We consider the complex integral equivalent
$$\int_0^{\infty}\frac{z^{\alpha}\ln(z)}{z^2+1}\,dz,$$
and make this part of a contour as follows (let $z=re^{i\theta}$):
\begin{align*}
\gamma_1:&\;\varepsilon<r\le R, \; \theta=0 \\
\gamma_2:&\;0<\theta<\pi, \; r=R \\
\gamma_3:&\;-R<r<-\varepsilon, \; \theta=\pi \\
\gamma_4:&\;r=\varepsilon, \; \pi >\theta >0.
\end{align*}
We define
\begin{align*}
z^{\alpha}&=r^{\alpha}e^{i\alpha \operatorname{arg}(z)}, \\
\ln(z)&=\operatorname{Log}(r)+i\operatorname{arg}(z),
\end{align*}
where
$$-\frac{\pi}{2}<\operatorname{arg}(z)<\frac{3\pi}{2}.$$
That is, we choose the branch cut for these two functions along the negative imaginary axis. Let
$$f(z)=\frac{z^{\alpha}\ln(z) }{z^2+1}.$$
Then the only pole of $f$ inside the contour described above is at $z=i$. Hence, we have that
$$\sum_{j=1}^4\int_{\gamma_j}f(z)\,dz=2\pi i \operatorname{Res}[f,i], $$
by the Residue Theorem. It follows that
$$\int_{\gamma_1}f(z)\,dz=2\pi i \operatorname{Res}[f,i]-\sum_{j=2}^4\int_{\gamma_j}f(z)\,dz=2\pi i\left(\frac{\pi e^{i\alpha\pi/2}}{4}\right)-\sum_{j=2}^4\int_{\gamma_j}f(z)\,dz=\frac{i\pi^2 e^{i\alpha\pi/2}}{2}-\sum_{j=2}^4\int_{\gamma_j}f(z)\,dz.$$
The next step is to show that
$$\int_{\gamma_2}f(z)\,dz=\int_{\gamma_4}f(z)\,dz=0.$$
Both of these will follow by the ML inequality. On $\gamma_4,$ we have that
$$|f(z)|\le \varepsilon^\alpha \sqrt{\ln^2(\varepsilon) + \pi^2}/(1 - \varepsilon^2)\lesssim \varepsilon^\alpha \ln(1/\varepsilon) \lesssim \varepsilon^{(\alpha - 1)}.$$
By ML, $\displaystyle\int_{\gamma_4}$ is $O(\varepsilon^{\alpha}),$ which goes to zero as $\varepsilon\to 0.$ Similarly, on $\gamma_2,$ $f$ is $O(R^{\alpha-1}),$ which goes to zero as $R\to\infty,$ since $\alpha<1.$ So we now have
$$\int_{\gamma_1}f(z)\,dz=\frac{i\pi^2 e^{i\alpha\pi/2}}{2}-\int_{\gamma_3}f(z)\,dz.$$
We can show (I'll omit the tedious calculations) that:
$$\int_{\gamma_1}f(z)\,dz=\int_{0}^{\infty}\frac{x^{\alpha}\ln(x)}{x^2+1}\,dx\quad\text{and}\quad\int_{\gamma_3}f(z)\,dz=e^{i\alpha\pi}\int_{0}^{\infty}\frac{x^{\alpha}\ln(x)}{x^2+1}\,dx+i\pi e^{i\alpha\pi}\int_{0}^{\infty}\frac{x^{\alpha}}{x^2+1}\,dx.$$
We'll pull the first one on the RHS to the LHS to obtain
$$(1+e^{i\alpha\pi})\int_{0}^{\infty}\frac{x^{\alpha}\ln(x)}{x^2+1}\,dx=\frac{i\pi^2 e^{i\alpha\pi/2}}{2}-i\pi e^{i\alpha\pi}\int_{0}^{\infty}\frac{x^{\alpha}}{x^2+1}\,dx. $$
Next we multiply by $e^{-i\alpha\pi}$ and take the real part to obtain
$$(1+\cos(\alpha\pi))\int_{0}^{\infty}\frac{x^{\alpha}\ln(x)}{x^2+1}\,dx=\frac{\pi^2\sin(\alpha\pi/2)}{2}. $$
Trig identities will yield the final result.
Note: many thanks to Euge for lots of help on this one. I learned not a few tricks!
