$\int_{[1+iR, 1+2iR]}\frac{e^{z^2}}{z^2}dz \to 0$ as $R \to \infty$

Question

Please help me with this problem,

$\int_{[1+iR, 1+2iR]}\frac{e^{z^2}}{z^2}dz \to 0$ as $R \to \infty$

My attempt : $r(t)= 1+ itR$, $1\leq t\leq 2$ Then the integral becomes $\int_1^2\frac{e^{(1+ itR)^2}}{(1+ itR)^2}(1+ iR)dt$. Then how can I proceed?

Answer 1

Develop the square $$\int_1^2\frac{e^{(1+ itR)^2}}{(1+ itR)^2}(1+ iR)dt = \int_1^2\frac{e^{1-t^2R^2+2irt}}{1+ itR}dt.$$ Remark that $$\left| \frac{e^{1-t^2R^2+2irt}}{1+ itR} \right| = \left|\frac{e^{1-t^2R^2}}{1+ itR}\right| \leqslant \frac{e^{1-t^2R^2}}{1- tR} \to 0,$$ if $R \to +\infty.$ Then conclude thanks to Lebesgue theorem.

Answer 2

Observe that the integration line is a vertical one from $\;(1, R)\;$ to $\;(1,2R)\;$ (identifying $\;\Bbb C\sim\Bbb R^2\;$) , so using Cauchy's Estimate (or something similar from real analysis):

$$\left|\int_{1+iR}^{1+2iR}\frac{e^{z^2}}{z^2}dz\right|\le 1\cdot\max_{z}\frac{|e^{z^2}|}{|z|^2}=\max_z\frac{e^{1-4R^2}}{1+R^2}\xrightarrow[R\to\infty]{}0$$