ML-estimate for complex integrals

Question

Let $f(z)=e^{iz^2}$ and $\gamma_R (t)=R+it,\,t\in [0,R].$ To show that $$\lim_{R \to +\infty}\int_{\gamma_R}f(z) \,dz=0 $$ we must find an upper bound for the integral and show that it goes to zero for big values of $R$. $$\Bigg|\int_{\gamma_R}f(z) \,dz \,\Bigg| \leq \int_{\gamma_R}\big|f(z)\big||dz|=M\cdot L$$ where $M$ is the upper bound of $f$ on $\gamma$ and $L$ the length of $\gamma$.

To find $M$, it's obvious that the complex number with the greatest modulus on $\gamma$ is $z_0=R+iR$ which gives $$M=\big| f(z_0)\big|=\big|e^{i(R+iR)^2}\big|=\big|e^{-2R}\big|=\frac{1}{e^{2R}} \quad\text{and}$$ $$L=R$$ Thus $$\lim_{R \to +\infty}\Bigg|\int_{\gamma_R}f(z) \,dz \,\Bigg| \leq \lim_{R\to +\infty} \frac{R}{e^{2R}}=\lim_{R\to +\infty} \frac{1}{2e^{2R}}=0 $$ $$\Rightarrow \lim_{R \to +\infty}\int_{\gamma_R}f(z) \,dz=0$$ Is this proof correct?

Answer 1

The general upper bound for the integral is not useful. $|f(R+it)|=e^{-2Rt}$. Hence what you have to show is $i\int_0^{R} e^{-2Rt}\, dt \to 0$. Make the substitution $s=2Rt$ and note that $\frac 1 {2R}\int_0^{2R^{2}} e^{-s} \, ds=\frac {1-e^{-2R^{2}}} {2R} \to 0$ as $R \to \infty$.

Answer 2

No, it is not correct. Indeed, although it is true that $R+iR$ is the element of $\gamma\bigl([0,1]\bigr)$ with the greatest absolute value, you gave no reason why you think you should have$$(\forall t\in[0,R]):\bigl\lvert f\bigl(\gamma(t)\bigr)\bigr\rvert\leqslant\bigl\lvert f(R+Ri)\bigr\rvert.$$By the way, this inequality doesn't hold.