solution of an improper integral. I was solving following improper integral:
$$
\int\limits_0^\frac{\pi}{2}\frac{log~x}{x^a}dx  
$$
where $a<1$.
My attempt:
$0$ is the only point of discontinuity. So, 
$\frac{log~x}{x^a}\leq \frac{x}{x^a} = \frac{1}{x^{a-1}}$. 
Now $f(x) =\frac{1}{x^{a-1}}$. And this is convergent if $a-1<1$.
So given integral converges if $a<2$.
Is my attempt correct? Are the steps to proceed to solution justified? Kindly rectify if I am wrong somewher. Thanks for the help. 
 A: Integral converges for all $ a \lt 1$ because $x=0$ is an integrable singularity of the log and the $x^{-a}$ term.
To evaluate, integrate by parts:
$$\begin{align}\int_0^{\pi/2} dx \, x^{-a} \log{x} &= \frac1{1-a} \left [ x^{1-a} \log{x} \right ]_0^{\pi/2} - \frac1{1-a} \int_0^{\pi/2} dx \, x^{1-a} x^{-1} \\&= \frac1{1-a} \left (\frac{\pi}{2} \right )^{1-a} \log{\left (\frac{\pi}{2} \right )} - \frac1{(1-a)^2} \left (\frac{\pi}{2} \right )^{1-a}\\ &= \frac1{1-a} \left (\frac{\pi}{2} \right )^{1-a} \left [\log{\left (\frac{\pi}{2} \right )} - \frac1{1-a} \right ]\end{align}$$
A: Integrate by parts using
\begin{align}
&&f &= \log(x)    \qquad &dg &= x^{-a}\\
&&df&=\frac{1}{x} \qquad &g &= \frac{x^{1-a}}{1-a}
\end{align}
Then
$$\int_{0}^{\frac{\pi}{2}} \frac{\log(x)}{x^a} dx = \frac{x^{1-a} \log(x)}{1-a} - \frac{1}{1-a} \int_{0}^{\frac{\pi}{2}} x^{-a}dx.$$
It yelds
$$\int_{0}^{\frac{\pi}{2}} \frac{\log(x)}{x^a} dx = x^{1-a} \left[\frac{ \log (x)}{1-a}-\frac{1}{(1-a)^2}\right] \Bigg{|}_{0}^{\frac{\pi}{2}} = \left(\frac{\pi}{2}\right)^{1-a} \left[\frac{ \log \left(\frac{\pi}{2}\right)}{1-a}-\frac{1}{(1-a)^2}\right] $$
