This is adapted from 'Analytic Number Theory' by Iwaniec and Kowalski. I think the result is off by a sign but I am not sure.
Let G(z) be an even holomorphic function bounded on $|Re(z)| < 4$ with $G(0) = 1$. Define
$$ \Lambda (s) = \pi^{-s/2}\Gamma(s/2)\zeta(s) $$
Then, by the functional equation $\Lambda(s) = \Lambda(1-s)$. Now for some $X>0$ define
$$ I(X,s) = \frac{1}{2\pi i}\int_{(3)} X^{u}\Lambda(s+u) G(u) \frac{du}{u} $$
where the integral is along the vertical line with real part 3, going from $3-i\infty$ to $3+i\infty$.
We now make the contour bounded by $\int_{(3)}$ and $\int_{(-3)}$. Using the functional equation and a change of variables,
$$ \int_{(-3)} X^{u} \Lambda(s+u) G(u) \frac{du}{u} = -I(X^{-1}, 1-s) $$
The poles inside the contour are at $u=0$ (given by the factor of $\frac{1}{u}$), $u=-s$ (given by $\Gamma (0)$) and $u=1-s$ (given by $\zeta(1)$). The residue at $u = 0$ is $\Lambda(s)$. Let $R$ be the sum of the residues at $-s$ and $1-s$.
The book claims that
$$ \Lambda(s) = I(X,s) + I(X^{-1}, 1-s) + R $$
But shouldn't it be the following?
$$ \Lambda(s) + R = I(X,s) + I(X^{-1}, 1-s) $$
That is, the residues having the same sign since they both have the same winding number with respect to this contour.