1
$\begingroup$

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?

$\endgroup$
  • 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
1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ @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
  • 1
    $\begingroup$ I updated my answer. It was simpler than I though. $\endgroup$ – Dylan Jan 4 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.