Vrftsjtryk

I'm trying to prove that $lim_{Rtoinfty}int_{C_R} dz expleft(iaz^2right) = 0$, where $a$ has a positive imaginary part and $C_R$ is an arc from $R$ to $frac{1+i}{sqrt{2}}R$ along the circle centered around the origin with radius $R$.

This is what I have so far:

But this proof does not go through unfortunately.

Consider the complex function
$$
f(z)=e^{ialpha z^2}, z=re^{itheta}, f(z=re^{itheta})=e^{ialpha r^2 e^{2 itheta}}.
$$
Thus we can analyze the term
$$
f(z=re^{itheta})=e^{ialpha r^2 e^{2 itheta}}
$$
and notice for $theta=pi/4$, this is just a real Gaussian integral $alpha >0$. If $theta=0$ we have $f(re^{itheta})=e^{ialpha r^2}$. Thus we split our contour into three pieces making an angle between the imaginary and real axis of $45^o$. The contour is shown here http://upload.wikimedia.org/wikipedia/commons/b/b9/Fresnel_Integral_Contour.svg . We know there are no poles inside the contour and $f(z)$ is holomorphic , thus by the Cauchy Goursat theorem we know we know
begin{equation}
0=oint f(z) dz=int_{0}^{infty} e^{ialpha x^2} dx +int_{0}^{pi/4} ire^{itheta} dtheta e^{ialpha r^2(cos(2theta)+isin(2theta))}+int_{infty}^{0}e^{-alpha r^2}e^{ipi/4}dr
end{equation}
where the integral is broken up into three contours shown in the illustration and I used $z=re^{itheta}, dz= e^{itheta }dr$. The difficult integral to evaluate is
$$
int_{0}^{pi/4} ire^{itheta} dtheta e^{ialpha r^2(cos(2theta)+isin(2theta))},
$$
however it vanishes as $r to infty$. We need to show that it vanishes, it is similar to Jordan's inequality, but that just places an upper bound on the integral,
$$
int_{0}^{pi} e^{-rsintheta}dtheta < frac{pi}{r} (r >0).
$$
We will prove this now by showing that
$$
lim_{rto infty} bigg| int_{0}^{pi/4} e^{ialpha r^2 e^{2itheta}} ir e^{itheta }dtheta bigg|=0.
$$

We know that
$$
bigg| int_{0}^{pi/4} e^{ialpha r^2 e^{2itheta}} ir e^{itheta }dtheta bigg| leq int_{0}^{pi/4} big|e^{ialpha r^2 e^{2itheta}}big| big|ir e^{itheta}dthetabig|=int_{0}^{pi/4} rdtheta big| e^{ialpha r^2cos(2theta)-alpha r^2sin(2theta)} big|=int_{0}^{pi/4}rdtheta big|e^{i alpha r^2cos(2theta)}big|big|e^{-alpha r^2 sin(2theta)}big|
$$
where I used $e^{2itheta}=cos 2theta +isin 2theta$. We can simplify this to obtain
$$
int_{0}^{pi/4} r dtheta big|e^{i alpha r^2cos(2theta)}big|big|e^{-alpha r^2sin(2theta)}big|=int_{0}^{pi/4} rdtheta e^{-alpha r^2sin(2theta)}.
$$
Thus we need to show that this vanishes as $r to infty.$ If we make the substitution $xi=2theta , dtheta=dxi/2$ and changing the bounds of integration we obtain
$$
int_{0}^{pi/2} frac{r}{2}dxi e^{-alpha r^2sin xi}.
$$
We can see that for $xi in [0,pi/2]$, $sinxi geq 2xi/pi$. Since $e^{-alpha r^2cdot 2xi/pi} geq e^{-alpha r^2 sin xi} $ (since exponential is bigger for a smaller exponent), we can write
$$
int_{0}^{pi/2} frac{r}{2}dxi e^{-alpha r^2sin xi} leq int_{0}^{pi/2} frac{r}{2}dxi e^{-alpha r^2 cdot 2/pi}=frac{pi}{4 alpha r}(1-e^{-alpha r^2}).
$$
Thus it is clear that
$$
lim_{rto infty}frac{pi}{4 alpha r}(1-e^{-alpha r^2})=0,
$$
thus we have shown that
$$
lim_{rto infty} bigg| int_{0}^{pi/4} e^{ialpha r^2 e^{2itheta}} ir e^{itheta }dtheta bigg|=0.
$$