A question about Jordan Lemma and estimation Lemma

Question

Let $|F(z)|\leq \frac{M}{R^k}$ for some positive numbers $M$ and $k$, where $z=Re^{i\theta}$. Prove that

$$\lim_{R\to\infty}\int_{|z|=R,\:\mathcal{R}(z)\geq0}e^{imz}F(z)\,dz=0$$

for any constant $m>0$.

The only thing I could think is using Jordan Lemma which I'm not sure how. Is it possible that the question be wrong? I appreciate any hint.

It's very similar to Proving a modified version of Jordan's lemma?. However here we have $e^{imz}$ instead of $e^{mz}$, which in this case, I couldn't prove it.

Answer 1

This time i will show a counter example that fully satisfy your hypothesis, but not the thesis. The main idea is the same of my previous answer. Please, correct me if wrong.

step 1 The real function $ \displaystyle f(x) = \frac{e^{-mx}}{x}$ is summable over $\mathbb{R^+}$ but not summable over $\mathbb{R^-}$, hence $$ \int_{-\infty}^{+\infty} \frac{e^{-mx}}{x} \; dx = - \infty $$

step 2 the complex function $\displaystyle f(z) = \frac{1}{z}$ is such that $\displaystyle |f(R e^{i \theta})| = \frac{1}{R}$.

step 3 We define four curves for $R > 1$ $$ \sigma_{1,R} : \left[ \frac{1}{R}, R \right] \to \mathbb{C} \qquad \sigma_{1,R}(t) = it \qquad \qquad \sigma_{2,R} : \left[ -R, -\frac{1}{R} \right] \to \mathbb{C} \qquad \sigma_{2,R}(t) = it $$ $$ C_{1,R}: \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \to \mathbb{C} \qquad C_{1,R}(t) = R e^{it} \qquad \qquad C_{2,R}: \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \to \mathbb{C} \qquad C_{2,R}(t) = \frac{e^{it}}{R} $$ To visualize it, the composition of this four path (noticing that $\sigma_{1,R}$ and $\sigma_{2,R}$ need to be reversed) is the boundary of a half donut

step 4 We now compute the integral of $\displaystyle \frac{e^{imz}}{z}$ over $\sigma_{1,R}, \sigma_{2,R}$ and $C_{2,R}$: $$ \int_{\sigma_{1,R}} \frac{e^{imz}}{z} \; dz = \int_{1/R}^R \frac{e^{-mt}}{t} dt \qquad \qquad \int_{\sigma_{2,R}} \frac{e^{imz}}{z} \; dz = \int_{-R}^{-1/R} \frac{e^{-mt}}{t} dt $$ Thus the limit $$ \lim_{R \to \infty} \int_{\sigma_{1,R}} \frac{e^{imz}}{z} \; dz + \int_{\sigma_{2,R}} \frac{e^{imz}}{z} \; dz = \int_{-\infty}^{\infty} \frac{e^{-mt}}{t} \; dt = \infty $$ Where $\infty$ is the complex compactification infinity. The other integral instead, by the small circle lemma - (if you need clarification i can explicitly prove it) is such that $$ \lim_{R \to \infty} \int_{C_{2,R}} \frac{e^{imz}}{z} \; dz = i \pi $$

step 4 Noticing that the composition of the aforementioned paths describe the boundary of a domain $D$ not containing $0$, the function $\displaystyle \frac{e^{imz}}{z}$ is olomorphic in this domain and in particular the integral over $\partial D$ is zero by the cauchy's integral theorem. Hence

$$ \lim_{R \to \infty} \int_{C_{1,R}} \frac{e^{imz}}{z} \; dz = \lim_{R \to \infty} \int_{\sigma_{1,R}} \frac{e^{imz}}{z} \; dz + \int_{\sigma_{2,R}} \frac{e^{imz}}{z} \; dz + \int_{C_{2,R}} \frac{e^{imz}}{z} \; dz = \int_{-\infty}^{+\infty} \frac{e^{-mt}}{t} \; dt + i \pi = \infty $$

In particular the LHS cannot be zero

Answer 2

I think that the following should be a counter example. I will build it in steps

1. The real function $\frac{ e^{-mx}}{x^2 + 1}$ is continuos and positive but not summable. So $$ \int_{-\infty}^{+\infty} \frac{ e^{-mx} }{x^2 + 1} \; dx = +\infty $$

2. The complex function $f(z) = \frac{1}{1 - z^2}$ has modulus smaller than $\frac{2}{|z|^2}$ for large enough $|z|$

3. Due to Residue theorem, since the function $\frac{e^{imz}}{1-z^2}$ has only one simple pole in the right half disk of radius R > 1, that is $1$ with residue $\frac{e^{im}}{2}$ , denoting with $\sigma_R$ the curve $$ \sigma_R : [-R,R] \quad \sigma_R(t) = it $$ then $$ \int_{|z|=R, \Re(z) > 0} e^{imz} f(z) \; dz - \int_{\sigma_R} e^{imz} f(z)\; dz = 2\pi i \frac{e^{im}}{2} $$

$$ \int_{|z|=R, \Re(z) > 0} e^{imz} f(z) \; dz = \pi e^{im} + \int_{\sigma_R} e^{imz} f(z) \; dz = i \int_{-R}^R \frac{e^{-mt}}{1 + t^2} \; dt $$

Point 1 and 3 togheter imply that the limit does not go to zero, while point 2 guarantee that all the hypothesis are satisfied