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Is it necessary to make use of the Gaussian integral and the complex exponential form of the cosine in evaluating the following integral?

$$\int_0^{\infty} \cos(x^2)\, \mathrm{d} x$$

Just curious - I can prove its convergence, but not evaluate it at the moment.

Anastasiya-Romanova 秀
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Eweler
  • 691

2 Answers2

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Is it necessary to use the Gamma function in integral form and then make the appropriate changes.... probably not, but is one of the shortest methods to choose.

Consider the integral \begin{align} I_{n} = \int_{0}^{\infty} \cos(x^{2n}) \, dx \end{align} and make the change of variables $t = x^{2n}$ which leads to \begin{align} I_{n} = \frac{1}{2n} \, \int_{0}^{\infty} \cos(t) \, t^{\frac{1}{2n} - 1} \, dt. \end{align} Now using the integral \begin{align} \int_{0}^{\infty} \cos(at) \, t^{p-1} \, dt = \frac{\Gamma(p)}{a^{p}} \, \cos\left( \frac{p \pi}{2} \right) \end{align} then $I_{n}$ becomes \begin{align} I_{n} = \cos\left( \frac{\pi}{4n} \right) \, \Gamma\left( \frac{1}{2n} + 1 \right). \end{align}

For the case of $n=1$ the result is \begin{align} \int_{0}^{\infty} \cos(t^{2}) \, dt = \frac{1}{2} \sqrt{\frac{\pi}{2}}. \end{align}

Leucippus
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$$ \begin{aligned} \int_0^{\infty}\left(\cos (x^2)-i \sin (x^2)\right) d x = & \int_0^{\infty} e^{-x^2 i} d x \\ = & \int_0^{\infty} e^{-\left(\frac{1+i}{\sqrt{2}} t\right)^2} d x \\ = & \frac{\sqrt{2}}{1+i} \int_0^{\infty} e^{-x^2} d x \\ = & \frac{\sqrt{2}}{1+i} \cdot \frac{\sqrt{\pi}}{2} \quad (*) \\ = & \sqrt{\frac{\pi}{8}} -i\sqrt{\frac{\pi}{8}} \end{aligned} $$ where (*) comes from the Gaussian integral.

Comparing their real and imaginary parts gives

$$ \boxed{\int_0^{\infty} \cos (x^2) d x=\int_0^{\infty} \sin (x^2) d x=\sqrt{\frac{\pi}{8}}} $$

Lai
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