Extensions and Shadows (11)

To continue our proof of Lemma A.4 and Corollary A.5, we see the integral given in (3.3) last time equals \[ \int_0^y g(\alpha) d \alpha = \int_0^y \left( 1 – \frac{t \alpha}{y} \right)^{-k} d \alpha = y \int_0^1 \left( 1 – t \beta \right)^{-k} d \beta, \] by changing \( \alpha \) into \( \beta […]