# Limit representation of Euler-Mascheroni constant

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
We have the following functional equation of digamma

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

It is then readily seen that

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

Prove the following

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

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#### ZaidAlyafey

##### Well-known member
MHB Math Helper

Prove that

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

#### chisigma

##### Well-known member

Prove that

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

Kind regards

$\chi$ $\sigma$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper

Kind regards

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

#### Random Variable

##### Well-known member
MHB Math Helper
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...epresentation-riemann-zeta-function-6398.html

Therefore,

$$\lim_{z \to 1} \Big( \zeta(s) - \frac{1}{z-1} \Big) = 2 \int_{0}^{\infty} \frac{t}{t^2+1} \frac{1}{e^{2 \pi t}-1}\ dt + \frac{1}{2}$$

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

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

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Hey RV , You are solving all my challenges . I should post extremely difficult questions . Here is a relatively elementary proof.

#### chisigma

##### Well-known member
Well, I won't count that as a solution . I already saw that but how would you prove that $$\displaystyle \gamma_0\equiv \gamma$$ ?... 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

$\chi$ $\sigma$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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 $$\displaystyle z=1$$ ?

$$\displaystyle \zeta(z)=\frac{1}{z-1}+\sum_{n=0}^\infty \frac{(-1)^n\, }{n! }\gamma_n(z-1)^n$$

#### chisigma

##### Well-known member
If we start by this definition , then how to prove the expansion of zeta function around the singularity $$\displaystyle z=1$$ ?

$$\displaystyle \zeta(z)=\frac{1}{z-1}+\sum_{n=0}^\infty \frac{(-1)^n\, }{n! }\gamma_n(z-1)^n$$

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...

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

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

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Hey chisigma , do you have a link or original proof of Stieltjes. Seems interesting for me .