0
$\begingroup$

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.

Thanks.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

$$f(z)=z^3 e^{-z^4}$$ 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 $\mathcal{C}$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.