This is not true, you cannot write the same thing in two different ways and then make the 2 mysteriously disappear. Yeah, I am worried because now we already have two mistakes (typos?) on one page.
Im sure of what Im saying, perhaps I wasnt clear enough. The first derived integral after the Mellin inversion and substitutions z->n*n*pi*z and s->s/2 was: \(\psi (z) = {1 \over 2 \pi i} \int_{c-i\infty}^{c+i\infty}\pi^{{-s \over 2}}\Gamma ({s \over 2}) \zeta (s) z^{{-s \over 2}} \frac{ds}{2}\) substituting \(\pi^{-s/2}\Gamma (s/2) \zeta (s)=\frac{2\xi(s)}{s(s-1)} \) we get \(\psi (z) = {1 \over 2 \pi i} \int_{c-i\infty}^{c+i\infty} \frac{2\xi(s)}{s(s-1)} z^{{-s \over 2}} \frac{ds}{2}= {1 \over 2 \pi i} \int_{c-i\infty}^{c+i\infty} \frac{\xi(s)}{s(s-1)} z^{{-s \over 2}} ds = {1 \over 2 \pi i} \int_{c-i\infty}^{c+i\infty}\pi^{{-s \over 2}}\Gamma ({s \over 2}) \zeta (s) z^{{-s \over 2}} ds\) now you see how the 2 cancelled. and you can see the last residue computation here: http://www.wolframalpha.com/input/?i=residue of pi^(-z/2)*gamma(z/2)*zeta(z) at z=1
From here Depending on normalization, there may or may not be an extra 1/2 in front of the expression.
You asked Wolfram to compute the residue for \(\pi^{{-s \over 2}}\Gamma ({s \over 2}) \zeta (s)\). Aren't you supposed to compute the residue for \(\pi^{{-s \over 2}}\Gamma ({s \over 2}) \zeta (s) z^{{-s \over 2}}\)? You should have stopped at: \(\psi (z) = {1 \over 2 \pi i} \int_{c-i\infty}^{c+i\infty} \frac{\xi(s)}{s(s-1)} z^{{-s \over 2}} ds\) because the residue is very easy to calculate (as you have actually already done at the beginning of the thread. The last equality only confuses things.