Consequence of Cauchy integral formula? The below is a theorem from a book and the author says that it is a consequence of Cauchy integral formula, though I am unable to prove it. I took the rectangular contour whose vertices were $\gamma-i \beta, \alpha - i \beta, \alpha + i \beta $ and $\gamma + i \beta$ and used the Cauchy integral formula. Now I would like to let $\alpha \to \infty$ but I am unable to proceed after that. I would greatly appreciate help on this front.

 A: This theorem is not true in general. First of all, the assumption that $\gamma \ne 0$ is weird, because you can always shift the function and consider the variable $w=z+t$ for some real $t$, which will replace $\gamma$ by $\gamma-t$. So if this would be true for some real $\gamma$, it is automatically true for all real $\gamma$. Now if you consider the function $f(z) = e^{-iz}$, $\gamma=0$, and $z_0=1$, then the integral in the formula is (with parametrization $z=it$, $dz=i \, dt$)
$$
\int_{-i\beta}^{i\beta} \frac{e^{-iz}}{z-1} \, dz = \int_{-\beta}^\beta \frac{e^t}{it-1} \, i \, dt = \int_{-\beta}^\beta \frac{e^t(t-i)}{1+t^2} \, dt,
$$
and it is pretty easy to see that both the real and imaginary part of this integral diverge as $\beta \to \infty$. (The exponential function grows too fast for positive $t$, and there is no cancellation because the real part is always positive, the imaginary part is negative for $t>0$.)
If you want an example with $\gamma \ne 0$, just pick $\gamma=2\pi$ and $z_0 = 2\pi+1$. By periodicity of the function, the integral you get is going to be exactly the same.
In order to make the statement true, you will need to assume something about the growth/decay of $f(z)$ as $z \to \infty$. Just curious, which book is this?
