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

Problem

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

Solution

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?

Solutions Collecting From Web of "Compute $\int_0^\infty \frac{dx}{1+x^3}$"

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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$\ds{\int_{0}^{\infty}{\dd x \over 1 + x^{3}}:\ {\large ?}}$

\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}

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

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
$$\int_0^\infty\frac{1}{x^3+1}\,dx=\lim_{r\to\infty}F(r)=\frac{\pi}{2\sqrt{3}}+\frac{\pi}{6\sqrt{3}}=\frac{2\pi}{3\sqrt{3}}.$$