I do not know how to prove the following proposition:
$$\frac{1}{\Gamma (\alpha)} \int_1^t \log(\frac{t}{x})^{\alpha -1} (\log (x)) ^{\beta-1} \frac{dx}{x} = \frac{\Gamma(\beta)}{\Gamma (\beta + \alpha)} (\log(t))^{\alpha +\beta -1}$$
"Theory and Application of Fractional Differential Equation" states it without a proof and says that it can verified directly.
By assuming $\alpha,\beta,t>0$ and letting $t=e^u$ the LHS is converted into $$ \frac{1}{\Gamma(\alpha)}\int_{0}^{\log t}\left(\log t-u\right)^{\alpha-1} u^{\beta-1}\,du $$ and by letting $u=v\cdot \log t$ this equals $$ \frac{\left(\log t\right)^{\alpha+\beta-1}}{\Gamma(\alpha)}\int_{0}^{1}(1-v)^{\alpha-1}v^{\beta-1}\,dv $$ where by Euler's Beta function $$ \int_{0}^{1}(1-v)^{\alpha-1}v^{\beta-1}\,dv = B(\alpha,\beta)=\frac{\Gamma(\alpha)\,\Gamma(\beta)}{\Gamma(\alpha+\beta)}.$$