Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

3 Answers 3

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 $$ \int_{0}^{\infty} \frac{x^{2}}{1+x^{5}}\;\mathrm{d}x=\frac{2\pi i}{5(e^{2\pi i/5}-e^{-2\pi i/5})}=\frac{\pi}{5\sin(2\pi/5)} $$

share|improve this answer
1  
$\csc (2\pi /5)=\sqrt{2-2/\sqrt{5}}$, 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
1  
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 $$ 1 + x^5 = (x + 1)(x - \omega)(x - \omega^3)(x - \omega^7)(x - \omega^9) $$

share|improve this answer

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

share|improve this answer
    
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 at 20:06
    
@ChrisK Thanks. There was a typo. It's the integration in the $\leftarrow$ segment. Fixed. –  Felix Marin Jun 24 at 20:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.