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The Riemann's Functional Equation as on wikipedia here:

Riemann's Functional Equation

describes the zeta function for all complex plane. But all integers are also in the complex plane. Thus,we can substitute for s=1 on the Left Hand Side of above equation.

If we substitute on the Left hand side s=1,

the Right hand side becomes:

= 21 * π0 * sin(π/2) * Factorial(0) * E(0)

= 2 * 1 * (1) * 1 * (-1/2)

= -1

Thus, E(1)=-1

Is this correct derviation or is there a problem in this with respect to meromorphic function theory related issue?

vinaych
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1 Answers1

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$$\zeta(s) \propto \Gamma(1-s) \implies \zeta(1) \propto \Gamma(0)$$

$\Gamma(z)$ does not exist for $z=0,-1,-2,\cdots$. This can be shown easily from the recursion property: $\Gamma(z+1)=z \Gamma(z)$. Let $z=0$ and solve for $\Gamma(0)$ to find a problem.

Note that $\Gamma(z+1) = z!$ when $z$ is a nonnegative integer, by the way. So if anything, $\Gamma(0) = (-1)!$ (were that to be defined).

In short, your derivation is incorrect and in turn pokes no holes in existing theory.

PrincessEev
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  • Sorry I completely misunderstood the gamma function, thinking it to be factorial of a number generator simply. But then if we go with this definition then E(0)=0 , as: R.H.S=2^0 * (1/pi) * 0 * 1 * E(1) = 0 , which disproves Riemann's Hypothesis! – vinaych Feb 12 '20 at 23:03
  • The gamma function was designed to be a function such that, for positive integer $n$ as an input, it gives you $n!$ as a result, and its definition lets you extend that notion to almost every complex and real number except nonpositive integers. $\Gamma(z+1) = z!$ by its construction. If you want a slightly more intuitive function for this respect, I refer you to the pi function, $\Pi(z)$, which has the property for nonnegative integers $\Pi(z)=z!$. – PrincessEev Feb 12 '20 at 23:07
  • As for the RHS, $\Gamma(0)$ is inherently undefined. There is no contradiction or problem you're introducing. You might as well say $1/0 = \pi$, it's equally inconsistent with mathematics. Truthfully, if you think disproving the Riemann hypothesis were so easy, wouldn't it have already been done decades ago and not been worthy of a million dollar prize? – PrincessEev Feb 12 '20 at 23:08
  • To reiterate that $\Gamma(0)$ is undefined, note that the recursion property of $\Gamma$, one of its defining properties, gives you

    $$\Gamma(z+1) = z \Gamma(z) \overset{?}{\implies} \Gamma(0) = \frac{\Gamma(1)}{0} = \frac 1 0$$

    Surely the problem is obvious.

     – PrincessEev Feb 12 '20 at 23:10
  • When I referred the ζ(0) derivation above, the Γ(0) does not come into scene only Γ(1-s) that is Γ(1)=1 comes into substitution above. I still am doubtful though. – vinaych Feb 12 '20 at 23:15