For $\Re(s) > 1$, $$\frac{\zeta(s)}{s} = \int_1^\infty (\sum_{1 \le n \le x} 1) x^{-s-1}dx, \qquad \frac{\zeta(s)}{s-1} = \int_1^\infty (x\sum_{1 \le n \le x} \frac1n) x^{-s-1}dx$$ Thus for $\Re(s) \in (0,1)$ $$\frac{\zeta(s)}{s}+\frac{\zeta(s)}{1-s}= \int_0^\infty ((\sum_{1 \le n \le x} 1-x)-x(\sum_{1 \le n \le x} \frac1n-\log x-\gamma)) x^{-s-1}dx$$

So that $\frac{\zeta(1/2+it)}{1/4+t^2}$ is the Fourier transform of $$g(u) = ((\sum_{1 \le n \le e^u} 1)-e^u-e^u(\sum_{1 \le n \le e^u} \frac1n) + e^u u+e^u\gamma)e^{-u/2}$$ and $$\int_{-\infty}^\infty \frac{\zeta(1/2+it)}{1/4+t^2} dt = 2\pi\lim_{u \to 0} \frac{g(u)+g(-u)}{2}=2\pi( \gamma-1)$$

The integral converges absolutely because $\zeta(1/2+it) = O(t^c),c < 1$, and for the same reason you can shift the contour to $1/2+i(-\infty,+\infty)$ to $+\infty+i(-\infty,+\infty)$) and evaluate the integral as $-2 \pi \text{Res}(\frac{\zeta(s)}{s(1-s)},1)$.

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