Show that $\int_0^{\infty} \frac{e^{2 \pi i \nu u}}{u^s} du =\Gamma(1-s) (\frac{2 \pi \nu}{i} )^{s-1}$

Question

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.

Answer 1

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 :)

Answer 2

When $\Re(\alpha)>0$ is fixed and $\beta>0$, a change of variable gives

$$ \int_0^{+\infty}u^{\alpha-1}e^{-\beta u}\mathrm du=\Gamma(\alpha)\beta^{-\alpha}. $$

Observe that the integral on the left hand side converges uniformly in any compact subset of $\Re(\beta)>0$ and any compact subset of $i\mathbb R\setminus\{0\}$, and the term on the right hand side is a one-valued analytic function in $A=\{\beta\in\mathbb C:\Re(\beta)\ge0\}\setminus\{0\}$, we conclude that this identity is valid in $A$ via analytic continuation.

Now, set $\alpha=1-s$ and $\beta=-2\pi\nu i$ so that we arrive at the following:

$$ \int_0^{+\infty}u^{-s}e^{2\pi i\nu u}\mathrm du=\Gamma(1-s)\left(2\pi\nu\over i\right)^{s-1}. $$