I'm trying to evaluate the following:

Came out as a mess... I'm thinking I should be using Cauchy’s Theorem but I'm not sure if it applies here and how to prove f(z) is analytic in this case.


  • $\begingroup$ You could try letting $z=\sin(t^2)-i\frac{2t^2}{\pi}$ and substitute into the integral. Use the limits $0$ and $\sqrt{\pi/2}$. But maybe you tried that already. $\endgroup$ – Pixel Aug 27 '15 at 13:28
  • $\begingroup$ You mean into the complex line integral formula or straight into i.imgur.com/RGsn44F.png? It comes out as HUGE, is that reasonable? Sure, I'll do that from now on. $\endgroup$ – Joe Aug 27 '15 at 13:31
  • $\begingroup$ Yes. But as you say it's rather ugly ! See Jack's answer below... $\endgroup$ – Pixel Aug 27 '15 at 13:34

$$f(z)=z^3 e^{-z^4}$$ is an entire function having $F(z)=-\frac{1}{4}e^{-z^4}$ has a primitive. That implies: $$ \int_{\mathcal{C}} f(z)\,dz = F(b)-F(a),$$ where $a=0$ and $b=1-i$ are the endpoints of $\mathcal{C}$.

  • $\begingroup$ Thanks :) Is there an obvious reason why you can just ignore the actual path and just use the endpoints like this? $\endgroup$ – Joe Aug 27 '15 at 13:40
  • $\begingroup$ @Joe: use both paths and exploit the fact that the integral of a well-behaved function over a closed path is zero to see they match. $\endgroup$ – Jack D'Aurizio Aug 27 '15 at 13:49

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