# 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.

• 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
• 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


\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}