Please prove that: for $x\in(0,1/2)$ and $a<0$,
$\int_{0}^{1}v^{-x}(1-v)^{-x}e^{a(1-v)M}dv$ tends to zero as $M$ goes to infinite.
Many thanks for your help!
Maybe we can use speicial functions
$\int _ { 0 } ^ { u } x ^ { \nu - 1 } ( u - x ) ^ { \mu - 1 } e ^ { \beta x } d x = \mathrm { B } ( \mu , \nu ) u ^ { \mu + \nu - 1 } \quad 1 F _ { 1 } ( \nu ; \mu + \nu ; \beta u )$
How to use this formula!
Many thanks for your help!
You already know (using special functions) that the integral is finite for each $M$. The integrand tends to $0$ pointwise as $M\to \infty$, it is decreasing in $M$ and the integrand is dominated by $v^{-x}(1-v)^{-x} e^{a(1-v)}$ for $M>1$. Since this last function is integrable the result follows immediately by Dominated Convergence Theorem.