Just to make sure I understand. The residue at $s=\frac{3}{2}$ is given by $I_f = \int_{\mathbb{R}_+^2} f(x,y) dx dy$, and at $s=\frac{1}{2}$ by $c_0=0$ since $f(0,0)=0$?

Me neither. Seems perfectly right. I will calculate the second term and maybe in the end they fit together properly. Must be so since the final result equals half the A-genus.

Thank you for that concise and precise answer. Unfortunately that is not the result as in the paper. Putting $a=\frac{\lambda^2}{2}$ and multiply by a factor 2 should give the value noted in the paper but it doesn't. Although, I came to the same computation and so does WolframAlpha. Did I miss something silly?

Ehm. Yeah. Silly me. It's always the sign. :D Thank you!

Of course. Linked it. It's on page 34.

There appears a $\zeta(1-s,a)$ which is not finite at $s=0$. :-/