I wish to prove that the following limit goes to zero:
$$\lim_{x\to \infty}\frac{B_\alpha(a+x,b)}{\alpha^x} \to 0, \qquad \forall 0 \leq \alpha \leq 1$$ where $B_\alpha(a+x,b) = \int_0^\alpha t^{a-1}(1-t)^{b-1}\mathrm{d}t$ is the incomplete Beta function.
Using L'Hôpital's rule, yields the following:
\begin{eqnarray*} \lim_{x\to \infty}\frac{B_\alpha(a+x,b)}{\alpha^x} &=& \lim_{x\to \infty}\frac{\int_0^\alpha \log(t) t^{a+x-1} (1-t)^{b-1}\mathrm{d}t}{\log(\alpha) \alpha^x} \end{eqnarray*} which seemly does not lead to an answer in a finite number of steps.
Trying to use the monotone convergence theorem, the derivative can be simplified to: $$\frac{\mathrm{d}}{\mathrm{d}x}\frac{B_\alpha(a+x,b)}{\alpha^x} = \frac{\int_0^\alpha \log\left(\frac{t}{\alpha}\right)t^{x+a-1}(1-t)^{b-a}\mathrm{d}t}{\alpha^x}$$ Which is negative since the $\log(\cdot)$ term is negative , but still I don't know how to show that it is bounded by zero at the limit.
Maybe a transformation of variables should be applied?
Any help and ideas are much appreciated.
Using identity eq. 8.17.7, and thus,
\begin{eqnarray*} \lim_{x \to \infty} \frac{B_\alpha(a+x,b)}{\alpha^x}&=& \lim_{x \to \infty}\frac{\alpha^a {}_2 F_1 (a+x,1-b;1+a+x;\alpha)}{a+x}, \end{eqnarray*} where ${}_2 F_1 (a+x,1-b;1+a+x;\alpha)$ is the hypergeometric function eq. 15.2.1.
Since the hypergeometric function converges for $\vert \alpha \vert < 1$, the enumerator converge, and the ratio goes to zero as $x\to \infty$.