CVRC9

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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}}}}