# Contour integration complex numbers

$$\int_{0}^{\infty} 1/(x^3+1) dx$$ 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 $$\int_{0}^{\infty} 1/(x^4+1) dx$$.

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

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

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 $$z=e^{\pi i/n}$$.

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$$\mathfrak{I}(\color{brown}{3})=\int\limits_0^{\infty}\frac {\mathrm dx}{1+x^{\color{brown}{3}}}=\frac {\pi}{\color{brown}{3}}\csc\left(\frac {\pi}{\color{brown}{3}}\right)\color{red}{=\frac {2\pi}{3\sqrt3}}$$