# Estimating Beta like integral

For $$\alpha \in (0,1)$$ define

$$p(x,z) =\frac{1}{(z-x) (z-1)^\alpha},\quad z>1,x \in (0,1),$$

I try to estimate the integral $$\int_1^\infty \left( \frac{z-1}{z} \right)^\gamma \frac{1}{(z-x) (z-1)^\alpha} d z$$

for $$\gamma = \alpha$$, I can factor out the $$(z-1)^\alpha$$ term and do the following \begin{align} &\int_1^\infty \frac{1}{z^\alpha (z-x)} dz \\ &= x^{-\alpha} \int_{1/x}^\infty \frac{1}{t^{\alpha}(t-1)} d t\\ &= x^{-\alpha} \int_{0}^x \frac{y^{\alpha-1}}{1-y} d y\\ &= x^{-\alpha} \sum_{k=0}^\infty \int_{0}^x y^k y^{\alpha-1} d y\\ &= x^{-\alpha} \sum_{k=0}^\infty \frac{x^{\alpha +k}}{a+k}\\ &= \sum_{k=0}^\infty \frac{x^{k}}{a+k} \end{align} The last term I can find upper and lower logarithmic bounds. For $$0<\gamma <\alpha$$ I hope to get something similar but I am not sure what to do with the $$(z-x)$$ term. I either get an unintegrable singularity or a $$(1-x)^{-1}$$ term.

• The indefinite integral is an Appell hypergeometric function of two variables. Commented Aug 1 at 13:53
• Mathematica gives a solution for the general case of the definite integral. Commented Aug 3 at 8:43
• @ClaudeLeibovici could you please post it as an answer? I don't have access to wolfram mathematica Commented Aug 3 at 12:21

This is not an answer but just the result from Mathematica $$I=\int_1^\infty \left( \frac{z-1}{z} \right)^\gamma \frac{dz}{(z-x) (z-1)^\alpha}$$
$$I=\pi \csc (\pi \alpha ) (-x)^{-\gamma} (1-x)^{\gamma -\alpha }-$$ $$\frac{\pi \csc (\pi \alpha ) \Gamma(\gamma-\alpha +1)}{(1-x) \Gamma (2-\alpha ) \Gamma (\gamma )}\, _2F_1\left(1,\gamma-\alpha +1;2-\alpha ;\frac{1}{1-x}\right)$$