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[br][br]\Rightarrow I = \displaystyle\int_0^1 \frac{(1-x^{\alpha})(1-x^{\beta})}{(1-x) \ln x}\ dx = \displaystyle\int_0^1 \frac{(1-x^{\alpha})}{\ln x} \displaystyle\sum_{i=0}^{\beta -1} x^i \ dx [br][br]\beginaligned I = \displaystyle\sum_{i=0}^{\beta -1} \displaystyle\int_0^1 \frac{(1-x^{\alpha})}{\ln x} x^i \ dx[br][br]\beginaligned I = \displaystyle\sum_{i=0}^{\beta -1} \displaystyle\int_0^1 \frac{(x^i-x^{\alpha +i})}{\ln x} \ dx[br]
\displaystyle \int_{-\infty}^0 \frac {e^{(i+1)t}- e^{(\alpha + i+1)t}}{t} \ dt = \displaystyle\int_{-\infty}^0 \displaystyle\int_{\alpha+i+1}^{i+1} e^{ty}\ dy\ dt = \displaystyle\int_{\alpha+i+1}^{i+1} \displaystyle\int_{-\infty}^0 e^{ty}\ dt\ dy[br][br]\Rightarrow \displaystyle\int_{\alpha+i+1}^{i+1} \displaystyle\int_{-\infty}^0 e^{ty}\ dt\ dy = \displaystyle\int_{\alpha+i+1}^{i+1} \frac{1}{y} \ dy = \ln {(\frac{i+1}{\alpha+i+1})}[br][br]\Rightarow \displaystyle\int_0^1 \frac{(x^i-x^{\alpha +i})}{\ln x} \ dx = \ln {(\frac{i+1}{\alpha+i+1})}[br][br]
[br][br]\beginaligned I = \displaystyle\sum_{i=0}^{\beta -1} \displaystyle\int_0^1 \frac{(x^i-x^{\alpha +i})}{\ln x} \ dx = \displaystyle\sum_{i=0}^{\beta -1} \ln {(\frac{i+1}{\alpha+i+1})}[br][br][br]\beginaligned I = \displaystyle\sum_{i=1}^{\beta} \ln {(\frac{i}{\alpha+i})}[br][br][br]\beginaligned I = \displaystyle \ln {\frac{\beta !}{(\alpha+1)...(\alpha+ \beta)}}[br][br][br]\beginaligned I = - \ln \binom{\alpha+\beta}{\alpha}[br]
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