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My question is about the proof of Theorem $3.1.3$ given in kubota's book, which shows how the function $\varphi(s)$ appearing in the Fourier expansion of eisenstein series can be continued holomorphically to the region determined by $Re(s)>1/2,$ and $s\notin]1/2;1]$.

So, how the formula $$ (\theta,\bar{\theta_{\zeta}})=\frac{1}{2\pi i}\int_{Re(s)=S_{0}}\frac{h(1-s)h(s)+h(s)^{2}\varphi(s)}{(s-\zeta)(s-(1-\zeta))} ds $$ where $Re(1-\zeta)<S_{0}<Re(\zeta)$, $S_{0}>1$ gives the formula $$ (\theta,\bar{\theta_{\zeta}})=\frac{1}{2\pi i}\int_{Re(s)=S_{1}}\frac{h(1-s)h(s)+h(s)^{2}\varphi(s)}{(s-\zeta)(s-(1-\zeta))} ds-\frac{(h(1-\zeta)h(\zeta)+h(\zeta)^{2}\varphi(\zeta))}{2\zeta-1} $$ where $S_1$ is any real number satisfying $Re(\zeta)<S_1$.

It is not clear for me how Kubota deduces the second formula from the first one (Theorem 3.1.1)!

Thank you.

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    $\begingroup$ Just use Cauchy's residue theorem, picking up a pole at $s = \zeta$. $\endgroup$
    – Peter Humphries
    Jan 22, 2015 at 0:19

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