3
$\begingroup$

I've been given a proof that shows the following:

If $f:[a,b]\to \mathbb C$ is a continuous function and $f(t)=u(t)+iv(t)$ then $$\left| \int_a^b f(t) dt \right| \leq \int_a^b \left| f(t) \right| dt$$ The proof begins by letting $\theta$ be the principle argument of the complex number $\int_a^b f(t)dt$ and there is one step in the proof I don't understand; can anyone explain to me why $$\int_a^b e^{-i\theta}f(t) dt=\Re\bigg(\int_a^be^{-i\theta}f(t)dt\bigg)$$

Would this not imply that the imaginary part of the left hand side is equal to $0$? I'm not sure why the LHS and RHS would be equal here.

$\endgroup$
  • $\begingroup$ The imaginary part of the RHS is $0$, and LHS = RHS.. $\endgroup$ – user258700 May 5 '16 at 8:33
  • $\begingroup$ I see that the imaginary part of the RHS is $0$, my question is why are the left and right hand side equal? $\endgroup$ – user330513 May 5 '16 at 8:42
  • $\begingroup$ Sorry, I misread the question. Well in this case the claim is of course not true for arbitrary $f$. Please write down all the assumptions and what exactly is being proved $\endgroup$ – user258700 May 5 '16 at 8:53
  • $\begingroup$ I've edited the question adding in the rest of the detail. $\endgroup$ – user330513 May 5 '16 at 9:06
  • $\begingroup$ Since $e^{-i\theta}\int_a^bf(t)dt$ is real and equal to $\int_a^be^{-i\theta}f(t)dt$, then $\int_a^be^{-i\theta}f(t)dt$ is equal to real part of itself. $\endgroup$ – vnd May 5 '16 at 9:16
3
$\begingroup$

If $A = \int_a^b f(t)dt$, notice that, since $\theta$ is the principal argument of $A$,

$$A = |A| e^{i \theta}$$

Which gives:

$$|A| = Ae^{-i\theta} = \left(\int_a^b f(t)dt \right) e^{-i\theta} = \int_a^b e^{-i\theta} f(t)dt$$

Hence $\int_a^b e^{-i\theta} f(t) dt = |A| \in \Bbb R$

$\endgroup$

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.