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Show that $\displaystyle \int_{\gamma_R}\exp(-w^2)\,\mathrm dw\to 0~(R\to\infty)$ along $\gamma_R(t)=R\exp(\mathrm it),t\in\left[0,\frac{\pi}{4}\right]$.

Hint. Use that $\cos 2t\geq 1-\frac{4}{\pi}t$ for $t\in\left[0,\frac{\pi}{4}\right]$.

My first naive attempt did not refer to the given hint and thus looked like this

$$ \begin{align*} \int_{\gamma_R}\exp(-w^2)\,\mathrm dw &= \int_{0}^{\pi/4}\exp(-(R\exp(\mathrm it))^2)\cdot \mathrm iR\exp(\mathrm it)\,\mathrm dt\\ &= \int_R^{R\exp(\mathrm i\pi/4)}\exp(-u^2)\,\mathrm du\\ &= \frac{\sqrt{\pi}}{2}(\operatorname{erf}(R\exp(\mathrm i\pi/4)) - \operatorname{erf}(R))\\ &\to \frac{\sqrt{\pi}}{2}(1-1) = 0 ~(R\to\infty) \end{align*} $$

which follows from $\lim_{x\to\pm\infty}\operatorname{erf}(x)=\pm 1$. This isn't relying on the given hint and I want to know whether I should apply Eulers formula in line 1 for $\exp(\mathrm it)$, in line 3 for $\exp(\mathrm i\pi/4)$ or do I have to do something completely different?

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  • $\begingroup$ Wouldn't you have to show that $\lim_{R \to \infty} \operatorname{erf}(R\exp(\mathrm i\pi/4)) = 1$ with your method? $\endgroup$ – Martin R May 8 '16 at 15:19
  • $\begingroup$ @MartinR I did the sloppy assumption that $\exp(\mathrm i\pi/4)$ behaves like a constant and $R\exp(\mathrm i\pi/4)\simeq R$ for $R\to\infty$. It now seems reasonable to perform the suggested way of finding the limit since my approach seems invalid. $\endgroup$ – Christian Ivicevic May 8 '16 at 15:29
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For $w = Re^{it}$, $0 \le t \le \frac \pi 4$ you have $$ \lvert \exp(-w^2) \lvert = \exp \left( \text{Re}(-w^2) \right) = \exp \left(- R^2 \cos(2t) \right) \le \exp \left(- R^2 (1 - \frac 4 \pi t) \right) $$ where the final step uses the given hint.

That should be good enough to estimate the absolute value of $\int_{\gamma_R}\exp(-w^2)\,\mathrm dw$ .

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  • $\begingroup$ Integrating this function yields $\frac{-\pi(\exp(-R^2)-1)}{4R^2}=\ldots\leq \frac{\pi}{4R^2}\to 0~(R\to\infty)$ if I am not wrong, isn't it? $\endgroup$ – Christian Ivicevic May 8 '16 at 14:39
  • $\begingroup$ @ChristianIvicevic: It should be $R$ in the denominator, not $R^2$. Note that $dw = iRe^{it} dt$. Apart from that, I got the same result. $\endgroup$ – Martin R May 8 '16 at 14:41
  • $\begingroup$ I have two questions - the first one might be obvious but I am struggling to see why $|\exp(-w^2)|=\exp(\operatorname{Re}(-w^2))$, why is that so? For the second one is it correct that you start (after your inequality) with the following expression: $$\begin{align*}\left|\int_{\gamma_R}\exp(-w^2)\,\mathrm dw\right|&\leq \int_0^{\pi/4}\exp\left(-R^2\left(1-\frac{4}{\pi}t\right)\right)\cdot\left|~ \mathrm{i} R \exp(\mathrm it)\right|\,\mathrm dt \\&\leq R\int_0^{\pi/4}\exp\left(-R^2\left(1-\frac{4}{\pi}t\right)\right)\end{align*}$$ $\endgroup$ – Christian Ivicevic May 8 '16 at 15:04
  • $\begingroup$ If the above is in fact true I now know where the missing $R$ you referred to comes from. $\endgroup$ – Christian Ivicevic May 8 '16 at 15:09
  • $\begingroup$ @ChristianIvicevic: Generally, for $z = x +iy$, $|e^z| = |e^x| \cdot |e^{iy}| = e^x$, hence $|e^z| = e^{\text{Re} \, z}$. – The second part looks correct. Integrating that gives $R^2$ in the denominator, but there is still the leading factor $R$. $\endgroup$ – Martin R May 8 '16 at 15:11

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