A definite integral with a real parameter Let $\alpha\in[0,1]$, and
$$\text{I}(\alpha)=\int_{0}^{1}\frac{\ln(t)}{(1-t)\sqrt{1-\alpha t}}\text{d}t$$
I need to find a general close form for the function $\text{I}(\alpha)$, but I'm really strugling on this one.
I can get the specific values for $\alpha=0$ and $1$ ($-\zeta(2)$ and $-4\ln(2)$ respectively), but not the general expression. None of the usual integration technique seem to apply here. Seeing those 2 values, I expect an expression involving logarithms and dilogarithms...
Do you have any insight ?
 A: With substitution $t=1-u$
$$\text{I}(\alpha)=\int_{0}^{1}\frac{\ln(1-u)}{u\sqrt{(1-\alpha)+\alpha u}}\ \mathrm{d}t =- \sum_{n=1}^\infty \int_{0}^{1}\frac{u^n}{nu\sqrt{(1-\alpha)+\alpha u}}\ \mathrm{d}u$$
now let $\dfrac{\alpha u}{1-\alpha}=\dfrac{w}{1-w}$ therefore
$$\text{I}(\alpha)= \sum_{n=1}^\infty \dfrac{-1}{n} \dfrac{(1-\alpha)^{n-1/2}}{\alpha^n} \int_{0}^{\alpha}w^{n-1}(1-w)^{-n-\frac12}\ \mathrm{d}w = \sum_{n=1}^\infty \dfrac{-1}{n} \dfrac{(1-\alpha)^{n-1/2}}{\alpha^n} \beta\left(\alpha;n,-n+\frac12\right)$$
where $\beta(x;a,b)$ is incomplete Beta function.
A: With the change of variable $t\mapsto 1-t$ the problem boils down to finding a closed form for
$$ \int_{0}^{1}\frac{\log(1-t)}{t}\cdot\frac{1}{\sqrt{1-\beta t}}\,dt=\sum_{n\geq 0}\frac{\binom{2n}{n}\beta^n}{4^n}\int_{0}^{1}t^{n-1}\log(1-t)\,dt=-\zeta(2)-\sum_{n\geq 1}\frac{\binom{2n}{n}\beta^n H_n}{n 4^n} $$
which should be
$$\sum_{n\geq 1}\frac{\binom{2n}{n}\beta^n H_n}{n 4^n} = -4\log^2\left(\frac{1}{\sqrt{\beta}}-\sqrt{\frac{1}{\beta}-1}\right)+\log^2\left(-\frac{1}{2}+\frac{1}{2\sqrt{1-\beta}}\right)-2\log\left(-\frac{1}{2}+\frac{1}{2\sqrt{1-\beta}}\right)\log\left(\frac{1}{2}+\frac{1}{2\sqrt{1-\beta}}\right)-2\,\text{Li}_2\left(\frac{1}{2}-\frac{1}{2\sqrt{1-\beta}}\right)$$
if I managed to simplify correctly the CAS-assisted manipulations through Fourier-Legendre series.
