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May 4th, 2013, 01:12 PM   #1
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Computation of an integral.

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This problem was encountered here.

Subject: computation of the integral $\int_{0}^{\infty}\frac{\cos(x)}{\sqrt[3]{x}}\;\mathbb{d}x$.

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We will integrate the function $f(z)=\frac{\mathbb{e}^{\mathbb{i}z}}{z^{p}}$ (0<p<1) over the next figure.

[attachment=0:29xm8yvr]ikk.png[/attachment:29xm8yvr]

$\oint_{\mathbb{T}}f(x)\;\mathbb{d}z=\int_{r}^{R}\f rac{\mathbb{e}^{\mathbb{i}x}}{x^p}\;\mathbb{d}x+\i nt_{0}^{1}\frac{\mathbb{e}^{\mathbb{i}\left(\mathb b{i}Rt+(1-t)R\right)}\left(\mathbb{i}R-R\right)}{\left(\mathbb{i}Rt+(1-t)R\right)^{p}}\;\mathbb{d}t+\mathbb{i}\int_{R}^{r }\frac{\mathbb{e}^{-y}}{\left(\mathbb{i}y\right)^{p}}\;\mathbb{d}y+\in t_{\frac{\pi}{2}}^{0}\frac{\mathbb{e}^{\mathbb{i}r }\mathbb{e}^{\theta \mathbb{i}}\mathbb{i}r\mathbb{e}^{\theta \mathbb{i}}}{r^{p}\mathbb{e}^{\mathbb{i}\theta p}}\;\mathbb{d}\theta=0$

Letting $R\to \infty$ and using the inequality $\left|\mathbb{i}Rt+(1-t)R\right|\geq \min_{0\leq t\leq 1 }\left|\mathbb{i}Rt+(1-t)R\right|=\frac{\sqrt{2}R}{2}$

$\hspace{250}\begin{eqnarray}\left|\int_{0}^{1}\fra c{\mathbb{e}^{\mathbb{i}\left(\mathbb{i}Rt+(1-t)R\right)}\left(\mathbb{i}R-R\right)}{\left(\mathbb{i}Rt+(1-t)R\right)^{p}}\;\mathbb{d}t\right|=&\leq=&\int_{0}^{1}\frac{\mathbb{e}^{-Rt}\sqrt{2}R}{\left(\mathbb{i}Rt+(1-t)R\right)^{p}}\;\mathbb{d}t\\=&\leq=&\int_{0}^{1}\frac{\mathbb{e}^{-Rt}\sqrt{2}R}{R^{p\frac{\sqrt{2}R}{2}}}\;\mathbb{d }t\\=&2R^{-p}\int_{0}^{1}\mathbb{e}^{-Rt}\;\mathbb{d}t\\=&\to=&0\end{eqnarray}=$

so $\lim_{R\to\infty}\int_{\sigma_{2}}f(z)\;\mathbb{d} z=0$. Since $z\frac{\mathbb{e}^{\mathbb{i}z}}{z^p}\to 0$ as $r\to 0$, $\lim_{r\to 0 }\int_{\sigma_{4}}f(z)\;\mathbb{d}z=0$. So as $R\to\infty$ we are left with,

$\textrm{P.V.}\int_{0}^{\infty}\frac{\cos(x)+\mathb b{i}\sin(x)}{x^p}\;\mathbb{d}x-\textrm{P.V.}\int_{0}^{\infty}\mathbb{i}\frac{\mat hbb{e}^{-y}}{\mathbb{i}^{p}y^{p}}\;\mathbb{d}y=0$

and since the last two integrals converge absolutely we can omit the P.V. (principal value) and knowing that $\int_{0}^{\infty}\mathbb{e}^{-y}y^{-p}\;\mathbb{d}y=\Gamma(1-p)$ we get

$\int_{0}^{\infty}\frac{\cos(x)+\mathbb{i}\sin(x)}{ x^{p}}\;\mathbb{d}x=\mathbb{i}\left(\cos\left(\fra c{\pi p }{2}\right)-\mathbb{i}\sin\left(\frac{ \pi p}{2}\right)\right)\Gamma(1-p)$.

So $\begin{tabular}{|c|c|}\hline \int_{0}^{\infty}\frac{\cos(x)}{x^p}\;\mathbb{d}x= \sin\left(\frac{\pi p}{2}\right)\Gamma(1-p)&\int_{0}^{\infty}\frac{\sin(x)}{x^{p}}\;\mathbb {d}x=\cos\left(\frac{\pi p}{2}\right)\Gamma(1-p)\\\hline\end{tabular}$ and for p=1/3 we get $\int_{0}^{\infty}\frac{\cos(x)}{\sqrt[3]{x}}\;\mathbb{d}x=\sin\left(\frac{\pi}{6}\right)\G amma\left(\frac{2}{3}\right)=\frac{1}{2}\Gamma\lef t(\frac{2}{3}\right)$.
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