$$\int_{0}^{\infty} 1/(x^3+1) dx$$ the question says use complex functions to answer this. I tried using Cauchy residue theorem and this $$1/2\int_{-\infty}^{\infty} 1/(x^3+1) dx=\int_{0}^{\infty} 1/(x^3+1) dx$$ So far i got that the poles are π/3,π.. but only π/3 matters as is it is inside the contour of integration. I then used cauchy residue theorem and did not get the answer of 2π/(3sqrt3). Is this the method i should be using? or is there an alternative way using complex functions. The intuition i used for this exercise was by looking at an example with $$\int_{0}^{\infty} 1/(x^4+1) dx$$.

  • $\begingroup$ BTW: if you do \infty you get $\infty$. $\endgroup$ – Dave Dec 26 '18 at 23:01

You made an error here:

$$ \frac12 \int^\infty_\infty \frac1{1 + x^3} \stackrel{??}{=} \int_0^\infty \frac1{1 + x^3}$$

This is not valid in this case because $\frac1{1 + x^3} \neq \frac1{1 + (-x)^3}$ in general (this is what was used for $1/(1 + x^4)$ - there it works because $x^4$ is even).

The trick I know for doing this integral is to integrate along the border of a circle segment with angle $2 \pi/3$ (see Frank's answer).


Let's generalize this by denoting$$\mathfrak{I}(n)=\int\limits_0^{\infty}\frac {\mathrm dx}{1+x^n}$$where $n\geq2$ and let$$f(z)=\frac 1{1+z^n}$$The trick for $\mathfrak{I}(n)$ is to integrate it along a sector instead of a circle or half circle. Let the radius of the sector be $R$ and the arc integral to be denoted as $\Gamma_R$. Since we only want to enclose one singularity, we'll have the central angle be $\theta=\frac {2\pi}n$ so the only singularity is at $z=e^{\pi i/n}$.

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Integrating about the contour, we have$$\begin{align*}\oint\limits_{\mathrm C}\mathrm dz\, f(z) & =\int\limits_0^R\mathrm dx\, f(x)+\int\limits_{\Gamma_R}\mathrm dz\, f(z)+\int\limits_R^0\mathrm dz\, f(z)e^{\pi i/n}\\ & =(1-e^{\pi i/n})\int\limits_0^R\mathrm dx\, f(x)+\int\limits_{\Gamma_R}\mathrm dz\, f(z)\end{align*}$$

The arc integral vanishes as $R$ tends towards infinity. This can be verified using the estimation lemma$$\left|\,\int\limits_{\Gamma_R}\mathrm dz\, f(z)\,\right|\leq ML$$where $L$ is the length of the contour and $|f(z)|$ is bounded by a maximum $M$. The length of the arc can be easily calculated to be $L=\tfrac {2\pi R}n$. And through the triangle inequality, the max can be found$$|z|^n=\left|z^n\right|=\left|z^n+1-1\right|\leq\left|z^n+1\right|+1$$Since $|z|=R$, then we have that$$\left|\,\int\limits_{\Gamma_{R}}\mathrm dz\, f(z)\,\right|\leq\frac {1}{R^n-1}\frac {2\pi R}n\xrightarrow{R\,\to\,\infty}0$$

Hence$$\oint\limits_{\mathrm C}\mathrm dz\, f(z)=(1-e^{\pi i/n})\int\limits_0^{\infty}\mathrm dx\, f(x)$$

The contour integral is equal to $2\pi i$ times the sum of its residues. Fortunately for us, due to our clever thinking in the beginning, we've managed to only encase one singularity of $f(z)$ within our contour: $z=e^{\pi i/n}$. Therefore, the residue can be calculated as$$\begin{align*}\operatorname*{Res}_{z\,=\, e^{\pi i/n}}f(z) & =\lim\limits_{z\to e^{\pi i/n}}\frac {z-e^{\pi i/n}}{1+z^n}\\ & =\lim\limits_{z\to e^{\pi i/n}}\frac z{nz^n}\\ & =-\frac {e^{\pi i/n}}n\end{align*}$$Note that above, I have used L Hopital's rule as a shortcut. Putting everything together$$\begin{align*}\int\limits_0^{\infty}\frac {\mathrm dx}{1+x^n} & =-\frac {2\pi i}n\frac {e^{\pi i/n}}{1-e^{2\pi i/n}}\\ & =\frac {\pi}n\frac {2i}{e^{\pi i/n}-e^{-\pi i/n}}\\ & \color{blue}{\,=\frac {\pi}n\csc\left(\frac {\pi}n\right)}\end{align*}$$

The integral under question is simply the case when $n=3$. Therefore$$\mathfrak{I}(\color{brown}{3})=\int\limits_0^{\infty}\frac {\mathrm dx}{1+x^{\color{brown}{3}}}=\frac {\pi}{\color{brown}{3}}\csc\left(\frac {\pi}{\color{brown}{3}}\right)\color{red}{=\frac {2\pi}{3\sqrt3}}$$


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