Contour Integration Limitations? So, I'm trying to evaluate the following integral by complex contour integration ONLY:
$$\int_0^\infty{\frac{x^\alpha}{x(x+1)}} dx$$ where alpha is real and not an integer.
Obviously, we need to use a key-hole contour to avoid the resulting branch cut, and I can carry out the analysis.  However, I am finding that, if alpha > 1, then the contour for R going from 0 to $2\pi$ becomes undefined (it basically blows up; I'm using an inequality arguement).  Is this correct, or can it actually be evaluated for alpha > 1?  Also, what about alpha < 0?
Please do not solve this out fully; I have already done so.  here is my result:
$$\int_0^\infty{\frac{x^\alpha}{x(x+1)}} dx=\frac{2\pi ie^{i\alpha\pi}}{e^{i2\pi\alpha}-1}$$ which gives me correct results.
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$\ds{\int_{0}^{\infty}{x^{\alpha} \over x\pars{x + 1}}\,\dd x:\ {\large ?}.\qquad
    0 < \alpha < 1}$.

\begin{align}&\color{#c00000}{%
\int_{0}^{\infty}{x^{\alpha} \over x\pars{x + 1}}\,\dd x}
=2\pi\ic\ \verts{-1}^{\alpha - 1}\ \expo{\ic\pi\pars{\alpha - 1}}
-\int_{\infty}^{0}{x^{\alpha}\expo{2\pi\pars{\alpha - 1}\ic} \over x\pars{x + 1}}\,\dd x
\end{align}

\begin{align}&\color{#44f}{\large%
\int_{0}^{\infty}{x^{\alpha} \over x\pars{x + 1}}\,\dd x}
=-2\pi\ic\,
{\expo{\ic\pi\alpha} \over 1 - \expo{2\pi\alpha\ic}}
=\pi\,{2\ic \over \expo{\ic\pi\alpha} - \expo{-\ic\pi\alpha}}
=\color{#44f}{\large{\pi \over \sin\pars{\pi\alpha}}}
\end{align}

