1
$\begingroup$

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 $$\int^1_0\left(\frac{\cosh(T)}{\sinh(T)}-\frac1T\right)\frac{dT}T+\int^{+\infty}_1\left(\frac{\cosh(T)}{\sinh(T)}-1\right)\frac{dT}T+\Gamma'(1)-1=-2\log(2)\zeta(0)+2\zeta'(0)$$

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Simply Beautiful Art Feb 28 '17 at 12:31
1
$\begingroup$

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 $F(s) = 2^{1-s} \zeta(s) \Gamma(s+1), F(0) = -1$ : $$ \int_0^\infty (\frac{\cosh(x)}{\sinh(x)}-1_{x > 1}-\frac{1_{x < 1}}{x}) \frac{ dx }{x} = \lim_{s \to 0}\int_0^\infty (\frac{\cosh(x)}{\sinh(x)}-1_{x > 1}-\frac{1_{x < 1}}{x}) x^{s-1} dx$$ $$ = \lim_{s \to 0}2^{1-s} \zeta(s) \Gamma(s)+ \frac{1}{s}-\frac{1}{s-1} = -1+\lim_{s \to 0}\frac{1}{s}(2^{1-s} \zeta(s) \Gamma(s+1)+ 1)$$ $$ = -1+ F'(0) = -1+ \log 2 + 2\zeta'(0)- \Gamma'(1) $$

$\endgroup$
  • $\begingroup$ (or something like that) $\endgroup$ – reuns Mar 2 '17 at 5:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.