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My lecturer simply stated the solution to the integral ($\alpha<1, x=O(1)$)

$$I = \int_0^\infty \frac{1}{x^\alpha}\frac{1}{1+x}\,{\rm d}x=\pi\,{\rm cosec}(\pi\alpha)$$

Now the reasoning she gave us that allows us to take this integral even though the integrand doesn't exist at $x=0$ is "we can integrate here since we are only taking leading order". She also said to get this solution she took the first term of the taylor series for $1/(1+x)$. Here is my attempt:

$$I = \int_0^\infty \frac{1}{x^\alpha}\frac{1}{1+x}\,{\rm d}x\approx \int_0^\infty \frac{1}{x^\alpha}(1)\, {\rm d}x=\left[\frac{x^{1-\alpha}}{1-\alpha}\right]_0^\infty$$

but this is unbounded so I don't really know how to get the solution. It would be great if someone could help tell me whats wrong or correct her suggestions. Thanks

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The reason this all works is because you are only considering the (integrable) singularity in the neighborhood of $x=0$. Thus, $\alpha \lt 1$ for the integral to converge there. Infinity is another story. There the integrand behaves as $x^{-(1+\alpha)}$ ; this leads to the requirement that $\alpha \gt 0$ for convergence. Thus, $\alpha \in (0,1)$.

To evaluate: evaluate the following integral in the complex plane:

$$\oint_C \frac{dz}{z^{\alpha} (1+z)} $$

where $C$ is a standard keyhole contour of outer radius $R$ and inner radius $\epsilon$ about the positive real axis. Thus the contour integral is equal to

$$\int_{\epsilon}^R \frac{dx}{x^{\alpha}(1+x)} + i R^{1-\alpha} \int_0^{2 \pi} d\theta \frac{e^{i (1-\alpha) \theta}}{1+R e^{i \theta}} \\ + e^{-i 2 \pi \alpha} \int_R^{\epsilon} \frac{dx}{x^{\alpha}(1+x)}+ i \epsilon^{1-\alpha} \int_{2 \pi}^0 d\phi \frac{e^{i (1-\alpha) \phi}}{1+\epsilon e^{i \phi}}$$

As $R \to \infty$, the magnitude of the second integral vanishes. As $\epsilon \to 0$, the magnitude of the fourth integral vanishes.

By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the residue of the pole at $z=-1$. Note that here we must use $-1=e^{i \pi}$ due to the choice of contour. Thus we have

$$\left (1-e^{-i 2 \pi \alpha} \right ) \int_0^{\infty} \frac{dx}{x^{\alpha} (1+x)} = i 2 \pi \, e^{-i \pi \alpha} $$

or

$$\int_0^{\infty} \frac{dx}{x^{\alpha} (1+x)} = \frac{\pi}{\sin{\pi \alpha}}$$

as asserted.

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  • $\begingroup$ Hi Ron, I was more concerned about how to get the solution rather than the convergence! Thanks for the info though. $\endgroup$ – user2850514 Oct 14 '16 at 20:33
  • $\begingroup$ @user2850514: oh. $\endgroup$ – Ron Gordon Oct 14 '16 at 20:33
  • $\begingroup$ Oh, all the things I would never think of. $\endgroup$ – Simply Beautiful Art Oct 14 '16 at 20:46
  • $\begingroup$ I think the $\epsilon$'s in the exponents in your fourth integral should be $\alpha$'s. $\endgroup$ – Michael Seifert Oct 14 '16 at 20:49
  • $\begingroup$ @MichaelSeifert: thanks for the catch. $\endgroup$ – Ron Gordon Oct 14 '16 at 20:50
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\begin{align} \int\limits_{0}^{\infty} \frac{x^{-\alpha}}{1+x} \mathrm{d}x &= \mathrm{B}(1-\alpha,\alpha) \\ &= \frac{\Gamma(1-\alpha)\Gamma(\alpha)}{\Gamma(1)} \\ &= \frac{\pi}{\sin(\pi \alpha)} \end{align}

Notes:

  1. From Volume 1 of Higher Transcendental Functions (Bateman Manuscript), Section 1.5, Equation 3: $$\mathrm{B}(x,y) = \int\limits_{0}^{\infty} v^{x-1} (1+v)^{-x-y} \mathrm{d}v$$ for $\Re x \gt 0$ and $\Re y \gt 0$. Thus for our problem, we have $0 \lt \alpha \lt 1$ if $\alpha$ is real.

  2. Euler reflection formula: $$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$$

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  • $\begingroup$ Oh, what a sweet treat! 3 steps beautiful to my eyes. $\endgroup$ – Simply Beautiful Art Oct 15 '16 at 0:01
  • $\begingroup$ @SimpleArt, thanks. $\endgroup$ – poweierstrass Oct 15 '16 at 6:13
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

The following integral converges whenever $\ds{0 < \Re\pars{\alpha} < 1}$.

\begin{align} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \int_{0}^{\infty}{1 \over x^{\alpha}}\,{1 \over 1 + x}\,\dd x & = \int_{0}^{1}{x^{-\alpha} \over 1 + x}\,\dd x + \int_{1}^{\infty}{x^{-\alpha} \over 1 + x}\,\dd x = \int_{0}^{1}{x^{-\alpha} + x^{\alpha - 1} \over 1 + x}\,\dd x \\[5mm] & = \int_{0}^{1}{x^{-\alpha} + x^{\alpha - 1} - x^{-\alpha + 1} - x^{\alpha} \over 1 - x^{2}}\,\dd x \\[5mm] & = {1 \over 2}\int_{0}^{1}{x^{-\alpha/2 - 1/2} + x^{\alpha/2 - 1} - x^{-\alpha/2} - x^{\alpha/2 - 1/2} \over 1 - x}\,\dd x \\[5mm] & = {1 \over 2}\bracks{% -\Psi\pars{-\,{\alpha \over 2} + {1 \over 2}} - \Psi\pars{\alpha \over 2} + \Psi\pars{-\,{\alpha \over 2} + 1} + \Psi\pars{{\alpha \over 2} + {1 \over 2}}}\label{1}\tag{1} \\[5mm] & = {1 \over 2}\braces{{% \bracks{\vphantom{\Huge A}\Psi\pars{{\alpha \over 2} + {1 \over 2}} - \Psi\pars{-\,{\alpha \over 2} + {1 \over 2}}}} + \bracks{\vphantom{\Huge A}\Psi\pars{-\,{\alpha \over 2} + 1} - \Psi\pars{\alpha \over 2}}} \\[5mm] & = {1 \over 2}\braces{\vphantom{\huge A}% \pi\cot\pars{\pi\bracks{-\,{\alpha \over 2} + {1 \over 2}}} + \pi\cot\pars{\pi\,{\alpha \over 2}}}\label{2}\tag{2} \\[5mm] & = {\pi \over 2}\braces{\vphantom{\huge A}\tan\pars{\pi\,{\alpha \over 2}} + \cot\pars{\pi\,{\alpha \over 2}}} = \bbx{\pi\csc\pars{\pi\alpha}} \end{align}


$$\bbox[20px,#ffe,border:1px groove navy]{% \left\{\begin{array}{rcl} \ds{\Psi}: && Digamma\ Function. \\[4mm] \ds{\Psi\pars{z + 1} + \gamma} & \ds{=} & \ds{\int_{0}^{1}{1 - t^{z} \over 1 - t}\,\dd t\,,\quad\Re\pars{z} > - 1\,,\quad \pars{~\mbox{see}\ \eqref{1}~}} \\ \ds{\gamma}: && Euler\!-\!Mascheroni\ Constant \\[4mm] \ds{\Psi\pars{1 - z} - \Psi\pars{z}} & \ds{=} & \ds{\pi\cot\pars{\pi z}\,, \quad \pars{~Euler\ Reflection\ Formula~}.\ \mbox{See}\ \eqref{2}.} \end{array}\right.} $$

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    $\begingroup$ Digamma function $\psi$ is our friend ;) (+1). $\endgroup$ – Olivier Oloa Oct 15 '16 at 0:09
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    $\begingroup$ @OlivierOloa Thanks. It's true. $\endgroup$ – Felix Marin Oct 15 '16 at 0:15
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Split up the integral $$ I(\alpha) = \int_0^\infty \frac{x^{-\alpha}}{1+x} \,dx = \int_0^1 \frac{x^{-\alpha}}{1+x} \,dx + \int_1^\infty \frac{x^{-\alpha}}{1+x} \,dx. $$ Define $f(\alpha) = \int_0^1 \frac{x^{-\alpha}}{1+x} \,dx$, for $\alpha < 1$. Then $\int_1^\infty \frac{x^{-\alpha}}{1+x} \,dx = f(1-\alpha)$ as can be seen by substituting $u = x^{-1}$, so $I(\alpha) = f(\alpha) + f(1-\alpha)$ when $0 < \alpha < 1$.

Now if $\alpha < 1$ then $$ f(\alpha) = \int_0^1 \frac{x^{-\alpha}}{1+x} \,dx = \int_0^1 x^{-\alpha} \Bigl(1 - \frac{x}{1+x}\Bigr) \,dx = -\frac{1}{\alpha-1} - f(\alpha-1). $$ By iterating this we see that $$f(\alpha) = \sum_{k=1}^{n} \frac{(-1)^k}{\alpha-k} + (-1)^n f(\alpha-n).$$ Taking the limit $n \to \infty$, one has $f(\alpha-n) \to 0$ (by the dominated convergence theorem and $x^{n-\alpha} \to 0$ on $x \in (0,1)$), so $$f(\alpha) = \sum_{k=1}^{\infty} \frac{(-1)^{k}}{\alpha-k}.$$ In fact this series converges for all $\alpha \in \mathbb{C} \smallsetminus \mathbb{N}$, because the terms go to $0$ as $k \to \infty$, and grouping terms together in pairs gives a sum of terms of order $k^{-2}$. So we may redefine $f$ to be this series, which is a meromorphic function on $\mathbb{C}$.

Next, $$f(1-\alpha) = \sum_{k=1}^{\infty} \frac{(-1)^{k}}{(1-\alpha)-k} = \sum_{k=-\infty}^0 \frac{(-1)^{k}}{\alpha-k}$$ so that $$\boxed{I(\alpha) = f(\alpha) + f(1-\alpha) = \sum_{k = -\infty}^\infty \frac{(-1)^k}{\alpha-k}.}$$ This is a meromorphic function of $\alpha$ on $\mathbb{C}$ with the same poles and residues as $\pi/\sin(\pi\alpha)$. I claim that in fact $I(\alpha) = \pi/\sin(\pi\alpha)$.

Clearly each function is periodic with period $2$, and it can be shown that each is bounded as $\operatorname{Im}(\alpha) \to \pm\infty$, so their difference is a bounded entire function. Thus by Liouville's theorem, $$I(\alpha) - \frac{\pi}{\sin(\pi\alpha)} = C$$ for some constant $C$. Putting $\alpha = 1/2$ and noting $$f(1/2) = -2 \sum_{k=1}^\infty \frac{(-1)^k}{2k-1} = 2 \arctan(1) = \frac{\pi}{2}$$ shows that $C = 0$.

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Too long for a comment:

$$\int\frac{x^{-\alpha}}{1+x}dx=\int(e^u-1)^{-\alpha}du\tag{$u=\ln(1+x)$}$$

Binomial expansion: $$=\int\sum_{k=0}^\infty\binom{-\alpha}ke^{-\alpha u-k}(-1)^kdu$$

Switching the sum and integral: $$=\sum_{k=0}^\infty\int\binom{-\alpha}ke^{-\alpha u-k}(-1)^kdu$$

Integrating: $$=c+\frac1{-\alpha}\sum_{k=0}^\infty\binom{-\alpha}ke^{-\alpha u-k}(-1)^k$$

Un-binomial expansion: $$=\frac1{-\alpha}(e^u-1)^{-\alpha}+c$$

Undoing substitution:

$$=\frac1{-\alpha}x^{-\alpha}+c$$

Which is strange, as this is not the correct answer, right? Putting this out for anyone to come up ideas with.

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  • $\begingroup$ Switch like this without to know if this is possible? Check that .. $\endgroup$ – ParaH2 Oct 14 '16 at 21:03
  • $\begingroup$ You are assuming that $\alpha$ is integer by using the binomial expansion. $\endgroup$ – Zaid Alyafeai Oct 14 '16 at 23:58
  • $\begingroup$ @ZaidAlyafeai Nah, its called generalized binomial expansion. nifty. $\endgroup$ – Simply Beautiful Art Oct 14 '16 at 23:59
  • $\begingroup$ what is the definition of $$\binom{-\alpha}k$$ $\endgroup$ – Zaid Alyafeai Oct 15 '16 at 0:09
  • $\begingroup$ @ZaidAlyafeai The same as usual. $$\binom{-\alpha}k=\frac{(-\alpha)(-\alpha-1)(-\alpha-2)\dots(-\alpha-k+1)}{k!}$$If you have further questions, try the Wikipedia. $\endgroup$ – Simply Beautiful Art Oct 15 '16 at 0:17

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