RELATION BETWEEN ZETA FUNCTION AND EULER MASCHEORNI CONSTANT

In this post we will show that


\lim_{s \to 1}\left(\zeta(s)-\frac{1}{s-1}\right)=\gamma(1)


In order to prove (1) we can use the Euler Maclaurin formula derived previously in this post.


The first order Euler Maclaurin Formula is given by:


\sum_{k=m}^nf(k)=\frac12\left(f(m)+f(n)\right)+\int_m^n f(x)\,dx+\int_m^n f^\prime(x)P_1(x)\,dx(2)


Where P_1(x)=x-\lfloor x \rfloor -\frac12, is the first Bernoulli Polynomial.


Letting f(x)=\frac{1}{x^s}  and  m=1, n=\infty in (2) we obtain


\zeta(s)=\frac{1}{s-1}+\frac12-s\int_1^\infty \frac{P_1(x)}{x^{s+1}}\,dx(3)


\zeta(s)-\frac{1}{s-1}=\frac12+\frac{s}2\int_1^\infty \frac{dx}{x^{s+1}} -s\int_1^\infty \frac{x-\lfloor x \rfloor}{x^{s+1}}\,dx(4)


Lets now use (4) to prove (1)


Proof:


\begin{aligned} \lim_{s \to 1}\left(\zeta(s)-\frac{1}{s-1}\right)&=\frac12+\frac{1}2\int_1^\infty \frac{dx}{x^{2}}-\int_1^\infty \frac{x-\lfloor x \rfloor}{x^{2}}\,dx\\ &=\frac12+\frac{1}2-\int_1^\infty \frac{x-\lfloor x \rfloor}{x^{2}}\,dx\\ &=1-\int_1^\infty \frac{x-\lfloor x \rfloor}{x^{2}}\,dx\\ &=1-\lim_{N \rightarrow \infty}\left(\int_1^{N} \frac{1}{x}\,dx-\int_1^{N}\frac{\lfloor x \rfloor}{x^2} \,dx \right)\\ &=1-\lim_{N \rightarrow \infty}\left(\ln(N)-\sum_{k=1}^{N-1} k\,\int_k^{k+1} \frac{1}{x^2}\,dx \right)\\ &=1-\lim_{N \rightarrow \infty}\left(\ln(N)-\sum_{k=1}^{N-1} k \left(\frac{1}{k}-\frac{1}{k+1} \right)\right)\\ &=1-\lim_{N \rightarrow \infty}\left(\ln(N)-\sum_{k=1}^{N-1} k \left(\frac{k+1-k}{k(k+1)} \right)\right)\\ &=1-\lim_{N \rightarrow \infty}\left(\ln(n)-\sum_{k=1}^{N-1} \left(\frac{1}{k+1} \right)\right)\\ &=1-\lim_{N \rightarrow \infty}\left(\ln(N)-\sum_{k=2}^{N} \frac{1}{k} \right)\\ &=1-\lim_{N \rightarrow \infty}\left(\ln(N)-\sum_{k=1}^{N} \frac{1}{k} +1\right)\\ &=1-1-\lim_{n \rightarrow \infty}\left(\ln(N)-H_N\right)\\ &=\gamma \qquad \blacksquare \end{aligned}

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