I wish to compute the integral $\int_0^\infty e^{-x^2}\text{cos}(x^2)\mathrm dx$ using complex integration techniques. I'm integrating the function $e^{-z^2}$ over the boundary of the circular sector $\{z:|z|<a, 0\leq \arg z \leq \pi/8\}$. I'm having difficulty integrating the said function over the arc, i.e. the second piece of the contour. I used the parametrization $\gamma_2(t)=ae^{it}, 0\leq t \leq \pi/8$. I think it reduces to bounding $|e^{-a^2e^{2it}}|$ but I can't seem to bound this. Any help is appreciated.
I've already looked at this:Integral of $e^{-x^2}\cos(x^2)$ using residues, but wish to proceed without considering the real parts.
Best Answer
You can use the famous Euler Integral formula related to the $\Gamma$-function. The derivation is pretty straight forward and can be completed by advanced high school students.
It states
$$\int_0^{\infty}u^{ns-1}\exp(-au^n)\cos(bu^n)du=\dfrac{\Gamma(s)}{n|p|^s}\cos (\alpha s),$$
in which $p=a+bi \implies |p|=\sqrt{a^2+b^2}$ and $\tan \alpha = \dfrac{b}{a}$.