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Consider the following integral: $$\int_{0}^{\infty} \frac{x^{2}}{1+x^{5}} \mathrm{d} x \>.$$

I did the following: Since $-1$ is a pole on the real axis, I took $z_{1}=e^{3\pi/5}$ then constructed an arc between $Rz_{1}$ and $R$ :

$f(e^{2\pi i /5}z)=f(z)$ it follows that : $(1-e^{2pi i/5})\int_{\alpha}f(z)dz + \int_{\beta}f(z)dz = 2\pi i \text{ Res } z_{2} f $ ,
where $\alpha$ is the way along the eral axis, $\beta$ is the way along the arc and $z_{2}=e^{i\pi/5}$

This gives: $\int_{0}^{\infty}...dx = 2\pi i (\frac{e^{2i\pi/5}}{-5})\cdot \frac{1}{(1-e^{2\pi i/5})}$

The real part of the right side in wolframalpha does not give the same result as the left side.

Does anybody see why? Please do tell.

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What is your $f$? $1+(e^{3\pi i/5}z)^5=1-z^5$ so I am not seeing $f(e^{3\pi i/5}z)= f(z)$. – robjohn Nov 19 '11 at 23:50
Imaginary part of $2\pi i \frac{e^{2\pi i/5}}{-5}\frac{1}{1-e^{3\pi i/5}}$ is not zero. – Norbert Nov 19 '11 at 23:54
up vote 3 down vote accepted

I think you want to use the curve $e^{2\pi i/5}x$ for the inbound curve circling the poie at $e^{\pi i/5}$. In that case we get $$ (1-e^{6\pi i/5})\int_{0}^{\infty} \frac{x^{2}}{1+x^{5}}\;\mathrm{d}x=2\pi i\frac{1}{5e^{2\pi i/5}} $$ Which gives $$ \int_{0}^{\infty} \frac{x^{2}}{1+x^{5}}\;\mathrm{d}x=\frac{2\pi i}{5(e^{2\pi i/5}-e^{-2\pi i/5})}=\frac{\pi}{5\sin(2\pi/5)} $$

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$\csc (2\pi /5)=\sqrt{2-2/\sqrt{5}}$, in case the OP was comparing this answer with the output from WolframAlpha. – Jonathan Gleason Nov 20 '11 at 0:31
@Jonathan: Thanks. I should have checked Wolfram's form of the answer. – robjohn Nov 20 '11 at 0:37
robjohn, why $(1-e^{6\pi i /5})$ and not $(1-e^{2\pi i /5})$ ? Why for the residue : $\frac{1}{5e^{2\pi i/5}}$ and not : $\frac{(z_{2}^{2}}{5(z_{2})^{4}}$ with $z_{2}=e^{\pi i/5}$ ? – VVV Nov 20 '11 at 0:42
@VVV: we get 2 factors of $e^{2\pi i/5}$ from the $x^2$ and 1 factor from the $\mathrm{d}x$; that gives $e^{6\pi i/5}$. The residue is $2\pi i$ times $$\begin{align}\lim_{z\to e^{\pi i/5}}\frac{z^2(z-e^{\pi i/5})}{1+z^5}&=\lim_{z\to e^{\pi i/5}}\frac{z^2}{5z^4}\\&=\frac{1}{5e^{2\pi i/5}}\end{align}$$ – robjohn Nov 20 '11 at 0:53
When you use the path $z=x$ and the path $z=e^{2pi i/5}x$ you substitute $x$ and $e^{2pi i/5}x$ for every occurrence of $z$. Thus, $\int_C\frac{z^2}{1+z^5}\;\mathrm{d}z$ becomes $\int_0^\infty\frac{x^2}{1+x^5}\;\mathrm{d}x$ on the curve $z=x$ and $\int_0^\infty\frac{e^{4\pi i/5}x^2}{1+x^5}\;e^{2\pi i/5}\;\mathrm{d}x$ on the curve $z=e^{2\pi i/5}x$ (but then we subtract since the curve is inbound). – robjohn Nov 20 '11 at 1:29

I ran out of patience, but the partial fraction approach can give an indefinite integral in closed form, as the tenth roots of 1 have real and imaginary parts that can be written with nothing worse than square root signs. I got $$ \frac{x^2}{1 + x^5} \; = \; \; \frac{A}{x + 1} + \frac{B_1 x + B_0}{x^2 - \frac{1}{2}(\sqrt 5 + 1)x + 1} + \frac{C_1 x + C_0}{x^2 + \frac{1}{2}(\sqrt 5 - 1)x + 1}.$$ Then I got $$ A = \frac{1}{3}, \; B_0 = B_1 = \frac{1}{6}(\sqrt 5 - 1), \; \; C_0 = C_1 = -\frac{1}{6}(\sqrt 5 + 1).$$ The two quadratic denominators terms are positive over the reals. The rest of the task is also cookbook, you rewrite, for example $B_1 x + B_0$ as $\frac{B_1}{2}$ times the derivative of $x^2 - \frac{1}{2}(\sqrt 5 + 1)x + 1$ plus a constant term , call it $B_3.$ So, for this part, you get the indefinite integral being $$ \frac{B_1}{2} \log(x^2 - \frac{1}{2}(\sqrt 5 + 1)x + 1) + B_3 \arctan(linear) $$ I imagine you can force your computer algebra system to show all the steps in this calculation while getting the specific numbers right. The beginning is to define $$ \omega = e^{\pi i / 5} = \frac{1}{4}(1 + \sqrt 5) + \frac{i}{4} \sqrt{10 - 2 \sqrt 5} $$ with $$ 1 + x^5 = (x + 1)(x - \omega)(x - \omega^3)(x - \omega^7)(x - \omega^9) $$

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ You can avoid the 5 poles by integrating \begin{align} \overbrace{\int_{0}^{\infty}{x^{2} \over 1 + x^{5}}\,\dd x} ^{\ds{\mbox{Set}\ x \equiv t^{1/5}}}&\ =\ \color{#66f}{{1 \over 5}\int_{0}^{\infty}{t^{-2/5} \over t + 1}\,\dd t} ={1 \over 5}\,2\pi\ic\expo{-2\pi\ic/5} -{1 \over 5}\int_{\infty}^{0}{t^{-2/5}\expo{-4\pi\ic/5} \over t + 1}\,\dd t \end{align}

such that \begin{align} \color{#66f}{\large\int_{0}^{\infty}{x^{2}\,\dd x \over 1 + x^{5}}} ={1 \over 5}\,2\pi\ic\,{\expo{-2\pi\ic/5} \over 1 - \expo{-4\pi\ic/5}} ={1 \over 5}\,2\pi\ic\,{1 \over \expo{2\pi\ic/5} - \expo{-2\pi\ic/5}} =\color{#66f}{\large{\pi \over 5\sin\pars{2\pi/5}}} \end{align} The integration contour is the following one: enter image description here

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should it not read $$+1/5\cdot \int_{0}^{\infty} \frac{t^{-2/5}e^{-4\pi i/5}}{t+1} dt$$ (as opposed to a "minus"). – Chris K Jun 24 '14 at 20:06
@ChrisK Thanks. There was a typo. It's the integration in the $\leftarrow$ segment. Fixed. – Felix Marin Jun 24 '14 at 20:29

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