Curious improper integral:$\displaystyle\int_0^\infty\dfrac{\sin(x^3)}{x^2}dx$

Question

I found the following integral, $\displaystyle\int_0^\infty\dfrac{\sin(x^3)}{x^2}dx$ which at first seemed quite curious to me, but although I tried several times I couldn't get anything. I know it's convergent thanks to WolframAlpha. I'm surprised that WolframAlpha tells me that the integral turns out to be $\dfrac{\Gamma(2/3)}{2}$. I have also tried to use Maz identity, without any success. Any ideas please.

Answer 1

Let $y=x^3$, then the integral is transformed into a Gamma function. $$ \begin{aligned} I & =\frac{1}{3} \int_0^{\infty} y^{-\frac{4}{3}} \sin y d y \\ & =-\frac{1}{3} \Im\int_0^{\infty} y^{-\frac{1}{3}-1} e^{-y i} d y \\ & =-\frac{1}{3} \Im\left[\frac{\Gamma\left(-\frac{1}{3}\right)}{i^{-\frac{1}{3}}}\right] \\ & =\Gamma\left(\frac{2}{3}\right) \operatorname{Im}\left(e^{\frac{\pi}{2} i}\right)^{\frac{1}{3}} \\ & =\frac{\Gamma\left(\frac{2}{3}\right)}{2} \end{aligned} $$ where the last line using $\Gamma\left(\frac{2}{3}\right)=\Gamma\left(1+\left(-\frac{1}{3}\right)\right)=-\frac{1}{3} \Gamma\left(-\frac{1}{3}\right).$

Answer 2

As @Lai commented, using integration by parts with $u=\sin(x^3)$

$$\int\dfrac{\sin(x^3)}{x^2}dx=-\frac{\sin \left(x^3\right)}{x}+3 \int x \cos \left(x^3\right)\,dx$$

Using the exponential integral function $$\int x \cos \left(x^3\right)\,dx=-\frac{1}{2} x^2 \left(E_{\frac{1}{3}}\left(+i x^3\right)+E_{\frac{1}{3}}\left(-i x^3\right)\right)$$

You could also use the incomplete gamma function.

Using the bounds, the result.