I need to evaluate the following integral: $$\int_{-\infty}^{\infty} \dfrac{\sin(x^2)}{1+x^4}dx.$$ I'm trying to use complex analysis. So, I define the function, $$f(z)=\dfrac{e^{iz^2}}{1+z^4},$$ and I consider the path $\gamma=[-R,R]+\gamma_R$, where $\gamma_R$ is the semi-circle of radious $r$ given by $\gamma_R(t)=Re^{it}$, $0\leq t\leq \pi$. I know that $f$ has four poles at $z=\pm\dfrac{1-i}{\sqrt{2}}$ and $z=\pm \dfrac{1+i}{\sqrt{2}}$ and just two of them are at the upper half plane: $\dfrac{1-i}{\sqrt{2}}$ and $\dfrac{1+i}{\sqrt{2}}$. My idea is to show that, $$\int_{\gamma_R}\dfrac{e^{iz^2}}{1+z^4}dz\to 0 \text{ when }R\to \infty.$$ However, I obtained that,
$$\left|\int_{\gamma_R}\dfrac{e^{iz^2}}{1+z^4}dz\right|\leq \int_0^{\pi}\dfrac{e^{-R^2\sin 2t}}{R^4-1}Rdt,$$ but I don't know how to deal with the term $e^{-R^2\sin 2t}$. Any suggestion?