$$\int_0^{\pi/2}{\log x\over x^a}\,\mathrm dx,\quad a<1$$
I tried solving using the $\mu-test$. so if I consider $\mu=1$ then
$\lim\limits_{x\rightarrow 0} {x\log x\over x^a}$ Solving further, I get a limit $0$, so the integral must be divergent. But according to the book it is convergent.
On
The integral of the above from 0 to $\frac{\pi}{2} is -\frac{\left( \pi \,\mathrm{log}\left( \frac{\pi }{2}\right) \,a-\pi \,\mathrm{log}\left( \frac{\pi }{2}\right) +\pi \right) \,{e}^{-\mathrm{log}\left( \frac{\pi }{2}\right) \,a}}{2\,{a}^{2}-4\,a+2}$ which is defined for all a<1
On
Let's consider the following integral with $n > 0$: \begin{equation} \int_{0}^{\pi/2}x^{n - a}\,{\rm d}x = \left.{x^{n - a + 1} \over n - a + 1}\right\vert_{0}^{\pi/2} = {\left(\pi/2\right)^{n - a + 1} \over n - a + 1} \tag{1} \end{equation}
Derive both members respect of $n$: $$ \int_{0}^{\pi/2}x^{n - a}\ln\left(x\right)\,{\rm d}x = {\left(\pi/2\right)^{n - a + 1}\ln\left(\pi/2\right)\left(n - a + 1\right) - \left(\pi/2\right)^{n - a + 1} \over \left(n - a + 1\right)^{2}} $$
Take the limit, in both members, $n \to 0^{+}$: $$ \int_{0}^{\pi/2}{\ln\left(x\right) \over x^{a}}\,{\rm d}x = {\left(\pi/2\right)^{-a + 1}\ln\left(\pi/2\right)\left(-a + 1\right) - \left(\pi/2\right)^{-a + 1} \over \left(-a + 1\right)^{2}} $$
\begin{equation} \color{#ff0000}{\large% \int_{0}^{\pi/2}{\ln\left(x\right) \over x^{a}}\,{\rm d}x \color{#000000}{\ =\ } \left[% {\ln\left(\pi/2\right) \over 1 - a} - {1 \over \left(1 - a\right)^{2}} \right] \left(\pi \over 2\right)^{1 - a}} \tag{2} \end{equation}
Notice that result $\left(1\right)$ requires $a < 1$ which keeps $n - a + 1 > 0\,,\ \forall\ n > 0$. Then, result $\left(2\right)$ is valid whenever $\color{#ff0000}{\large a < 1}$.
Hint: For any $\varepsilon>0$ we have $x^{\varepsilon}\log x \to 0$ as $x \to +0$. Let $\varepsilon := \dfrac{1-a}{2}$, write $\dfrac{\log x}{x^a} = \dfrac{x^\varepsilon \log x}{x^{a+\varepsilon}}$ and compare with $\int_0^{\pi/2} \dfrac{dx}{x^{a+\varepsilon}}$.