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

Question

How can I calculate the following integral:

$$\int_0^\infty x ^{-\frac{1}{3}}\sin x \, dx$$

WolframAlpha gives me $$ \frac{\pi}{\Gamma\Big(\frac{1}{3}\Big)}$$

How does WolframAlpha get this?

I don't understand how we can rearrange the formula in order to apply the gamma-function here. Any helpful and detailed hint/answer is appreciated.

Answer 1

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 $$ \begin{align} \int_0^\infty x^{-1/3}\sin(x)\,\mathrm{d}x &=\sin\left(\frac\pi3\right)\Gamma\left(\frac23\right)\\ &=\sin\left(\frac\pi3\right)\frac{\pi\csc\left(\frac\pi3\right)}{\Gamma\left(\frac13\right)}\\ &=\frac{\pi}{\Gamma\left(\frac13\right)}\tag{3} \end{align} $$

Answer 2

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