MSE

---

Homework #3
Let $\gamma_1$ denote the straight line along the real line from 0 to $R, \gamma_2$ denote the eighth of a circle from $R$ to $R e^{i \frac{\pi}{4}}$, and $\gamma_3$ denote the line from $R e^{i \frac{\pi}{4}}$ to 0 . Then by Cauchy's theorem, \[ \int_{\gamma_1+\gamma_2+\gamma_3} e^{-z^2} d z=0 \] We can calculate \begin{aligned} -\int_{\gamma_3} e^{-z^2} d z &=\int_0^R e^{-\left(e^{i \pi / 4}\right)^2} e^{i \pi / 4} d t \\ &=e^{i \pi / 4} \int_0^R e^{-i t^2} d t \\ &=e^{i \pi / 4} \int_0^R \cos \left(-t^2\right) d t+i \sin \left(-t^2\right) d t \\ &=e^{i \pi / 4} \int_0^R \cos \left(t^2\right) d t-i \sin \left(t^2\right) d t \end{aligned} So we can calculate the Fresnel integrals by calculating $\int_{\gamma_1+\gamma_2} e^{-z^2} d z$, taking $R \rightarrow \infty$, dividing by $e^{i \pi / 4}$, and looking at the real and negative imaginary parts. First we show the integral over $\gamma_2$ goes to zero: