find the limit in distributions(space of generalized functions)

Question

How to find limit in $ \mathcal{D}^{'}$ $$ \exp(itx)(x+i0)^{-1}, t \rightarrow \infty $$ I try to use Sokhotsky's formula $(x+i0)^{-1} = -i\pi\delta(x) + \rho\frac{1}{x} $ , but did not come to a meaningful answer.

Answer 1

By Sokhotsky's formula, what we want to calculate is $$ e^{itx} \left( -i\pi\delta(x) + \operatorname{pv}\frac{1}{x} \right) = -i\pi \left( e^{itx} \delta(x) + e^{itx} \operatorname{pv}\frac{1}{x} \right) . $$

The first term is easy, using $f\delta=f(0)\delta$: $$ e^{itx} \delta(x) = e^{it0} \delta(x) = \delta(x) . $$

For the second term we first use a couple of definitions: $$ \langle e^{itx} \operatorname{pv}\frac{1}{x}, \phi(x) \rangle = \langle \operatorname{pv}\frac{1}{x}, e^{itx} \phi(x) \rangle = \lim_{\epsilon\to 0} \int_{|x|>\epsilon} \frac{1}{x} e^{itx} \phi(x) \, dx . $$ Then we note that because of symmetry only the odd part of $e^{itx}\phi(x)$ matters. The odd part equals $$ \cos tx\ \phi_{\text{odd}}(x) + i \sin tx\ \phi_{\text{even}}(x) . $$

For the first of these terms we use the fact that there exists $\psi\in C_c^\infty(\mathbb R)$ such that $\phi_{\text{odd}}(x) = x\psi(x).$ This gives $$ \lim_{\epsilon\to 0} \int_{|x|>\epsilon} \frac{1}{x} \cos tx\ \phi_{\text{odd}}(x) \, dx = \lim_{\epsilon\to 0} \int_{|x|>\epsilon} \frac{1}{x} \cos tx\ x\psi(x) \, dx = \lim_{\epsilon\to 0} \int_{|x|>\epsilon} \cos tx\ \psi(x) \, dx \\= \int \cos tx\ \psi(x) \, dx \to 0 \text{ as }t\to\infty. $$

For the second term we have $$ \lim_{\epsilon\to 0} \int_{|x|>\epsilon} \frac{1}{x} \sin tx\ \phi_{\text{even}}(x) \, dx = \lim_{\epsilon\to 0} \int_{|x|>\epsilon} \frac{\sin tx}{x} \ \phi_{\text{even}}(x) \, dx = \int_{|x|>\epsilon} \frac{\sin tx}{x} \ \phi_{\text{even}}(x) \, dx \\ \to \pi\phi_{\text{even}}(0) \text{ as } t\to\infty. $$ But $\phi_{\text{even}}(0) = \phi(0)$ so this just becomes $\pi\phi(0).$

I leave it to you to summarize the result.