# Help on contour integral on another answer on this site

Do you mind expanding on the part along the diagonal in the first answer by Robjohn to this question proof? Particularly how to achieve (3).

I am trying use a parameterization for $z=xe^{i \pi/4}$ for x from $0$ to R and I am not getting the correct result.

$$\int_{diagonal}e^{-z^2}=\int_0^R e^{-(xe^{i \pi/4})^2}e^{i\pi/4}dx=\int_0^R e^{-x^2e^{i \pi/2}}e^{i\pi/4}dx$$

• $e^{i\pi/2}=i$ and the integral is from $Re^{i\pi/4}$ to 0 so you have to change the sign? – Bob Jul 20 '18 at 20:14
• Oh you might be right. Let me take a look at that in a few. Thanks. :) – MathIsHard Jul 20 '18 at 20:39
• Thank you. I see it now. I appreciate the help. – MathIsHard Jul 20 '18 at 20:50
• One of you should post that as an answer so that the question doesn't remain unanswered. – joriki Jul 20 '18 at 23:06
• Yes, please post that Bob or let me know if you want me to do it. Thanks! – MathIsHard Jul 20 '18 at 23:34

## 1 Answer

Parametrize the segment from $0$ to $R e^{i\pi / 4}$ by: $$\gamma :[0,R]\rightarrow \mathbb{C}, t\mapsto te^{i(\pi/4)}.$$ Since you want to integrate from $Re^{i\pi/4}$ to $0$ and not from $0$ to $Re^{i\pi/4}$, you have to switch the orientation of this curve, with the result that the integral changes sign, so the value of the integral you're looking for is $$-\int_\gamma e^{-z^2}dz= -\int_0 ^ R e^{-\gamma(t)^2}\gamma'(t)dt = -\int_0 ^{R} e^{-(te^{i\pi/4})^2}e^{i\pi/4}dt=\\-e^{i\pi/4}\int_0 ^{R} e^{-t^2e^{i\pi/2}}dt=-e^{i\pi/4}\int_0 ^{R} e^{-it^2}dt,$$ where we used the well known fact that $e^{i\pi/2}=i.$ Now, letting $R\rightarrow\infty$ you get the result claimed in the linked answer.

• Thanks Bob :) appreciate it. – MathIsHard Jul 21 '18 at 5:03