Is there any special convergence factor like $\frac{1}{x}\to\int \limits_0^\infty e^{-xy}\mathrm{d}y$ that can transform the $\frac{1}{x^{1/3}}$? I don't mean $\int \limits_0^\infty e^{-x^{1/3}y}\mathrm{d}y$.
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Are you asking if it converges or for the sum it tends to? – Rab Aug 29 '17 at 15:39
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Mathematica gives: $\frac{\pi }{\Gamma \left(\frac{1}{3}\right)}$. – David G. Stork Aug 29 '17 at 15:43
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the value is $$\Gamma(2/3)\sin(\frac{\pi}{6})$$ – Guy Fsone Aug 29 '17 at 15:48
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@GuyFsone: I believe it should be $\Gamma!\left(\frac23\right)\sin\left(\frac\pi3\right)$. – robjohn Aug 29 '17 at 16:00
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you are right thanks for the remarks – Guy Fsone Aug 30 '17 at 06:11
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Use the contour $\gamma=[0,R]\cup Re^{i[0,\pi/2]}\cup[iR,0]$ as $R\to\infty$. There are no singularities inside $\gamma$ so $$ \int_\gamma e^{iz}\,z^{-1/3}\,\mathrm{d}z=0 $$ Furthermore, setting $z=Re^{it}$, we get $$ \begin{align} \left|\,\int_{Re^{i[0,\pi/2]}} e^{iz}\,z^{-1/3}\,\mathrm{d}z\,\right| &\le\int_0^{\pi/2}e^{-R\sin(t)}R^{2/3}\,\mathrm{d}t\\ &\le\int_0^{\pi/2}e^{-2Rt/\pi}R^{2/3}\,\mathrm{d}t\\ &\le R^{2/3}\int_0^\infty e^{-2Rt/\pi}\,\mathrm{d}t\\[6pt] &=\frac\pi2R^{-1/3} \end{align} $$ which tends to $0$ as $R\to\infty$.
Therefore, since the integral along the curve at infinity vanishes, we get $$\newcommand{\Im}{\operatorname{Im}} \begin{align} \int_0^\infty\sin(x)\,x^{-1/3}\,\mathrm{d}x &=\Im\left(\int_0^\infty e^{ix}\,x^{-1/3}\,\mathrm{d}x\right)\\ &=\Im\left(\int_\gamma e^{iz}\,z^{-1/3}\,\mathrm{d}z+\int_0^{i\infty}e^{ix}\,x^{-1/3}\,\mathrm{d}x\right)\\ &=\Im\left(0+e^{i\pi/3}\int_0^\infty e^{-x}\,x^{-1/3}\,\mathrm{d}x\right)\\[3pt] &=\sin(\pi/3)\,\Gamma\!\left(\frac23\right)\\[3pt] &=\frac{\sqrt3}2\,\Gamma\!\left(\frac23\right) \end{align} $$
robjohn
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You can't use $\gamma$ hitting zero, of course, but that can be worked around easily enough. – Thomas Andrews Aug 29 '17 at 16:21
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@ThomasAndrews: It can be worked around easily, but since $x^{-1/3}$ is integrable near $0$, is it really necessary? I guess for Cauchy's Theorem, it might be. – robjohn Aug 29 '17 at 16:23
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In general, for $a$, $s>0$ we have $$\int_{0}^{\infty}e^{- a x} t^s \frac{ d x}{x}= \frac{\Gamma(s)}{a^s} $$ This can be extended to $a$,$s \in \mathbb{C}$, $\Re a (,\Re s) >0$ by considering $a^s = e^{s \log a}$, with $\log$ the principal branch on the right half plane. Therefore we have $$\int_{0}^{\infty} e^{-(t+i)x} x^{\frac{2}{3}}\frac{dx}{x}= \frac{\Gamma(\frac{2}{3})}{\exp(\frac{2}{3}\log(t+i))}$$ As $t\to 0_{+}$, LHS converges to $\int_{0}^{\infty}(\cos x -i \sin x) x^{-\frac{1}{3}} d x$ ( this has to be analyzed with some care), while the RHS clearly converges to $\frac{\Gamma(\frac{2}{3})}{\exp(\frac{2}{3}\cdot i \frac{\pi}{2})}=\Gamma(\frac{2}{3})(\cos \frac{\pi}{3} -i \sin \frac{\pi}{3})$
orangeskid
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