# Gamma function formula

For $s\in\mathbb C,~\Re(s)>1$, then we have \begin{eqnarray} \frac1{\Gamma(s)}\int^{+\infty}_0T^{s-1}\left(\frac{\cosh(T)}{\sinh(T)}-1\right)dT&=&2^{1-2}\zeta(s).\\ \frac1{\Gamma(s)}\int^{+\infty}_0T^{s-1}\left(\frac{\cosh(T)}{\sinh(T)}-1\right)dT &=&\frac1{\Gamma(s)}\int^1_0T^{s-1}\left(\frac{\cosh(T)}{\sinh(T)}-\frac1T\right)dT\\ &&+\frac1{\Gamma(s)}\int^{+\infty}_1T^{s-1}\left(\frac{\cosh(T)}{\sinh(T)}-1\right)dT\\ &&-\frac1{\Gamma(s+1)}+\frac1{\Gamma(s)(s-1)}. \end{eqnarray}

Q: From the above two formulas, how to get the following $$\int^1_0\left(\frac{\cosh(T)}{\sinh(T)}-\frac1T\right)\frac{dT}T+\int^{+\infty}_1\left(\frac{\cosh(T)}{\sinh(T)}-1\right)\frac{dT}T+\Gamma'(1)-1=-2\log(2)\zeta(0)+2\zeta'(0)$$

• I imagine this can be done by expanding a geometric series, integrating term by term, and simplifying with the zeta function. Just a quick sketch of an idea. – Simply Beautiful Art Feb 28 '17 at 12:31

For $Re(s) > 1$ : $\frac{\cosh(x)}{\sinh(x)}-1 = \frac{2e^{-2x} }{1-e^{-2x}} = 2\sum_{n=1}^\infty e^{-2nx}$ so $$\int_0^\infty (\frac{\cosh(x)}{\sinh(x)}-1) x^{s-1} dx = 2 \sum_{n=1}^\infty \int_0^\infty e^{-2nx}x^{s-1}dx$$ $$= 2^{1-s} \sum_{n=1}^\infty n^{-s} \int_0^\infty e^{-x}x^{s-1}dx =2^{1-s} \zeta(s) \Gamma(s)$$ and $$\int_0^\infty (\frac{\cosh(x)}{\sinh(x)}-1_{x > 1}-\frac{1_{x < 1}}{x}) x^{s-1} dx =2^{1-s} \zeta(s) \Gamma(s)+\int_0^1 (1-\frac{1}{x}) x^{s-1}dx$$ $$= 2^{1-s} \zeta(s) \Gamma(s)+ \frac{1}{s}-\frac{1}{s-1}$$ converges for $Re(s) > -1$. Thus, with $F(s) = 2^{1-s} \zeta(s) \Gamma(s+1), F(0) = -1$ : $$\int_0^\infty (\frac{\cosh(x)}{\sinh(x)}-1_{x > 1}-\frac{1_{x < 1}}{x}) \frac{ dx }{x} = \lim_{s \to 0}\int_0^\infty (\frac{\cosh(x)}{\sinh(x)}-1_{x > 1}-\frac{1_{x < 1}}{x}) x^{s-1} dx$$ $$= \lim_{s \to 0}2^{1-s} \zeta(s) \Gamma(s)+ \frac{1}{s}-\frac{1}{s-1} = -1+\lim_{s \to 0}\frac{1}{s}(2^{1-s} \zeta(s) \Gamma(s+1)+ 1)$$ $$= -1+ F'(0) = -1+ \log 2 + 2\zeta'(0)- \Gamma'(1)$$