Show that $\displaystyle \int_{\gamma_R}\exp(-w^2)\,\mathrm dw\to 0~(R\to\infty)$ along $\gamma_R(t)=R\exp(\mathrm it)$

Question

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?

Answer 1

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 $\int_{\gamma_R}\exp(-w^2)\,\mathrm dw$ .