Evaluating the integral $ \int_0^{\infty} \cos(x^2)\, \mathrm{d} x$?

Question

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.

Answer 1

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}