$\int_0^\infty \frac{\ln(x^2+1)}{x^{\alpha+1}}dx, 0<\alpha <2 $ using Residue and IBP I am practicing the examples in alfors and I've been stuck in this problem for a quite few days $$\int_0^\infty \frac{\ln(x^2+1)}{x^{\alpha+1}}dx, 0<\alpha <2$$
What I've done so far was 
\begin{eqnarray*}
\int_0^\infty \frac{\ln(x^2+1)}{x^{\alpha+1}}dx\
&=&\frac{1}{-\alpha} \frac{\ln(x^2+1)}{x^{\alpha}}|_0^\infty 
+\frac{1}{\alpha} \int_0^\infty \frac{2x}{(x^2+1)x^{\alpha}}dx\\
&=^L& \frac{1}{(-\alpha)^2}[0-\lim_{x\to 0}\frac{2x}{(x^2+1)x^{\alpha +1}}] +\frac{2}{\alpha(1+e^{-i\pi (\alpha-1)})} \int_0^\infty \frac{(1+e^{i\pi (\alpha-1)})}{(x^2+1)x^{\alpha-1}}dx\\
&=&-\frac{1}{\alpha^2}\lim_{x\to 0}\frac{2}{(x^2+1)x^{\alpha}} +\frac{2}{\alpha(1+e^{-i\pi (\alpha-1)})}[\int_0^\infty \frac{1}{(x^2+1)x^{\alpha-1}}dx +\int_0^\infty \frac{e^{-i\pi (\alpha-1)}}{(x^2+1)x^{\alpha-1}}dx]\\
&=&-\frac{1}{\alpha^2}\lim_{x\to 0}\frac{2}{(x^2+1)x^{\alpha}} +\frac{2}{\alpha(1+e^{-i\pi (\alpha-1)})}[\int_0^\infty \frac{1}{(x^2+1)x^{\alpha-1}}dx +\int_0^\infty \frac{1}{(x^2+1)(xe^{i\pi})^{(\alpha-1)}}dx]\\
&=&-\frac{1}{\alpha^2}\lim_{x\to 0}\frac{2}{(x^2+1)x^{\alpha}} +\frac{2}{\alpha(1+e^{-i\pi (\alpha-1)})}[\int_0^\infty \frac{1}{(x^2+1)x^{\alpha-1}}dx +\int_0^\infty \frac{1}{((-x)^2+1)(-x)^{(\alpha-1)}}dx]\\
&=&-\frac{1}{\alpha^2}\lim_{x\to 0}\frac{2}{(x^2+1)x^{\alpha}} +\frac{2}{\alpha(1+e^{-i\pi (\alpha-1)})}[\int_0^\infty \frac{1}{(x^2+1)x^{\alpha-1}}dx +\int_0^{-\infty} \frac{1}{((x)^2+1)x^{\alpha-1}}(-dx)]\\
&=&-\frac{1}{\alpha^2}\lim_{x\to 0}\frac{2}{(x^2+1)x^{\alpha}} +\frac{2}{\alpha(1+e^{-i\pi (\alpha-1)})}\int_{-\infty}^\infty \frac{1}{(x^2+1)x^{\alpha-1}}dx\\
&=&-\frac{1}{\alpha^2}\lim_{x\to 0}\frac{2}{(x^2+1)x^{\alpha}} +\frac{2}{\alpha(1+e^{-i\pi (\alpha-1)})}[\oint-\int_{|z|=R,R\to \infty\\{Im(z>0)}\\Counterclockwise} -\int_{|z|=r,r\to 0\\{Im(z>0)}\\Clockwise}]\frac{1}{(z^2+1)z^{(\alpha-1)}}dz\\
\end{eqnarray*}
Then, I'am stucked........
Any help would be appreciated.
 A: Denote 
$$
I(a)=\int_{0}^{\infty}\frac{\log(x^2+1)}{x^{a+1}}
$$
Now integrate by parts to get (boundary terms equal zero)
$$
I(a)=\frac{1}{a}\int_{0}^{\infty}\frac{x^{-a+1}}{(x^2+1)}
$$
Next we introduce the function the complex valued function
$$
f(z)=\frac{z^{-a+1}}{a(z^2+1)}
$$
and integrate it 
around a keyhole with slit at the positive real axis. 
This implies a choice for the complex logaritm ($\delta\rightarrow 0_+$,$x\in\mathbb{R_{\geq0}}$) which induces  $\log(x+i\delta)=\log(|x|)$ and $\log(x-i \delta)=\log(|x|)+2\pi i$. 
Since for our choice of $a$ the big arc as well as the small arc of the keyhole contour vanish, we obtain according to the Resiude theorem
$$
(1-e^{-2i\pi a})I(a)=2\pi i[\text{Res}(f(z),z=i)+\text{Res}(f(z),z=-i)]
$$
keeping in mind that $\arg(- i)=\frac{3\pi}{2}$ and $\arg(+ i)=\frac{\pi}{2}$ it is easy to see that $\text{Res}(f(z),z=i)=-\frac{e^{-i a \pi/2}}{2a}$ and $\text{Res}(f(z),z=-i)=-\frac{e^{-i 3 a \pi/2}}{2a}$

$$
I(a)=\frac{\pi}{a\sin(\frac{\pi a}2)}
$$

