Let $c>1$ and $\lambda\in\mathbb{C}$ with $\Re(\lambda)\geq1 $. Show that $\lim_{r\to\infty}\int_{c-ir}^{c+ir}z^{-\lambda}(\log(z))^{-1}dz=0$, where the path of integration is the straight line and $\log$ denotes the principle branch of the logarithm.
I had the following idea: Closing the path with the semicircle $\gamma(t)=c+re^{it}$ for $t\in[-\pi/2,\pi/2]$. Then for a fixed $r$ the integral over this simple closed contour is $0$ by Cauchy's Theorem because the function is holomorphic inside and on the contour. So I am left to show that $\lim_{r\to\infty}\int_\gamma z^{-\lambda}(\log(z))^{-1}dz=0$. I think at this point I have to use the definition of a path integral using the parametrization of $\gamma$ and then estimate the absolut value of the integral hoping that I get something like $1/r$, but I have difficulties with estimating the absolut value of $\dfrac{1}{e^{\lambda\log(c+re^{it})}\log(c+re^{it})}$. The derivative of $\gamma$ just gives me a factor $r$ and then I also have to multiply by the length of the curve which is $\pi r$.
Am I on the right way? Can anyone help me with the estimate?
Yes, you are (just keep in mind that even a rough estimate will do).
You can go even a simpler way: let $c+ir=\rho e^{i\phi}$, choose $\gamma(t)=\rho e^{it}$ for $t\in[-\phi,\phi]$ instead, and see that $z=\rho e^{it} \implies |z^{-\lambda}\log^{-1}z|=\rho^{-\sigma}\Big/\sqrt{t^2+\log^2\rho}\leqslant\rho^{-\sigma}/\log\rho$ with $\sigma=\Re\lambda$, so that the integral over $\gamma$ is at most $$\frac{2\phi\rho^{1-\sigma}}{\log\rho}<\frac{\pi\rho^{1-\sigma}}{\log\rho}.$$