# Verify that $\left|\int_{\gamma} \exp(iz^2)dz\right| \leq \frac{\pi\big(1-\exp(-r^2)\big)}{4r}$ where $\gamma(t)=re^{it}$, for $0\leq t \leq \pi/4$.

Verify that $$\left|\int_{\gamma} \exp(iz^2)dz\right| \leq \frac{\pi\big(1-\exp(-r^2)\big)}{4r}$$ where $$\gamma(t)=re^{it}$$, for $$0\leq t \leq \pi/4$$ and $$r > 0$$.

I'm stuck. here is my attempt:

$$|\int_{\gamma} e^{i\cdot z^2}dz|\leq \int_{\gamma} |e^{i\cdot z^2}|dz=\int_{\gamma}|(e^{z^2})^i|dz$$. As $$t > 0$$ on $$0\leq t \leq \pi/4$$.
$$\Rightarrow|e^{r^2 e^{2it}}|=|e^{r^2}e^{e^{2it}}|>0$$ $$\Rightarrow\int_{0}^{\pi/4} e^{r^2}e^{2it}rie^{it}dt=e^{r^2}ri\int_{0}^{\pi/4} exp(e^{2it}+it)dt$$
Let $$\alpha=e^{r^2}ri$$
$$\Rightarrow \alpha\int_{0}^{\pi/4}e^{\cos2t}e^{i\cdot \sin2t}(\cos(t)+i\cdot \sin(t)dt)$$

Am I on track?

• What is $sen(t)$? Did you mean to write $\sin(t)$? – LoveTooNap29 Jan 3 at 19:23
• yeah, I corrected that. tks – Lincon Ribeiro Jan 3 at 19:25
• Hints: $|e^{i(x+iy)^2}|=e^{-2xy}$ and $\sin\phi\geqslant 2\phi/\pi$ for $0\leqslant\phi\leqslant\pi/2$. – metamorphy Jan 4 at 2:29

You have

$$\int_\gamma e^{iz^2}dz = \int_0^{\pi/4} e^{ir^2(\cos 2t + i\sin 2t)}ire^{it} dt$$

Taking the magnitude

$$\left\vert \int_\gamma e^{iz^2}dz \right\vert \le \int_0^{\pi/4} \left\vert e^{ir^2(\cos 2t + i\sin 2t)}ire^{it} \right\vert dt = \int_0^{\pi/4} re^{-r^2\sin 2t} dt$$

Since $$\sin 2t$$ curves upwards on the interval $$(0,\pi/4)$$, it always lies above its secant line from $$t=0$$ to $$t=\pi/4$$, therefore

$$\sin 2t \ge \frac{4t}{\pi}, \quad \forall t \in \left(0,\frac{\pi}{4}\right)$$

And

$$\int_0^{\pi/4} re^{-r^2\sin 2t} dt \le \int_0^{\pi/4} re^{-4r^2t/\pi} dt = \frac{\pi}{4r}\left(1 - e^{-r^2} \right)$$

• @LinconRibeiro Can you remove your check mark? I made a mistake in my answer – Dylan Jan 4 at 12:17
• Ok, done. How did you get that term on the right side of the last inequality $\frac {1-e^{-r^2}} {r^2}$? btw, I forgot to add that r must be greater than zero. I don't know if it changes anything. – Lincon Ribeiro Jan 4 at 12:37
• I updated my answer. It was simpler than I though. – Dylan Jan 4 at 13:47