How to integrate $ \int_0^\infty \sin x \cdot x ^{-1/3} dx$ (using Gamma function)

Solution 1:

Hint: Use the integral expression for the $\Gamma$ function in conjunction with Euler's formula.

Solution 2:

As a generalization, for $0\lt a\lt1$, $$ \begin{align} \int_0^\infty x^{a-1}\sin(x)\,\mathrm{d}x &=\frac1{2i}\left(\int_0^\infty x^{a-1}e^{ix}\,\mathrm{d}x-\int_0^\infty x^{a-1}e^{-ix}\,\mathrm{d}x\right)\\ &=\frac1{2i}\left(e^{ia\pi/2}\int_0^\infty x^{a-1}e^{-x}\,\mathrm{d}x-e^{-ia\pi/2}\int_0^\infty x^{a-1}e^{-x}\,\mathrm{d}x\right)\\ &=\sin\left(\frac{a\pi}{2}\right)\Gamma(a)\tag{1} \end{align} $$ The changes of variables $x\mapsto ix$ and $x\mapsto-ix$ used in the second equation above, are justified because $$ \int_{\gamma_k} z^{a-1}e^{-z}\,\mathrm{d}z=0\tag{2} $$ where $\gamma_1=[0,R]\cup Re^{i\frac\pi2[0,1]}\cup iR[1,0]$ and $\gamma_2=[0,R]\cup Re^{-i\frac\pi2[0,1]}\cup-iR[1,0]$ contain no singularities.

Plugging $a=\frac23$ into $(1)$ and using Euler's reflection formula, gives $$ \begin{align} \int_0^\infty x^{-1/3}\sin(x)\,\mathrm{d}x &=\sin\left(\frac\pi3\right)\Gamma\left(\frac23\right)\\ &=\sin\left(\frac\pi3\right)\frac{\pi\csc\left(\frac\pi3\right)}{\Gamma\left(\frac13\right)}\\ &=\frac{\pi}{\Gamma\left(\frac13\right)}\tag{3} \end{align} $$

Solution 3:

Considering $R> 0$ and $C_R$ the path from $0 $ to $iR$ we have

$$\int_{0}^{\infty} x^{-1/3} \sin x \, dx = \lim_{R \to \infty}\mathfrak{Im}\Bigg(\int _0^R e^{iz}z^{-1/3}dz\Bigg) = \lim_{R \to \infty} \mathfrak{Im}\Bigg(\int _{C_R} e^{iz}z^{-1/3}dz\Bigg)$$

Making $-w = iz$ we have that $- dw = i dz \implies i dw = dz$ then

$$\int _{C_R} e^{iz}z^{-1/3}dz = e^{\frac{i\pi}{3}} \int_{C_R} e^{-w} w^{2/3 -1} dw = e^{\pi i / 3} \Gamma \Big(\frac{2}{3}\Big) = e^{\pi i / 3}\frac{\pi}{\sqrt{3}\,\,\Gamma \Big(\frac{1}{3}\Big)}$$

Thus

$$\int_{0}^{\infty} x^{-1/3} \sin x \, dx = {\sqrt{3}\frac{\pi}{\sqrt{3}\,\,\Gamma \Big(\frac{1}{3}\Big)}} = \color{red}{\frac{\pi}{\,\,\Gamma \Big(\frac{1}{3}\Big)}}$$