MHB Limit representation of Euler-Mascheroni constant

[alyafey22]

We have the following functional equation of digamma

$$\psi(x+1)-\psi(x)=\frac{1}{x}$$

It is then readily seen that

$$-\gamma= \lim_{z\to 0} \left\{ \psi(z) +\frac{1}{z} \right\}$$
​

Prove the following

$$-\gamma = \lim_{z \to 0} \left\{ \Gamma(z) -\frac{1}{z} \right\}$$​

 

[alyafey22]

Additional exercise

Prove that

$$\gamma =\lim_{z \to 1}\left\{\zeta(z)-\frac{1}{z-1} \right\}$$​

 

[chisigma]

Additional exercise

Prove that

$$\gamma =\lim_{z \to 1}\left\{\zeta(z)-\frac{1}{z-1} \right\}$$​

Spoiler

Stieltjes Constants -- from Wolfram MathWorld

Kind regards

$\chi$ $\sigma$

 

[alyafey22]

Spoiler

Stieltjes Constants -- from Wolfram MathWorld

Kind regards

$\chi$ $\sigma$

Well, I won't count that as a solution (Nerd). I already saw that but how would you prove that $$\gamma_0\equiv \gamma$$ ? There is an analytic way to prove the above limit.

 

[polygamma]

Spoiler

The Taylor expansion of $\Gamma(z+1)$ at the origin is

$$ \Gamma(z+1) = \Gamma(1) + \Gamma'(1) z + \mathcal{O}(z^{2}) = 1- \gamma z + \mathcal{O}(z^{2}) $$

$$ \implies \Gamma(z) = \frac{1}{z} - \gamma + \mathcal{O}(z) $$Therefore,

$$ \lim_{z \to 0} \left( \Gamma(z) -\frac{1}{z} \right) = \lim_{z \to 0} \Big( - \gamma + \mathcal{O}(z) \Big) = - \gamma $$

For all complex values $z$, the Riemann zeta function has the integral representation $$ \zeta(s) = 2 \int_{0}^{\infty} \frac{\sin (z \arctan t)}{(1+t^{2})^{z/2} (e^{2 \pi t} - 1)} \ dt + \frac{1}{2} + \frac{1}{z-1} $$

http://mathhelpboards.com/challenge-questions-puzzles-28/another-integral-representation-riemann-zeta-function-6398.htmlTherefore,

$$ \lim_{z \to 1} \Big( \zeta(s) - \frac{1}{z-1} \Big) = 2 \int_{0}^{\infty} \frac{t}{t^2+1} \frac{1}{e^{2 \pi t}-1}\ dt + \frac{1}{2}$$Differentiating Binet's log-gamma formula,

Binet's Log Gamma Formulas -- from Wolfram MathWorld $$-2 \int_{0}^{\infty} \frac{t}{t^2+z^{2}} \frac{1}{e^{2 \pi t}-1}\ dt = \psi(z) -\log z -1 + \frac{1}{2z} + 1 $$

$$ \implies 2 \int_{0}^{\infty} \frac{t}{t^2+1} \frac{1}{e^{2 \pi t}-1}\ dt = -\psi(1) - \frac{1}{2} = \gamma - \frac{1}{2}$$So

$$ \lim_{z \to 1} \Big( \zeta(s) - \frac{1}{z-1} \Big)= \gamma$$

 

[alyafey22]

Hey RV , You are solving all my challenges . I should post extremely difficult questions . Here is a relatively elementary proof.

 

[chisigma]

Well, I won't count that as a solution (Nerd). I already saw that but how would you prove that $$\gamma_0\equiv \gamma$$ ?... There is an analytic way to prove the above limit.

The Stieltjes Constants are defined as follows...

$\displaystyle \gamma_{n} = \lim_{m \rightarrow \infty} (\sum_{k=1}^{m} \frac{\ln^{n} k}{k} - \frac{\ln^{n+1} m} {n + 1})\ (1)$

... so that for n=0 is...

$\displaystyle \gamma_{0} = \lim_{m \rightarrow \infty} (\sum_{k=1}^{m} \frac{1}{k} - \ln m) = \gamma\ (2)$

Kind regards

$\chi$ $\sigma$

 

[alyafey22]

The Stieltjes Constants are defined as follows...

$\displaystyle \gamma_{n} = \lim_{m \rightarrow \infty} (\sum_{k=1}^{m} \frac{\ln^{n} k}{k} - \frac{\ln^{n+1} m} {n + 1})\ (1)$

If we start by this definition , then how to prove the expansion of zeta function around the singularity $$z=1$$ ?$$ \zeta(z)=\frac{1}{z-1}+\sum_{n=0}^\infty \frac{(-1)^n\, }{n! }\gamma_n(z-1)^n$$

 

[chisigma]

If we start by this definition , then how to prove the expansion of zeta function around the singularity $$z=1$$ ?$$ \zeta(z)=\frac{1}{z-1}+\sum_{n=0}^\infty \frac{(-1)^n\, }{n! }\gamma_n(z-1)^n$$

About 130 years ago the Dutch Mathematician Thomas Joannes Stieltjes has arrived to the following Laurent series expansion…

$\displaystyle \zeta (s) = \frac{1}{s - 1} + \sum_{n = 0}^{\infty} \frac{\gamma_{n}}{n!}\ (1 - s)^{n}\ (1)$

... where the constants $\gamma_{n}$ are given by...

$\displaystyle \gamma_{n} = \frac{(-1)^{n}\ n!}{2\ \pi}\ \int_{0}^{2\ \pi} e^{- i\ n\ x}\ \zeta (1 + e^{i\ x})\ d x\ (2)$

<sub>... and are obtained applying the Cauchy integral formulas to the circular path that is indicated with $\lambda$ in the figure...

http://www.123homepage.it/u/i82933278._szw380h285_.jpg.jfif

Kind regards

$\chi$ $\sigma$

P.S. because it is not pretty elegant to say 'zeta of zeta' for the Riemann Zeta Function the independent complex variable is usually indicated with the letter s...</sub>

 

[alyafey22]

Hey chisigma , do you have a link or original proof of Stieltjes. Seems interesting for me .

 

[chisigma]

Hey chisigma , do you have a link or original proof of Stieltjes. Seems interesting for me .

Although a little 'old' [1972] this paper supplies a good tutorial presentation of the argument...

https://archive.org/details/jresv76Bn3-4p161

Kind regards

$\chi$ $\sigma$