A real integral (may be requires contour integration)? The integral I have in mind is 
$$\int^\infty_0 x^{r}(x + \lambda)^{-1}dx$$
where $r \in (-1, 0)$, and $\lambda$ is a non-negative constant. I apologize if this is really easy and I am missing some simple trick.
 A: Denote your integral by $I$. Note that $f(z) = z^r(z+\lambda)^{-1}$ has branch points at $z=0$ and $z=\infty$ (from the $z^r$, given that $r\in (-1,0)$), and that two cases arise depending on the value of $\lambda$:
(1) a simple pole at $z=-\lambda$ if $\lambda >0$
(2) no poles if $\lambda = 0$
But note that the real integral does not converge for case (2), so we may ignore it.

Consider (1):
Take a branch cut along the positive real axis s.t. that $arg (0^+) = 0$, $arg(0^-) = 2\pi$, and consider $\displaystyle\int_L f(z) dz$, where $L$ is the standard keyhole contour traversed anti-clockwise about the origin. Then $\displaystyle\int_L f = I_1+I_R+I_2+I_{\epsilon}$ where (as $R\to \infty, \epsilon \to 0$):
$I_1 \to \displaystyle\int_0^{\infty} x^r(x+\lambda)^{-1}dx = I$
$I_2 \to \displaystyle\int_{\infty}^0 e^{2r\pi i}x^r(x+\lambda)^{-1}dx = - e^{2r\pi i}I$
$|I_R|\to 0$ since $|zf(z)| \sim |z|^r \to 0$ as $|z|\to \infty$
$|I_{\epsilon}|\to 0$ since $|zf(z)| \sim |z|^{r+1}\to 0$ as $|z|\to 0$ 
So $\displaystyle\int_Lf \to (1-e^{2r\pi i})I$
Furthermore, by the residue theorem:
$\displaystyle\int_L f = 2\pi i\text{Res}[f,z=e^{i\pi}\lambda] = 2\pi ie^{r\pi i}\lambda^r$
So it follows that $\displaystyle\int_0^{\infty} x^r(x+\lambda)^{-1}dx = \dfrac{2\pi i}{e^{-r\pi i}-e^{r\pi i}}\lambda^r = - \dfrac{\pi}{\sin (\pi r)}\lambda^{r}$
