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How can I calculate the integral $\displaystyle{\int_{0}^{\infty} \frac{1}{x^5+1} dx}$?

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at it currently stands, this question does not mean the standard of questions asked on this site as it lacks context i.e. what you have tried yourself. –  Lost1 Jun 25 at 22:50
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Do you need to evaluate the integral, or just check whether it converges? The first question can be answered by an appropriate contour integral - this is an exercise in Lang's complex analysis, IIRC. –  user61527 Jun 25 at 22:54
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@Lost1: Alpha does the indefinite integral, but it isn't pretty. The definite integral comes out $ 1/5 \Gamma(1/5) \Gamma(4/5)\approx 1.06896$ –  Ross Millikan Jun 25 at 23:08
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@RossMillikan $$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin \pi z},$$ so the two expressions are equal. –  Daniel Fischer Jun 25 at 23:35
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I believe it is definitely possible to compute the indefinite integral by partial fraction decomposition into complex factor (involving the quintic roots of (-1)). But it'll be very tedious. –  Deepak Jun 25 at 23:42

4 Answers 4

Consider the contour integral of the integrand from 0 to R on the real axis, then a circle segment of radius R from R to R exp(2 pi i/5) and then along a straight line back from there to 0. The integral along this line is proportional to the one you want to evaluate. There is one pole at z = exp(pi i/5) located inside the contour, the residue is trivial to compute.

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The change of variable $\frac{1}{1+x^{5}}=u$ gives $dx=-\frac{u^{-1-\frac{1}{5}}(1-u)^{-\frac{4}{5}}}{5}du$

Hence our integral reduces to $$\frac{1}{5}\int_{0}^{1}{u^{\frac{4}{5}-1}(1-u)^{\frac{1}{5}-1}du}=\frac{1}{5}\beta(1-\frac{1}{5},\frac{1}{5})=\frac{1}{5}\Gamma(1-\frac{1}{5})\Gamma(\frac{1}{5})=\frac{1}{5}\frac{\pi}{\sin(\frac{\pi}{5})}$$ we can actually simplify things a bit more by considering the following identity, deduced by De'Moivres formula $$\sin(5x)=16\sin^{5}(x)-20\sin^{3}(x)+5\sin(x)$$ pluggin in $x=\frac{\pi}{5}$ combined with the fact that $\sin(\frac{\pi}{5})>0$ We get $$16\sin^{4}(\frac{\pi}{5})-20\sin^{2}(\frac{\pi}{5})+5=0$$ which we can begin by solving for $\sin^{2}(\frac{\pi}{5})$ and then taking the positive root. We will end up with $\sin(\frac{\pi}{5})=\sqrt{5-\sqrt{5}}$ hence your integral will finally evaluate at $$\frac{1}{5}\frac{\pi}{\sqrt{5-\sqrt{5}}}$$

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In this answer, it is shown, using contour integration, that $$ \frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right)=\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x $$ Plug in $n=0$ and $m=5$ to get $$ \int_0^\infty\frac1{1+x^5}\,\mathrm{d}x=\frac\pi5\csc\left(\frac\pi5\right) $$

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would the downvoter care to comment? –  robjohn Jun 26 at 22:01
    
@BillDubuque: communications on the net, especially between users of different native languages, can easily be misconstrued. Egos can easily be bruised. –  robjohn Aug 6 at 4:39

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \overbrace{\int_{0}^{\infty}{\dd x \over x^{5} + 1}} ^{\ds{\mbox{Set}\ x^{5} \equiv t}\ \imp\ x = t^{1/5}}&= \color{#00f}{{1 \over 5}\int_{0}^{\infty}{t^{-4/5} \over t + 1}\,\dd t} ={1 \over 5}\,2\pi\ic\expo{-4\pi\ic/5} -{1 \over 5}\int_{\infty}^{0}{t^{-4/5}\expo{-8\pi\ic/5} \over t^{5} + 1}\,\dd t \\[3mm]&={1 \over 5}\,2\pi\ic\expo{-4\pi\ic/5} + \expo{-8\pi\ic/5}\, \color{#00f}{{1 \over 5}\int_{0}^{\infty}{t^{-4/5} \over t^{5} + 1}\,\dd t} \end{align}

\begin{align} \int_{0}^{\infty}{\dd x \over x^{5} + 1}& =\color{#00f}{{1 \over 5}\int_{0}^{\infty}{t^{-4/5} \over t^{5} + 1}\,\dd t} ={1 \over 5}\,2\pi\ic\expo{-4\pi\ic/5}\,{1 \over 1 - \expo{-8\pi\ic/5}} ={\pi \over 5}\,{2\ic \over \expo{4\pi\ic/5} - \expo{-4\pi\ic/5}} \\[3mm]&={\pi \over 5}\,{2\ic \over 2\ic\sin\pars{4\pi/5}} \end{align}

$$ \color{#66f}{\large\int_{0}^{\infty}{\dd x \over x^{5} + 1}} ={\pi \over 5}\,\csc\pars{4\pi \over 5} =\color{#66f}{\large{\root{50 + 10\root{5}} \over 25}\,\pi}\approx {\tt 1.0690} $$

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