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comment Is there a deep reason why replacing $\cos(x)$ with $e^{ix}$ and taking the real part often makes a contour integral work out?
Can you say a bit more about the intuitive meaning of this formula? I'm sure it's obvious to an analyst, but it's not quite so clear to me. Should I think of it as a mobius transformation and a factor that dies off at infinity?
Jul
31
comment Is there a deep reason why replacing $\cos(x)$ with $e^{ix}$ and taking the real part often makes a contour integral work out?
For my own future reference: apparently if I take $$g(t) = \frac{\cos(a t)}{(t^2 + b^2)^2}$$ then I get $$H(t) = \frac{e^{-ab} [(a b +1) z^3 + (3b^2-ab^3) z]}{2b^3(b^2+z^2)^2} + i \frac{e^{i a z}}{(b^2 + z^2)^2}$$
Jul
31
comment Is there a deep reason why replacing $\cos(x)$ with $e^{ix}$ and taking the real part often makes a contour integral work out?
Thanks. It'll take me a while to digest this, but have an accept in the meantime.
Jul
31
accepted Is there a deep reason why replacing $\cos(x)$ with $e^{ix}$ and taking the real part often makes a contour integral work out?
Jul
31
comment Is there a deep reason why replacing $\cos(x)$ with $e^{ix}$ and taking the real part often makes a contour integral work out?
Well, I included the word "deep" in the question title, and in case that was too vague I included four more precise questions, two of which (at least) have well-defined answers. If you can think of a way to be even more explicit, please suggest it.
Jul
31
comment Is there a deep reason why replacing $\cos(x)$ with $e^{ix}$ and taking the real part often makes a contour integral work out?
I know that -- it's implicit in how I solved the problem. But is it simply a coincidence that, given the function $\cos(z)$, there happens to be a function bounded in the upper half plane that matches it on the real axis?
Jul
31
revised Trying to solve a pair of trigonometric simultaneous equations
edited tags
Jul
31
answered Trying to solve a pair of trigonometric simultaneous equations
Jul
31
comment Is there a deep reason why replacing $\cos(x)$ with $e^{ix}$ and taking the real part often makes a contour integral work out?
I understand that. I even used it in the question. That's not what I asked for, though.
Jul
31
comment Is there a deep reason why replacing $\cos(x)$ with $e^{ix}$ and taking the real part often makes a contour integral work out?
@Dr.MV: The only thing I see there is that the integral along $C_R$ vanishes in the limit if $\lim_{R \to \infty} M_R = 0$. But that's just the case of Jordan's lemma where it establishes that the contour vanishes. It's not an if-and-only-if condition, and it's also not a "nice" condition, in the sense of being, say, local at $\infty$ or something.
Jul
31
comment Is there a deep reason why replacing $\cos(x)$ with $e^{ix}$ and taking the real part often makes a contour integral work out?
@wltrup: I don't think you can use a different contour, at least not in the sense here where the contour splits up into two pieces with the integral along one vanishing in the limit. The function $$f(z)= \frac{\cos(az)}{(z^2+b^2)}$$ has poles at $z = \pm bi$, and the residues are $$\operatorname{res}_{z = \pm bi} f(z) = \pm \frac{e^{ab}(ab-1) - e^{-ab}(ab+1)}{8 b^3} i.$$ So the integral isn't equal to $2 \pi i$ times some combination of the residues.
Jul
31
comment Is there a deep reason why replacing $\cos(x)$ with $e^{ix}$ and taking the real part often makes a contour integral work out?
@Dr.MV: I read through the article, but I don't see any condition mentioned there other than Jordan's lemma itself.
Jul
31
comment Is there a deep reason why replacing $\cos(x)$ with $e^{ix}$ and taking the real part often makes a contour integral work out?
@wltrup: Sorry, I meant to say that $\oint_{C_R} \frac{\cos(az) \, dz}{(z^2+b^2)^2}$ doesn't vanish as $R \to \infty$, not that it diverges. I've fixed that sentence in the question.
Jul
31
revised Is there a deep reason why replacing $\cos(x)$ with $e^{ix}$ and taking the real part often makes a contour integral work out?
added 24 characters in body
Jul
31
asked Is there a deep reason why replacing $\cos(x)$ with $e^{ix}$ and taking the real part often makes a contour integral work out?
Jul
29
accepted What's wrong with this calculation involving pullbacks of divisors on surfaces?
Jul
29
comment What's wrong with this calculation involving pullbacks of divisors on surfaces?
Why don't you add that as an answer to get this off the open questions list?
Jul
29
comment What's wrong with this calculation involving pullbacks of divisors on surfaces?
Ah, so is the degree of every generically finite map $\mathbb{P}^2 \to \mathbb{P}^2$ a perfect square?
Jul
29
asked What's wrong with this calculation involving pullbacks of divisors on surfaces?
Jul
26
answered Geometric interpretation of cubic curve?