Proving that the line integral $\int_{\gamma_{2}} e^{ix^2}\:\mathrm{d}x$ tends to zero

Question

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?

Answer 1

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.

Answer 2

You can do this as follows. Start as you did with $$\left\vert\int_{\gamma_2} e^{iz^2}dz\right\vert\leq R\,\int_0^{\frac\pi4} e^{-R^2\sin 2\theta} d\theta\, . $$ Then observe that on $[0\frac\pi4]$ you have $$\sin 2\theta \geq \frac4\pi \, \theta\, ,$$ thanks to the concavity of the sine function on $[0\frac\pi2]$.

It follows that $$\left\vert\int_{\gamma_2} e^{iz^2}dz\right\vert\leq R\,\int_0^{\frac\pi4} e^{-\frac{4R^2}\pi \theta} d\theta=R\times\frac{\pi}{4R^2}(1-e^{-R^2})\leq \frac{\pi}{4R}\, ,$$ which gives the result.