MHB What is the integral of cos(x^3)?

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Here is this week's POTW:

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Compute the integral

$$\int_{-\infty}^\infty \cos(x^3)\, dx.$$

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This week's problem was solved correctly by Opalg and chisigma. Here are their solutions.

1. Opalg's solution

Spoiler

View attachment 4334

By Cauchy's theorem, $$ \oint_C e^{-z^3}dz = 0,$$ where $C$ is the wedge-shaped contour consisting of the interval $C_1 = [0,R]$ of the real axis, $C_2$ is the arc $\{Re^{i\theta}:0\leqslant \theta \leqslant \pi/6\}$ and $C_3$ is the line from $Re^{i\pi/6}$ to the origin.

Taking the limit as $R\to\infty$, $$\int_{C_1}e^{-z^3}dz \to \int_0^\infty e^{-x^3}dx.$$ Substitute $y=x^3$ to get $$\int_0^\infty e^{-y}\frac{dy}{3y^{2/3}} = \frac13\int_0^\infty y^{\frac13-1}e^{-y}dy = \tfrac13\Gamma\bigl(\tfrac13\bigr) = \Gamma\bigl(\tfrac43\bigr).$$

Along $C_2$ the integral goes to zero as $R\to\infty.$ On $C_3$, make the substitution $z=re^{i\pi/6}$, getting $$\int_{\infty}^0 e^{-ir^3}e^{i\pi/6}\,dr = -\bigl(\cos\tfrac\pi6 + i\sin\tfrac\pi6\bigr) \int_0^\infty \bigl(\cos(r^3) - i\sin(r^3)\bigr)\,dr.$$ Now add the three integrals and take the real and imaginary parts to get $$-\frac{\sqrt3}2 \int_0^\infty \cos(x^3)\,dx - \frac12 \int_0^\infty \sin(x^3)\,dx + \Gamma\bigl(\tfrac43\bigr) = 0,$$ $$-\frac12\int_0^\infty \cos(x^3)\,dx + \frac{\sqrt3}2\int_0^\infty \sin(x^3)\,dx = 0.$$ Solve those two simultaneous equations for the two integrals, getting $$\int_0^\infty \cos(x^3)\,dx = \tfrac{\sqrt3}2\Gamma\bigl(\tfrac43\bigr).$$

Finally, $x^3$ is an odd function and $\cos x$ is even, so $ \cos(x^3)$ is even, and $$\int_{-\infty}^\infty \cos(x^3)\,dx = 2\int_0^\infty \cos(x^3)\,dx = \sqrt3\Gamma\bigl(\tfrac43\bigr).$$

2. chisigma's solution

Spoiler

Let's define...

$\displaystyle g(t) = \int_{0}^{\infty} \cos(tx^3) \ dx\ (1)$

... and compute the L.T. of (1)...

$\displaystyle \mathcal {L}\ \{g(t)\} = \int_{0}^{\infty} e^{- s t} \int_{0}^{\infty} \cos (tx^{3})\ d x\ d t= \int_{0}^{\infty} d x\ \int_{0}^{\infty} e^{- s t}\ \cos(t x^{3})\ d t = \int_{0}^{\infty} \frac{s}{s^{2} + x^{6}} = \frac{\pi}{3\ s^{\frac{2}{3}}}\ (2) $

Taking the inverse L.T. of (2)...

$\displaystyle \mathcal{L}^{-1} \left\{ \frac{\pi}{3\ s^{\frac{2}{3}}} \right\} = \frac{\pi}{3\ \sqrt{t}\ \Gamma (\frac{2}{3})}$

... so that for t=1 we obtain...

$\displaystyle \int_{0}^{\infty} \cos(x^3) \ dx = \frac{\pi}{3\ \Gamma(\frac{2}{3})}\ \implies \int_{- \infty}^{+ \infty} \cos(x^3) \ d x = \frac{2\ \pi}{3\ \Gamma (\frac{2}{3})} $