In Theorem 4.13. in Titchmarsh's book The Theory of the Riemann Zeta-Function the following equality is assumed without proof: $$\displaystyle\int_0^{\infty} \dfrac{e^{2 \pi i \nu u}}{u^s} du = \Gamma(1-s) \bigg(\dfrac{2 \pi \nu}{i} \bigg)^{s-1}.$$ How is this derived?
Writing the integral definition of $\Gamma(1-s)$ I couldn't reach from RHS to LHS. Neither by change of variable nor by $\Gamma(s) e^{\frac{i \pi s}{2}} = \int_0^{\infty} y^{s-1} e^{iy} dy$ (Rademacher's book), from LHS to RHS is also not achievable.
Best Answer
You change the path of integration to the upper vertical axis. To do this consider the contour going from $\epsilon$ to $R$ then from $R$ to $iR$ around the arc in the positive direction, then from $iR$ to $i\epsilon$ and from there around the arc to $\epsilon$ in the negative direction. By Chauchy's theorem the integral around this contour is zero. I think you can take it from here :)