The formula $$\zeta(s) = - \dfrac{e^{i \pi s}}{2 \pi i} \int_{c - i \infty}^{c+ i \infty} \bigg( \dfrac{\Gamma'(1+z)}{\Gamma(1+z)} - \log z \bigg) z^{-s} dz \ \ \ (*)$$ can be proved by the calculus of residues for $\sigma > 1$ and the formula holds for $\sigma > 1$ by analytic continuation.
My aim is prove Eq 2.9.2. of Titchmarsh's book The Theory of the Riemann Zeta-Function which states $$\zeta(s) = - \dfrac{\sin \pi s}{\pi} \int_0^{\infty} \bigg( \dfrac{\Gamma'(1+x)}{\Gamma(1+x)} - \log x \bigg) x^{-s} dx.$$
We consider two counters $C$ and $D$. $C=C_1 + C_2 + C_3 + C_4$: $C_1$ is the line from $\infty+i0$ to $\rho+i0$; $C_2$ is the arc from $\rho+i0$ to $\rho e^{\frac{3 \pi i}{4}}$; $C_3$ is the line from $\rho e^{\frac{3 \pi i}{4}}$ to $\rho e^{\frac{3 \pi i}{4}} + i \infty$; and $C_4$ is the arc from $\rho e^{\frac{3 \pi i}{4}} + i \infty$ to $\infty+i0$. Integration on $C$ is zero. $D=D_1 + D_2 + D_3 + D_4$: $D_1$ is the line from $\rho+i0$ to $\infty+i0$; $D_2$ is the arc from $\infty+i0$ to $\rho e^{\frac{-3 \pi i}{4}} - i \infty$; $D_3$ is the line from $\rho e^{\frac{-3 \pi i}{4}} - i \infty$ to $\rho e^{\frac{-3 \pi i}{4}}$; and $D_4$ is the arc from $\rho e^{\frac{-3 \pi i}{4}}$ to $\rho+i0$. Integration on $D$ is also zero.
Summing $C$ and $D$ and letting $\rho \to 0$ gives $$\zeta(s) : = - \dfrac{e^{i \pi s}}{2 \pi i} \int_{c - i \infty}^{c+ i \infty} \bigg( \dfrac{\Gamma'(1+z)}{\Gamma(1+z)} - \log z \bigg) z^{-s} dz = - \dfrac{e^{i \pi s}}{2 \pi i} \int_0^{\infty} \bigg( \dfrac{\Gamma'(1+x)}{\Gamma(1+x)} - \log x \bigg) x^{-s} dx - \dfrac{e^{i \pi s}}{2 \pi i} \int_{\infty}^0 \bigg( \dfrac{\Gamma'(1+xe^{2 \pi i})}{\Gamma(1+xe^{2 \pi i})} - \log xe^{2 \pi i} \bigg) (xe^{2 \pi i})^{-s} d(xe^{2 \pi i}) = - \dfrac{e^{i \pi s}}{2 \pi i} \int_0^{\infty} \bigg( \dfrac{\Gamma'(1+x)}{\Gamma(1+x)} - \log x \bigg) x^{-s} dx - \dfrac{e^{i \pi s}}{2 \pi i} \int_{\infty}^0 \bigg( \dfrac{\Gamma'(1+xe^{2 \pi i})}{\Gamma(1+xe^{2 \pi i})} - \log x - \log e^{2 \pi i} \bigg) (xe^{2 \pi i})^{-s} d(xe^{2 \pi i}) = - \dfrac{\sin \pi s}{\pi} \int_0^{\infty} \bigg( \dfrac{\Gamma'(1+x)}{\Gamma(1+x)} - \log x \bigg) x^{-s} dx - \dfrac{e^{i \pi s}}{2 \pi i} \int_0^{\infty} (- 2 \pi i) (xe^{2 \pi i})^{-s} dx.$$
How that the last term above is zero so that we conclude validity of Eq 2.9.2. ?