gamma(s)+\Gamma(1-s)$ approximation of zero appears $\rho \approx -0.60940537628997711023 - 0.82913081575572747216 i$ checked to $5000$ digits of precision.

With 100 digits:

 -0.6094053762899771102337308158313839002012166649163876907688596366808893391382113824494098816671945331 - 0.8291308157557274721587141536678087800797120641344787653174391388417832472543392187032283839972409848*I


Edit In comments juan suggested using x-ray to investigate the zeros.

The primary reference for x-ray I found is X-Ray of Riemann zeta-function, J. Arias-de-Reyna

AFAICT x-ray are the plots of Re(f(s))=0 and Im(f(s))=0. The zeros are the intersection.

The x-ray and juan's comments suggest the above quadruples of zeros are indeed zeros off $\frac12$ and possibly there are no more complex zeros zeros off the line.

Here is the x-ray of $\Gamma(s)-\Gamma(1-s)$. Blue is $\Re(\Gamma(s)-\Gamma(1-s))=0$ and red is $\Im(\Gamma(s)-\Gamma(1-s))=0$. (source) (source)

• @Joro, you are the "Master of the Counter Example" :-) As you pointed out, the zeros of $\Gamma(s) \pm \Gamma(1-s)$ are very small, so precision of the calculation can be an issue, however with 5000 digits accuracy, your two counter examples could also easily fall in the category: (...)this reduces the claim to z=s+it with |s|≤15 and |t|≤4 where the claim can be checked directly(...). If your counter examples are correct, then my only escape is to restrict the claim to the critical strip only (similar to the $\zeta(s)$ and $\zeta^{(k)}(s)$ equivalents).
– Agno
Apr 9, 2012 at 11:04

Since we have

$\Gamma(1/2+it)=\sqrt{\pi/\cosh(\pi t)}\exp[i(2 \vartheta(t)+t \log(2\pi)+\arctan(\tanh(\pi t/2)))]$

where $\vartheta(t)$ is the Riemann Siegel function. The zeros on the critical line have ordinates the zeros of

the cosine or sine of the real function

$2 \vartheta(t)+t \log(2\pi)+\arctan(\tanh(\pi t/2))$

But there are real zeros, for example there is one at $s = 4.0260426340124070065475\dots$

X-ray: • juan, how do you explain the approximations of zeros in my answer with precision 100 checked in pari, sage and maple? Have I misunderstood the question?
– joro
Apr 9, 2012 at 9:26
• With mpmath I check also this zero. I think now that my parenthesis "(with some effort)" contains an error. :-/ What it is clear is that the real zeros are those of the function given. But the behavior of this function for t complex is really not simple. I will try to do an X-ray of this function. Later we will try to post it. If I know how to do it.
– juan
Apr 9, 2012 at 19:39
• Apparently I can not post here a plot. The x-ray gives little doubt that the zeros of $\Gamma(s)-\Gamma(1-s)$ are the ones with real part $1/2$ that appeared computed in this question. The complex at $-1.69-0.30 i$ its complex conjugate the symmetrical of this with respect the critical line $2.69+0.30i$ and its complex conjugate. And then the real zeros one at $0.5$. The others real zeros can be obtained best from a real plot of the function. In the x-ray this zeros, that are very near the poles at $4$, $5$, $\dots$, can not be seen since they are contained in very short lines.
– juan
Apr 10, 2012 at 16:11
• Of course the real zeros are symmetric with respect to 0.5 so that there are zeros near $-3$, $-4$, $\dots$
– juan
Apr 10, 2012 at 16:24
• juan, you can post images (possibly unless you don't have enough reputation). The format is HTML, i.e. write <img src="server/file.gif"> in your answer. The x-rays i found are at drememi.ludost.net/gamma1.png and drememi.ludost.net/gamma2.png
– joro
Apr 11, 2012 at 5:19

Given that $\Gamma(s)$ and $\Gamma(1-s)$ are complex conjugates when $\Re(s)=1/2$, it is not surprising that $\Gamma(s)+\theta\Gamma(1-s)$ has an infinitude of zeros on the line $\Re(s)=1/2$, as long as $|\theta|=1$. The monotonicity argument given in the first answer then shows that there are no other zeros with $0<\Re(s)<1$. With the possible exception when the imaginary part of $s$ is small, the zeros for two different $\theta$ should interlace (if $\theta$ goes around the unit circle once, a zero is carried to an adjacent zero).