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if $f \sim g$ is $\text{Im}(f) \sim \text{Im}(g)$?

I wanted to find an asymptotic expansion to $\int_0^\infty e^{-xt} \sin(t) dt$ as $x \to \infty$

My thought was to look at $\int_0^\infty e^{-xt}e^{it}dt$ and then apply Watson's Lemma to show that

$$\int_0^\infty e^{-xt}e^{it}dt \sim \sum_{n=0}^\infty \frac{i^n}{x^{n+1}} \ \ \ \text{ as } \ x \to \infty$$ And then take the Imaginary parts of both sides to get:

$$\int_0^\infty e^{-xt} \sin(t)dt \sim \frac1{x^2}\sum_{n=0}^\infty \frac{(-1)^n}{x^{2n}} \ \ \ \text{ as } \ x \to \infty$$

Is this valid? I would like to say it is, as the imaginary parts and real parts should be independent of each other. I could aslo rephrase this as:

If $(f, g) \sim (F, G)$ is $f \sim F$ and $g \sim G$?

Since complex numbers (and functions) can be thought of as vectors. Again, I would believe this to be true for the same reasons as before, but don't know a real way to convince myself in either case.

infinitylord
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1 Answers1

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Consider

$$ a_n = n + i \qquad \text{and} \qquad b_n = n + i\sqrt{n}. $$

Then $a_n \sim b_n$ but $\operatorname{Im} a_n = 1 \not\sim \sqrt{n} = \operatorname{Im} b_n$ as $n \to \infty$.

Antonio Vargas
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  • Thank you for the counterexample. Is there any condition I can add so that the statement becomes true? I am fairly certain the method I used provides a correct asymptotic expansion, though I am not sure how to justify why – infinitylord Jan 15 '18 at 10:05
  • You get the right asymptotic in your case because Watson's lemma basically replaces $e^{it}$ with its Maclaurin series and integrates term-by-term; the imaginary part of the Maclaurin series for $e^{it}$ is the Maclaurin series for $\sin t$, so you get the same answer that you would if you had just replaced $\sin t$ by its Maclaurin series then integrated term-by-term. – Antonio Vargas Jan 16 '18 at 06:56
  • I see, thank you. I ended up using the fact that if $f(x) \sim \sum a_n \phi_n(x)$ and $g(x) \sim \sum b_n \phi_n(x)$ then $\alpha f(x) + \beta g(x) \sim \sum (\alpha a_n + \beta b_n) \phi_n(x)$, but that's effectively what your argument uses as well. – infinitylord Jan 16 '18 at 14:29