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?

  • 1
    $\begingroup$ What is $sen(t)$? Did you mean to write $\sin(t)$? $\endgroup$ – LoveTooNap29 Jan 3 at 19:23
  • $\begingroup$ yeah, I corrected that. tks $\endgroup$ – Lincon Ribeiro Jan 3 at 19:25
  • 1
    $\begingroup$ Hints: $|e^{i(x+iy)^2}|=e^{-2xy}$ and $\sin\phi\geqslant 2\phi/\pi$ for $0\leqslant\phi\leqslant\pi/2$. $\endgroup$ – 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

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


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

  • $\begingroup$ @LinconRibeiro Can you remove your check mark? I made a mistake in my answer $\endgroup$ – Dylan Jan 4 at 12:17
  • $\begingroup$ 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. $\endgroup$ – Lincon Ribeiro Jan 4 at 12:37
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    $\begingroup$ I updated my answer. It was simpler than I though. $\endgroup$ – Dylan Jan 4 at 13:47

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