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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.

Incognito
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    The integral is finite only for $0 < \alpha < 1$. For $\alpha \leqslant 0$, you have a non-integrable singularity in $0$, and for $\alpha \geqslant 1$, it decays too slowly. Note that you can simplify your result to $$\frac{\pi}{\sin \pi\alpha}.$$ – Daniel Fischer Mar 23 '14 at 21:15
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    All integrals of the form $\displaystyle\int_0^\infty\frac{x^n}{(1+x^m)^p}dx$ are solved by substituting $t=\dfrac1{(1+x^m)^p}$, and then recognizing the expression of the beta function in the new integral, followed by employing Euler's reflection formula for the $\Gamma$ function. In this case, $n=a-1$. – Lucian Mar 23 '14 at 22:08

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{x^{\alpha} \over x\pars{x + 1}}\,\dd x:\ {\large ?}.\qquad 0 < \alpha < 1}$.

\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}

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Felix Marin
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