Compute $$\displaystyle \int_0^\infty \frac{dx}{1+x^3}.$$


I do partial fractions $$\frac{1}{x^3+1}= \frac{2-x}{3 \left( x^{2}-x+1 \right)}+\frac{1}{3 \left( x+1 \right)}.$$

But we could simplify the left one $$\frac{2-x}{3\left( x^{2}-x+1 \right)} = \frac{2}{3}\cdot \frac{1}{x^{2}-x+1}-\frac{x}{x^{2}-x+1}$$

From here, I do this see images.

But I find the wrong primitive functions. Why am I wrong, and how do I find the correct one?

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    $\begingroup$ Hint: you can use the $\beta$ functon. See here. $\endgroup$ – Mhenni Benghorbal Nov 7 '13 at 22:33
  • $\begingroup$ I posted this hint as an answer. Why has it been changed a comment? $\endgroup$ – Mhenni Benghorbal Nov 7 '13 at 22:34
  • $\begingroup$ Hint: you can use the $\beta$ functon. See here. $\endgroup$ – Mhenni Benghorbal Nov 7 '13 at 22:38
  • $\begingroup$ This is not a trivial answer!! It is a real answer. $\endgroup$ – Mhenni Benghorbal Nov 7 '13 at 22:39
  • $\begingroup$ How do you judge the answers? Look at some of the answers already have been posted. I think you need to distinguish between real short answers and trivial answers. $\endgroup$ – Mhenni Benghorbal Nov 7 '13 at 22:42

For the integrand $\frac{1}{x^3+1}$, use partial fractions $\frac{2-x}{3 (x^2-x+1)}+\frac{1}{3 (x+1)}$. Integrate the sum term by term and factor out constants: $$ I=\int \frac{1}{x^3+1} \operatorname{d}x =\frac{1}{3} \int \frac{2-x}{ x^2-x+1} \operatorname{d}x+\frac{1}{3} \int \frac{1}{x+1}\operatorname{d}x $$ Rewrite the integrand $\frac{2-x}{ x^2-x+1}$ as $\frac{3}{2( x^2-x+1)}-\frac{2x-1}{2( x^2-x+1)}$ $$I= \frac{1}{3} \int \frac{3}{2( x^2-x+1)}\operatorname{d}x-\frac{1}{3}\int\frac{2x-1}{2( x^2-x+1)} \operatorname{d}x+\frac{1}{3} \int \frac{1}{x+1}\operatorname{d}x=I_1+I_2+I_3$$ Integrate the sum term by term and factor out constants. For the integral $I_1=\frac{1}{3} \int \frac{3}{2( x^2-x+1)}\operatorname{d}x$, substitute $u = x^2-x+1$ and $\operatorname{d}u = (2 x-1)\operatorname{d}x$ $$I_1 = -\frac{1}{6} \int \frac{1}{u} \operatorname{d}u=-\frac{1}{6}\ln u = -\frac{1}{6}\ln(x^2-x+1)$$ For the integral $I_2=\frac{1}{3}\int\frac{2x-1}{2( x^2-x+1)} \operatorname{d}x$, complete the square $$ I_2 = \frac{1}{2} \int \frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}} $$ and substitute $s = x-1/2$ and $ \operatorname{d}s = \operatorname{d}x$ $$I_2= \frac{1}{2} \int\frac{1}{s^2+3/4} \operatorname{d}s=\frac{2}{3} \int \frac{1}{\frac{4s^2}{3}+1} \operatorname{d}s$$ Substitute $p = 2 s/\sqrt 3$ and $\operatorname{d}p = \frac{2}{\sqrt 3} \operatorname{d}s$ so that $$I_2= \frac{1}{\sqrt 3} \int\frac{1}{p^2+1} dp=\frac{1}{\sqrt 3}\arctan p=\frac{1}{\sqrt 3}\arctan \left(\frac{2x-1}{\sqrt 3} \right)$$ For the integral $I_3=\frac{1}{3} \int \frac{1}{x+1}\operatorname{d}x$, substitute $w = x+1$ and $dw = dx$: $$I_3 = \frac{1}{3} \int \frac{1}{w}\operatorname{d}w=\frac{1}{3}\ln w=\frac{1}{3}\ln (x+1)$$ Finally
$$ I(x)= -\frac{1}{6}\ln(x^2-x+1)+\frac{1}{\sqrt 3}\arctan \left(\frac{2x-1}{\sqrt 3} \right) +\frac{1}{3}\ln (x+1)+\textrm{constant}=\frac{1}{3}\ln\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)+\frac{1}{\sqrt 3}\arctan \left(\frac{2x-1}{\sqrt 3} \right) +\textrm{constant} $$ taking into account that for the logarithms: $$ -\frac{1}{6}\ln(x^2-x+1)=\frac{1}{3}\frac{1}{2}\ln\frac{1}{(x^2-x+1)}=\frac{1}{3}\ln\frac{1}{(x^2-x+1)^{\frac{1}{2}}}=\frac{1}{3}\ln\left(\frac{1}{\sqrt{x^2-x+1}}\right) $$ so that $$-\frac{1}{6}\ln(x^2-x+1)+\frac{1}{3}\ln (x+1)=\frac{1}{3}\ln\left(\frac{1}{\sqrt{x^2-x+1}}\right)+\frac{1}{3}\ln (x+1)=\frac{1}{3}\ln\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)$$ The integral is $$ \int_0^{+\infty}\frac{1}{x^3+1}\operatorname{d}x=\left[\lim_{x\to\infty}I(x)\right]-I(0)= \frac{2\pi}{3\sqrt 3} $$ Infact $$ \lim_{x\to\infty}I(x)=\lim_{x\to\infty}\frac{1}{3}\ln\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)+\lim_{x\to\infty}\frac{1}{\sqrt 3}\arctan \left(\frac{2x-1}{\sqrt 3} \right) $$ and the first limit is $0$ because for $x\rightarrow\infty$ the ratio $\frac{x+1}{\sqrt{x^2-x+1}}\sim 1$ and then $\ln(1)=0$; the second limit is $\frac{1}{\sqrt 3}\frac{\pi}{2}$ because for $x\rightarrow\infty$ the function $\arctan \left(\frac{2x-1}{\sqrt 3} \right)\sim\arctan(x)$ and goes to $\pi/2$.

For $I(0)$ you have $$ I(0)=\frac{1}{3}\ln\left(\frac{0+1}{\sqrt{0^2-0+1}}\right)-\frac{1}{\sqrt 3}\arctan \left(\frac{2\cdot 0-1}{\sqrt 3} \right) =\frac{1}{3}\ln(1)-\frac{1}{\sqrt 3}\arctan \left(\frac{2\cdot 0-1}{\sqrt 3} \right)=-\frac{1}{\sqrt 3}\arctan \left(\frac{-1}{\sqrt 3} \right) $$ observing that $\arctan(1/x)+\arctan(x)=\pi/2$ and that $\arctan(-x)=-\arctan(x)$ and that $\arctan(\sqrt 3)=\pi/3$ one has $$ I(0)=-\frac{1}{\sqrt 3}\arctan \left(\frac{-1}{\sqrt 3} \right)=\frac{1}{\sqrt 3}\left(\frac{\pi}{2}-\frac{\pi}{3}\right) $$ Finally putting all together $$ \int_0^{+\infty}\frac{1}{x^3+1}\operatorname{d}x=\left[\lim_{x\to\infty}I(x)\right]-I(0)=\frac{\pi}{2\sqrt 3}+\frac{1}{\sqrt 3}\left(\frac{\pi}{2}-\frac{\pi}{3}\right)=\frac{2\pi}{3\sqrt 3} $$

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  • $\begingroup$ This is very well written! Thank you. Just from the beginning, where you write "Rewrite the integrand" ang get I1, I2, and I3 it really smells like arctan, ln and ln – respectively. Questions: Q1. WolframAlpha gets $I_1$ being arctan... and the thing I noticed was that i got $\frac{1}{u} \frac{du}{2x-1}$ Q2. When you write "complete the square" the numerator has no $2x-1$ suddenly. Q3. After writing this, I realise that maybe you just swapped the order and first solved I2 then I1 then I3. $\endgroup$ – jacob Nov 8 '13 at 20:40
  • $\begingroup$ Ok, I have done the algebra and I see why you are correct. But how do you do the last step, when you go from $$\frac{1}{3}\ln\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)+\frac{1}{\sqrt 3}\arctan \left(\frac{2x-1}{\sqrt 3} \right)$$ to $$\int_0^{\infty}\frac{1}{x^3+1}\operatorname{d}x=\frac{2\pi}{3\sqrt 3}$$ I can't see how those limits would play out. $\endgroup$ – jacob Nov 9 '13 at 17:09
  • $\begingroup$ I added more explanations. $\endgroup$ – alexjo Nov 11 '13 at 17:20

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{\dd x \over 1 + x^{3}}:\ {\large ?}}$

\begin{align} &\color{#c00000}{\int_{0}^{\infty}{\dd x \over 1 + x^{3}}} =\int_{0}^{\infty}\int_{0}^{\infty}\expo{-\pars{1 + x^{3}}t}\,\dd t\,\dd x =\int_{0}^{\infty}\expo{-t}\ \overbrace{\int_{0}^{\infty}\expo{-t\,x^{3}}\,\dd x} ^{\ds{tx^{3}\equiv\xi\imp x=t^{-1/3}\xi^{1/3}}}\ \,\dd t \\[3mm]&=\int_{0}^{\infty}\expo{-t} \int_{0}^{\infty}\expo{-\xi}\,t^{-1/3}\,{1 \over 3}\,\xi^{-2/3}\,\dd\xi\,\dd t \\[3mm]&={1 \over 3}\pars{\int_{0}^{\infty}t^{-1/3}\expo{-t}\,\dd t} \pars{\int_{0}^{\infty}\xi^{-2/3}\expo{-\xi}\,\dd\xi} ={1 \over 3}\,\Gamma\pars{2 \over 3}\Gamma\pars{1 \over 3} \end{align} where $\ds{\Gamma\pars{z}}$ is the Gamma Function ${\bf\mbox{6.1.1}}$.

With Euler Reflection Formula ${\bf\mbox{6.1.17}}$: \begin{align} &\color{#c00000}{\int_{0}^{\infty}{\dd x \over 1 + x^{3}}} ={1 \over 3}\,{\pi \over \sin\pars{\pi/3}} ={1 \over 3}\,{\pi \over \root{3}/2} \end{align}

$$\color{#00f}{\large% \int_{0}^{\infty}{\dd x \over 1 + x^{3}} = {2\root{3} \over 9}\,\pi} \approx 1.2092 $$

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Since $$ \frac{1}{x^3+1}=\frac{1}{(x+1)(x^2-x+1)}, $$ we can find some $a,b,c \in \mathbb{R}$ such that $$ \frac{1}{x^3+1}=\frac{a}{x+1}+\frac{bx+c}{x^2-x+1}. $$ A simple computation shows that $$ a=-b=\frac{c}{2}=\frac13, $$ i.e. \begin{eqnarray} \frac{1}{x^3+1}&=&\frac13\cdot\frac{1}{x+1}-\frac13\cdot\frac{x-2}{x^2-x+1}\\ &=&\frac13\cdot\frac{1}{x+1}-\frac16\cdot\frac{2x-1}{x^2-x+1}+\frac12\cdot\frac{1}{x^2-x+1}\\ &=&\frac13\cdot\frac{1}{x+1}-\frac16\cdot\frac{2x-1}{x^2-x+1}+\frac12\cdot\frac{1}{\left(x-\frac12\right)^2+\frac34}. \end{eqnarray} It follows that \begin{eqnarray} F(r)&=&\int_0^r\frac{1}{x^3+1}\,dx=\frac13\ln(1+r)-\frac16\ln(r^2-r+1)+\frac{1}{\sqrt{3}}\arctan\frac{2r-1}{\sqrt{3}}+\frac{1}{\sqrt{3}}\arctan\frac{1}{\sqrt{3}}\\ &=&\frac16\ln\frac{(r+1)^2}{r^2-r+1}+\frac{1}{\sqrt{3}}\arctan\frac{2r-1}{\sqrt{3}}+\frac{\pi}{6\sqrt{3}}\\ &=&\frac16\ln\frac{r^2+2r+1}{r^2-r+1}+\frac{1}{\sqrt{3}}\arctan\frac{2r-1}{\sqrt{3}}+\frac{\pi}{6\sqrt{3}}. \end{eqnarray} Thus $$ \int_0^\infty\frac{1}{x^3+1}\,dx=\lim_{r\to\infty}F(r)=\frac{\pi}{2\sqrt{3}}+\frac{\pi}{6\sqrt{3}}=\frac{2\pi}{3\sqrt{3}}. $$

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