# CVRC9

Evaluate $\int _{0}^{\infty }{\frac {dx}{x^{3}+1}}\,$

Consider the following four curves where $R>\epsilon >0$:

$\gamma _{1}:[0,R]\to {\mathbb {C}}$ where $\gamma _{1}(t)=t$

$\gamma _{2}:[0,{\frac {2\pi }{3}}]\to {\mathbb {C}}$ where $\gamma _{2}(t)=Re^{{it}}$

and

$\gamma _{3}:[0,R]\to {\mathbb {C}}$ where $\gamma _{3}(t)=te^{{{\frac {2\pi i}{3}}}}$

We want to integrate ${\frac {1}{z^{3}+1}}$ over the closed contour $C=\gamma _{1}+\gamma _{2}-\gamma _{3}$.

By the residue theorem we know that

$\int _{C}{\frac {dz}{z^{3}+1}}=2\pi \;res_{{z=e^{{{\frac {\pi i}{3}}}}}}\left({\frac {1}{z^{3}+1}}\right)$


A very useful lemma is if $f\,g$ are holomorphic functions defined on an open set that contains the point $z_{0}$, and $z_{0}$ is a simple zero of $g$ then $res_{{z=z_{0}}}{\frac {f}{g}}={\frac {f(z_{0})}{g^{\prime }(z_{0})}}$.

So we have $\int _{C}{\frac {dz}{z^{3}+1}}=2\pi i{\frac {1}{3{\left(e^{{{\frac {\pi i}{3}}}}\right)}^{2}+0}}={\frac {2\pi i}{3}}e^{{-{\frac {2\pi i}{3}}}}$

Now we integrate over each part of the contour, and then take the limit as $R$ goes to infinity.

$\int _{{\gamma _{1}}}{\frac {dz}{z^{3}+1}}=\int _{0}^{R}{\frac {dt}{t^{3}+1}}$

The $p$-test from single variable calculus tells us that this integral converges as $R$ goes to infinity. Therefore we simply have

$\lim _{{R\to \infty }}\int _{{\gamma _{1}}}{\frac {dz}{z^{3}+1}}=\int _{0}^{\infty }{\frac {dt}{t^{3}+1}}$.

Next we see that

$\left|\int _{{\gamma _{2}}}{\frac {dz}{z^{3}+1}}\right|=\left|\int _{0}^{{{\frac {2\pi }{3}}}}{\frac {Rie^{{it}}dt}{{\left(Re^{{it}}\right)}^{3}+1}}\right|\leq \int _{0}^{{{\frac {2\pi }{3}}}}\left|{\frac {Rie^{{it}}}{R^{3}e^{{3it}}+1}}\right|dt$

For all $R$ large enough we have $R^{3}-1>{\frac {1}{2}}R^{3}$. Then for a $R$ large enough it's easy to see $|R^{3}e^{{3it}}+1|=|R^{3}+e^{{-3it}}|\geq |R^{3}-1|>{\frac {1}{2}}R^{3}$. Then getting back to out integral we have

$\left|\int _{{\gamma _{2}}}{\frac {dz}{z^{3}+1}}\right|\leq \int _{0}^{{{\frac {2\pi }{3}}}}{\frac {|Ri||e^{{it}}|dt}{|R^{3}e^{{3it}}+1|}}<\int _{0}^{{{\frac {2\pi }{3}}}}{\frac {Rdt}{{\frac {1}{2}}R^{3}}}={\frac {2}{R^{2}}}\int _{0}^{{{\frac {2\pi }{3}}}}dt={\frac {4\pi }{3R^{2}}}$

Taking the limit as both as $R$ tend to infinity

$\lim _{{R\to \infty }}\left|\int _{{\gamma _{2}}}{\frac {dz}{z^{3}+1}}\right|\leq 0\Rightarrow \lim _{{R\to \infty }}\int _{{\gamma _{2}}}{\frac {dz}{z^{3}+1}}=0$

Now we have just one more curve to integrate.

$\int _{{\gamma _{3}}}{\frac {dz}{z^{3}+1}}=\int _{0}^{R}{\frac {e^{{{\frac {2}{3}}\pi i}}\ dt}{t^{3}+1}}$

Again taking the limit as $R$ goes to infinity we get

$\lim _{{R\to \infty }}\int _{{\gamma _{3}}}{\frac {dz}{z^{3}+1}}=e^{{{\frac {2}{3}}\pi i}}\int _{0}^{\infty }{\frac {dt}{t^{3}+1}}$.

i.e. $\ e^{{{\frac {2}{3}}\pi i}}\$ times the original integral

Therefore

$\int _{{C}}{\frac {dz}{z^{3}+1}}=(1-e^{{{\frac {2}{3}}\pi i}})\int _{0}^{\infty }{\frac {dx}{x^{3}+1}}=(1-e^{{{\frac {2}{3}}\pi i}})I$

inserting the result of the integral on the left hand side we get

${\frac {2\pi i}{3}}e^{{-{\frac {2\pi i}{3}}}}=(1-e^{{{\frac {2}{3}}\pi i}})I=e^{{{\frac {1}{3}}\pi i}}(e^{{-{\frac {1}{3}}\pi i}}-e^{{{\frac {1}{3}}\pi i}})I=e^{{{\frac {1}{3}}\pi i}}\ 2i\ \sin(-{\frac {\pi }{3}})I$

so that

${\frac {\pi }{3}}e^{{-{\frac {2\pi i}{3}}}}=e^{{{\frac {1}{3}}\pi i}}\ \sin(-{\frac {\pi }{3}})I$

${\frac {\pi }{3}}e^{{-\pi i}}=-{\frac {{\sqrt {3}}}{2}}I$

Eventually,

$I=\int _{0}^{\infty }{\frac {dx}{x^{3}+1}}={\frac {2\pi }{3{\sqrt {3}}}}$