I am trying to evalute the two following integrals

$$ I_1 = \int_0^\infty x \cos (x^3) \, \mathrm{d}x \quad \text{and} \quad I_2 = \int_0^\infty x \sin (x^3) \, \mathrm{d}x$$

I already know the numerical values, they are respectively

$$ I_1 = \frac{1}{6}\Gamma\left( \frac{2}{3} \right) \quad \text{and} \quad I_2 = \frac{1}{2\sqrt{3}}\Gamma\left( \frac{2}{3} \right) $$

Any ideas? I saw this on another forum and tried a few contours, but alas nothing worked. Thanks in advance =)

  • $\begingroup$ I'm getting $I_1=\frac16\Gamma\left(\frac23\right)$... $\endgroup$ – J. M. is a poor mathematician Apr 24 '12 at 16:19
  • $\begingroup$ @J.M. That's what I got as well. $\endgroup$ – Ragib Zaman Apr 24 '12 at 16:30

Here's one approach:

Consider the integral $$I_1+i I_2 = \int_0^{\infty} x e^{i u^3} du$$ With a change of variables $t = u^3$,

$$I_1+i I_2 = \frac{1}{3} \int_0^{\infty} t^{-\frac{1}{3}} e^{i t} \; dt$$ Since $\lim_{\sigma \downarrow 0} e^{(-\sigma + i)t} = e^{i t}$, we can write $$I_1+i I_2 = \lim_{\sigma \downarrow 0} \frac{1}{3} \int_0^{\infty} t^{-\frac{1}{3}} e^{(-\sigma + i)t} \; dt.$$

However, since the inner integral is just the Laplace transform of $t \mapsto t^{-\frac{1}{3}}$ evaluated at $s=\sigma - i$, we have $${I_1} + i{I_2} = \mathop {\lim }\limits_{\sigma \downarrow 0} \frac{1}{3}\frac{{\Gamma \left( {1 - \frac{1}{3}} \right)}}{{{{(\sigma - i)}^{1 - \frac{1}{3}}}}} = \frac{1}{3}\frac{{\Gamma \left( {\frac{2}{3}} \right)}}{{{{( - i)}^{\frac{2}{3}}}}} = \frac{1}{3}\Gamma \left( {\frac{2}{3}} \right)\left( {\frac{1}{2} + i\frac{{\sqrt 3 }}{2}} \right)$$ from which you get the desired result (with appropriate correction for $I_1)$.

  • 1
    $\begingroup$ Of course, this avoids the hard grind by using a table of Laplace transforms... $\endgroup$ – copper.hat Apr 24 '12 at 16:35
  • $\begingroup$ Great solution. I'll tweak the $\LaTeX$ a bit. $\endgroup$ – Pedro Tamaroff Apr 24 '12 at 16:36

Consider $\displaystyle \int_{C_R} z^{-1/3} e^{iz} dz $ where $C_R$ is the quadrant of radius $R$ in upper-right quarter plane oriented counter-clockwise. There are no poles enclosed in the contour so its value is $0.$

As $R\to \infty:$

  • The integral of the part along the real axis tends to $\displaystyle I=\int^{\infty}_0 z^{-1/3} e^{iz} dz.$
  • The integral along the curved part tends to $0$ by Jordan's lemma.
  • The integral along the imaginary axis tends to $$ -i \int^{\infty}_0 (it)^{-1/3} e^{-t} dt .$$

So $$I = i^{2/3} \int^{\infty}_0 t^{-1/3} e^{-t} dt = i^{2/3} \Gamma(2/3).$$

Letting $z=x^3$ and taking real/imaginary parts gives

$$\int^{\infty}_0 x \cos(x^3) dx = \frac{1}{6} \Gamma(2/3) \ \text{ and } \int^{\infty}_0 x \sin(x^3) dx = \frac{1}{2\sqrt{3}} \Gamma(2/3).$$


I think that just converting $cos(x)=\frac{e^{ix}+e^{-ix}}{2}$, then doing your integration by parts would probally do it for that one. Making the same change for sine would likly get your your integral.

  • $\begingroup$ By doing that you are technically transforming the integral into a line integral on the complex plane. Then you would need to show how that $\int _C z^{-\frac{1}{3}}e^{-z}dz=\int_0^{\infty}t^{-\frac{1}{3}}e^{-t}dt$. Which atleast for me is not straight forward $\endgroup$ – N3buchadnezzar Apr 24 '12 at 16:05

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