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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.

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

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

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

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