Computing generalized integral with parameters Assume $0<\lambda<1$ and $k>0$, compute the integral 
$$ \displaystyle I=\int_{0}^{+\infty}\frac{x^{-\lambda}}{1+(x+k)^2}{\rm d}x $$
I tried to express it in the form of $\beta$ function or $\varGamma$ function but failed, though I know it should be computed in that way.
My question:
(1) How to compute it?
(2) By the way, why here $\lambda$ can't $\geq1$?
Thanks!
 A: To answer the second question, expand around $x=0$ to get
$$\frac 1{1+(x+k)^2}=\frac{1}{k^2+1}-\frac{2 k x}{\left(k^2+1\right)^2}+O\left(x^2\right)$$
$$\frac {x^{-\lambda}}{1+(x+k)^2}=\frac{x^{-\lambda}}{k^2+1}-\frac{2 k x^{1-\lambda}}{\left(k^2+1\right)^2}+\cdots$$
What would happen if you integrate the first term close to $x=0$ ?
A: Since:
$$
\small
\begin{aligned}
\int \frac{x^{-\lambda}}{1 + (x + \mu)^2}\,\text{d}x
& = \frac{\text{i}}{2} \int \frac{x^{-\lambda}}{(\text{i} - \mu) - x}\,\text{d}x +
\frac{\text{i}}{2} \int \frac{x^{-\lambda}}{(\text{i} + \mu) + x}\,\text{d}x \\
& = \frac{\text{i}}{2} \int \frac{(\text{i} - \mu)^{-\lambda}\,y^{-\lambda}}{(\text{i} - \mu) - (\text{i} - \mu)\,y}\,(\text{i} - \mu)\,\text{d}y -
\frac{\text{i}}{2} \int \frac{(-\text{i} - \mu)^{-\lambda}\,z^{-\lambda}}{(\text{i} + \mu) - (\text{i} + \mu)\,z}\,(\text{i} + \mu)\,\text{d}z \\
& = \frac{\text{i}\,(\text{i} - \mu)^{-\lambda}}{2} \int y^{-\lambda}\,(1 - y)^{-1}\,\text{d}y - \frac{\text{i}\,(-\text{i} - \mu)^{-\lambda}}{2} \int z^{-\lambda}\,(1 - z)^{-1}\,\text{d}z \\
& = \frac{\text{i}\,(\text{i} - \mu)^{-\lambda}}{2}\,\mathcal{B}\left(y;\,1-\lambda,\,0\right) - \frac{\text{i}\,(-\text{i} - \mu)^{-\lambda}}{2}\,\mathcal{B}\left(z;\,1-\lambda,\,0\right) \\
& = \frac{\text{i}\,(\text{i} - \mu)^{-\lambda}}{2}\,\mathcal{B}\left(\frac{x}{\text{i}-\mu};\,1-\lambda,\,0\right) - \frac{\text{i}\,(-\text{i} - \mu)^{-\lambda}}{2}\,\mathcal{B}\left(\frac{x}{-\text{i}-\mu};\,1-\lambda,\,0\right) ;
\end{aligned}
$$
if $\lambda,\,\mu \in \mathbb{R}$ and $|\lambda| < 1$, we get:
$$
\int_0^{+\infty} \frac{x^{-\lambda}}{1 + \left(x + \mu\right)^2}\,\text{d}x =
\begin{cases}
\frac{\pi}{2} - \arctan(\mu) & \text{if} \; \lambda = 0 \\
\frac{\pi\,\text{i}}{2}\,\frac{\left(\mu + \text{i}\right)^{-\lambda} - \left(\mu - \text{i}\right)^{-\lambda}}{\sin(\lambda\,\pi)} & \text{if} \; \lambda \ne 0
\end{cases}\,;
$$
i.e.
$$
\int_0^{+\infty} \frac{x^{-\lambda}}{1 + \left(x + \mu\right)^2}\,\text{d}x =
\begin{cases}
\frac{\pi}{2} - \arctan(\mu) & \text{if} \; \lambda = 0 \\
\frac{\pi}{2 \cos\left(\lambda\,\frac{\pi}{2}\right)} & \text{if} \; \mu = 0 \\
\frac{\pi}{\left(1 + \mu^2\right)^{\lambda/2} \sin(\lambda\,\pi)}\,\sin\left(\lambda\left(\pi+\text{arccot}(\mu)\right)\right) & \text{if} \; \lambda \ne 0 \land \mu < 0 \\
\frac{\pi}{\left(1 + \mu^2\right)^{\lambda/2} \sin(\lambda\,\pi)}\,\sin\left(\lambda\,\text{arccot}(\mu)\right) & \text{if} \; \lambda \ne 0 \land \mu > 0
\end{cases}\,;
$$
if $|\lambda| \ge 1$ the integral doesn't converge.
