# Calculating $\int_0^\infty \frac{1}{x^\alpha}\frac{1}{1+x}\,{\rm d}x$

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

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.

• Hi Ron, I was more concerned about how to get the solution rather than the convergence! Thanks for the info though. – user2850514 Oct 14 '16 at 20:33
• @user2850514: oh. – Ron Gordon Oct 14 '16 at 20:33
• Oh, all the things I would never think of. – Simply Beautiful Art Oct 14 '16 at 20:46
• I think the $\epsilon$'s in the exponents in your fourth integral should be $\alpha$'s. – Michael Seifert Oct 14 '16 at 20:49
• @MichaelSeifert: thanks for the catch. – Ron Gordon Oct 14 '16 at 20:50

\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: $$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$$

• Oh, what a sweet treat! 3 steps beautiful to my eyes. – Simply Beautiful Art Oct 15 '16 at 0:01
• @SimpleArt, thanks. – poweierstrass Oct 15 '16 at 6:13

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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}

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

• Digamma function $\psi$ is our friend ;) (+1). – Olivier Oloa Oct 15 '16 at 0:09
• @OlivierOloa Thanks. It's true. – Felix Marin Oct 15 '16 at 0:15

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 $C = 0$.

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.

• Switch like this without to know if this is possible? Check that .. – ParaH2 Oct 14 '16 at 21:03
• You are assuming that $\alpha$ is integer by using the binomial expansion. – Zaid Alyafeai Oct 14 '16 at 23:58
• @ZaidAlyafeai Nah, its called generalized binomial expansion. nifty. – Simply Beautiful Art Oct 14 '16 at 23:59
• what is the definition of $$\binom{-\alpha}k$$ – Zaid Alyafeai Oct 15 '16 at 0:09
• @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. – Simply Beautiful Art Oct 15 '16 at 0:17