a question from Sbornik Zadach (Vladimirov, UMF)

[berbas]

Hello,
I am sorry for writing in english. My russian is not enough to write in russian.
I have a question from the sbornik zadach of Vladimirov. It is this:

Evaluate the limit : $ \frac{e^{i x t}}{x-i 0} $ as $ t \to \infty $ in the space of generalized functions. The result is $ 2 \pi i \delta(x) $. As a hint, they say to use Sohotskii forumla, which is given in Vladimirov's uravneniya mat. fiz. book. of which I read all relevant sections and understand how it goes but, I would like a detailed solution if possible, since i would like to understand his (Vladimirov, and others) reasoning.

Sorry for the english. I can read your replies in russian, and thanks in advance.

[myra_panama]

berbas в сообщении #458229 писал(а):

$ \frac{e^{i x t}}{x-i 0} $

sorry what's the $x-i 0$ ?

[berbas]

$\frac{1}{x- i 0} $ term is given by the Sohotskii formula as
$\lim_{\epsilon \to 0 } \frac{1}{x-i \epsilon} = \pi i \delta(x) + \mathcal{P} \frac{1}{x} $
where
$\mathcal{P} \frac{1}{x} $ acts according to the formula $ (\mathcal{P} \frac{1}{x} , \phi) = Vp. \displaystyle \int \frac{\phi(x)}{x} \, dx $.

[ewert]

Как с Сохоцким -- не знаю (там в любом случае придётся как-то бороться с этим главным значением), а вот как в лоб:

$$\int\limits_{-R}^R\dfrac{e^{ixt}\varphi(x)}{x-i\varepsilon}\,dx=\varphi(0)\int\limits_{-R}^R\dfrac{1}{x-i\varepsilon}\,dx+\varphi(0)\int\limits_{-R}^R\dfrac{e^{ixt}-1}{x-i\varepsilon}\,dx+\int\limits_{-R}^R\dfrac{e^{ixt}(\varphi(x)-\varphi(0))}{x-i\varepsilon}\,dx\ \mathop{\longrightarrow}\limits_{\varepsilon\to+0}$$

$$\mathop{\longrightarrow}\limits_{\varepsilon\to+0}\ \varphi(0)\cdot i\pi+\varphi(0)\int\limits_{-R}^R\dfrac{e^{ixt}-1}{x}\,dx+\int\limits_{-R}^Re^{ixt}\dfrac{\varphi(x)-\varphi(0)}{x}\,dx\,.$$

Теперь при $t\to+\infty$ третье слагаемое стремится к нулю по лемме Римана, а второе -- это

$$i\,\varphi(0)\int\limits_{-R}^R\dfrac{\sin(xt)}{x}\,dx=i\,\varphi(0)\int\limits_{-Rt}^{Rt}\dfrac{\sin(y)}{y}\,dy\ \mathop{\longrightarrow}\limits_{t\to+\infty}\ i\,\varphi(0)\int\limits_{-\infty}^{+\infty}\dfrac{\sin(y)}{y}\,dy=i\,\varphi(0)\cdot\pi\,.$$

Да, а при $t\to-\infty$ общий ответ получится, соответственно, нулевым.

[berbas]

спасибо большое. you have been very helpful.