Show $\int_0^{\infty} \frac{e^{-x}-e^{-xw}}{x} dx = \ln{w}$ for $\operatorname{Re}({w})>0$

Question

I want to show that for $\operatorname{Re}({w})>0$, $$\int_0^{\infty} \frac{e^{-x}-e^{-xw}}{x} dx = \ln{w}.$$ I've tried setting the problem up as: $$\int_\gamma \frac{e^{-z}}{z} dz = 0,$$ where $\gamma$ is the path around the quadrilateral with vertices $a,b,bw,aw$ for some $a,b \in \mathbb{R}$ where $0<a<b<\infty$, but I'm not sure if I am parametrizing the paths between these points correctly.

Answer 1

Consider the integral \begin{align} \int_{1}^{w} e^{-x u} \, du = \left[ -\frac{1}{x} \, e^{-u} \right]_{1}^{w} = \frac{e^{-x} - e^{-w x}}{x}. \end{align} Now, \begin{align} I &= \int_{0}^{\infty} \frac{e^{-x} - e^{-w x}}{x} \, dx = \int_{0}^{\infty} \, \int_{1}^{w} e^{-x u} \, du \, dx \\ &= \int_{1}^{w} \left[ \int_{0}^{\infty} e^{-x u} \, dx \right] \, du = \int_{1}^{w} \frac{du}{u} = [ \ln(u) ]_{1}^{w} \\ &= \ln(w). \end{align} Hence \begin{align} \int_{0}^{\infty} \frac{e^{-x} - e^{-w x}}{x} \, dx = \ln(w). \end{align}