Proving that the line integral $\int_{\gamma_{2}} e^{ix^2}\:\mathrm{d}x$ tends to zero Let $f(z) = e^{iz^2}$ and $\gamma_2 = \{ z : z = Re^{i\theta}, 0 \leq \theta \leq \frac{\pi}{4} \} $. All the sources I have found online, says that the line integral 
$$ \left| \int_{\gamma_2} e^{iz^2}\mathrm{d}z \right| $$
tends to zero as $R \to \infty$. 

By using the ML-inequality one has
$$
      \left| \int_{\gamma_2} e^{iz^2}\mathrm{d}z \right| 
 \leq \frac{R\pi}{4} \max_{\theta \in [0,\pi/4]} \left| e^{iR\exp(i2\theta)} \right|
 \leq \frac{R\pi}{4} \max_{\theta \in [0,\pi/4]} e^{-R^2 \sin 2 \theta}
$$
The problem is now that this is a decreasing function, and if one inserts 
$\theta=0$, then the inequality becomes$\pi R e^{0}/4 = \pi R / 4$, which does not tend to zero. If one instead looks at the interval $(0,\pi/4)$ then everything works out. 
This document instead tells us to look at Jordans lemma, which one can use by rewriting the function as $e^{ix^2} = e^{ix^2-ix}e^{ix}$. However I run into exactly the same problems here if one studies $[0,\pi/4]$, as then one get $\pi/4 \cdot \cos R(R-1)$. 
Both Jordan's lemma and the $ML$-inequality clearly states that one should include the endpoints, and clearly this does not work here? What does one do instead?
Why is it wrong to look at $[0,\pi/4]$ and correct to ignore the endpoints? 

Split the path at theta=r^(-3/2), say. Then you can bound both contributions in such a way that each goes to zero as r->infinity.
