The value of ${\int }_0^1 \frac{x^{2\alpha }-1}{\log x}\text{ d }x\text{, if }\alpha =\frac{2n-1}{2}$ is

Let $I={\int }_0^1 \frac{x^{2\alpha }-1}{\log x}dx$

$\begin{array}{l}{\left.\Rightarrow \frac{dI}{d\alpha }={\int }_0^1 \frac{2x^{2\alpha }\log x}{\log x}dx=\frac{2}{2\alpha +1}{x}^{2\alpha +1}\right]}_0^1\\ =\frac{2}{2\alpha +1}\\ \Rightarrow I=\log (2\alpha +1)+C\end{array}$

When $\alpha =0,I=0\Rightarrow C=0$

Thus $I=\log (2\alpha +1)=\log \left(2n\right)$