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Feb 24, 2015 at 15:42 history edited Antonio Vargas CC BY-SA 3.0
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Feb 24, 2015 at 10:49 vote accept Olivier Oloa
Feb 24, 2015 at 10:47 comment added Olivier Oloa Thank you Antonio. +1 and accepted. There is a typo in your result, the factor $a-b$ should appear in your expansion. Observe that, using the notations in my previous comment, we get $$\gamma_0(a,b)=-\psi(b+1)$$ due to the classic evaluation $$\sum_{n=1}^\infty \frac{a-b}{(n+a)(n+b)} =\psi(a+1)-\psi(b+1).$$ I'm looking for an 'interesting' evaluation for $\gamma_1(a,b)$.
Feb 24, 2015 at 4:20 history edited Antonio Vargas CC BY-SA 3.0
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Feb 24, 2015 at 4:12 review First posts
Feb 24, 2015 at 4:17
Feb 24, 2015 at 4:12 history answered Antonio Vargas CC BY-SA 3.0