MHB How can you evaluate the integral of cosine raised to the fifth power?

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Here is this week's POTW:

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Evaluate the integral $$\int_0^\infty \cos\!\big(x^5\big)\, dx$$
You may express your answer in terms of the Gamma function.
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No one answered this problem. You can read my solution below.

Spoiler

Consider the contour integral \[\oint_{\Gamma(R)} e^{iz^5}\, dz\] over a sector of radius $R$ in the first quadrant subtended by $\frac{\pi}{10}$. The integral along the arc of the sector is dominated by \[\int_0^{\pi/10} e^{-R^5\sin(5\theta)}\, d\theta = \int_0^{\pi/10} e^{-10R^5 \theta/\pi}\, d\theta = \frac{\pi}{10R^5}\bigl(1 - e^{-R^5}\bigr)\] which is $O(R^{-5})$ as $R \to \infty$. By Cauchy's theorem $\int_0^\infty e^{ix^5}\, dx = \lim\limits_{R\to \infty} \int_{L(R)} e^{iz^5}\, dz$ where $L(R)$ is the edge of $\Gamma(R)$ elevated $\frac{\pi}{10}$. On $L(R)$, $z = xe^{i\pi/10}$ where $0 \le x \le R$. Thus \[\lim_{R\to \infty} \int_{L(R)} e^{iz^5}\, dz = \lim_{R\to \infty} e^{i\pi/10} \int_0^R e^{i(x^5 e^{i\pi/2})}\, dx = e^{i\pi/10} \int_0^\infty e^{-x^5}\, dx = e^{i\pi/10}\, \Gamma\left(\frac{6}{5}\right)\] Comparing real parts of the equation \[\int_0^\infty e^{ix^5}\, dx = e^{i\pi/10}\Gamma\left(\frac{6}{5}\right)\] results in the solution \[\int_0^\infty \cos(x^5) = \Gamma\left(\frac{6}{5}\right) \cos\left(\frac{\pi}{10}\right)\]