I am trying to show the following:
Assume that the only singularities of a function $h(z)$ are finitely many poles that lie away from the origin and the negative real axis. Show that integration of the function $f(z) = h(z)\ln z$, with $\arg z \in [-\pi , \pi)$, around an appropriate keyhole contour, leads to being able to find the value of $\int^{\infty}_{0}h(-x)\space dx$.
I am quite unsure where to begin with this in terms of explaining the poles and residues for $h(x)\ln(x)\space dx$, any help would be greatly appreciated. Thanks in advance.
Take $C$ the contour $-\infty\to \epsilon \to -\infty$ enclosing $(-\infty,0]$ counterclockwise $$\int_C h(z)\log zdz=\int_{-\infty}^0 h(x)\log xdx-\int_{-\infty}^0 h(e^{2i\pi }x )\log (e^{2i\pi } x)d(e^{2i\pi } x)$$ $$ = \int_{-\infty}^0 h(x)(\log x-\log e^{2i\pi}x)dx=-2i\pi \int_{-\infty}^0 h(x)dx$$