Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 1 / 14E / d / Solution

For real $s>0$, collapse the Hankel contour onto the negative axis and use the jump in the argument of $t^{s-1}$. The reflection formula for the gamma function yieldsso $K(z,s,t)=z{t}^{s-1}/[\Gamma (s)(e^t-z)]$. At positive integer $s$, the original Hankel expression is interpreted by analytic continuation: the pole of $\Gamma (1-s)$ cancels the corresponding zero of the contour integral.

When $z$ crosses the slit at $x>1$, the pole of the real-integral kernel at $t=\log x$ crosses the integration path. The two boundary values differ by $2\pi i$ times its residue. Since $d(e^t-z)/dt=x$ there, that residue is $(\log x{)}^{s-1}/\Gamma (s)$, giving the jump