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Using Sokhotsky's formulae prove the following limits when $t \rightarrow + \infty$

$$\frac{e^{ixt}}{x - i0} \rightarrow 2 \pi i \delta(x)$$

and

$$\frac{e^{ixt}}{x + i0} \rightarrow 0.$$

Where Sokhotsky's formulae are

$$\frac{1}{x + i0} = -i \pi \delta(x) + p.v.(\frac{1}{x}),$$

$$\frac{1}{x - i0} = i \pi \delta(x) + p.v.(\frac{1}{x}).$$

1 Answers1

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We start with expanding the expressions: $$ \frac{e^{ixt}}{x \pm i0} = e^{ixt} ( \mp i\pi\,\delta(x) + \text{pv}\frac{1}{x} ) = \mp i\pi\,e^{ixt}\delta(x) + (\cos xt)\text{pv}\frac{1}{x} + i (\sin xt)\text{pv}\frac{1}{x} . $$

For the first term we use $f(x)\,\delta(x) = f(0)\,\delta(x)$ to get $\mp i\pi\,e^{ixt}\delta(x) = \mp i\pi\,\delta(x).$

I haven't been able to prove it yet, but I am sure that the second term tends to $0$ as $t \to \pm\infty$ since the quick oscillations make things cancel, even at $x=0$.

The third term tends to $i\pi\,\delta(x)$. See for example your own question about this limit. Here the oscillations do not cancel at $x=0$ but instead give a spike of height $t$.

Thus the limits are $\mp i\pi\,\delta(x) + i\pi\,\delta(x).$

md2perpe
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  • But i.e. $cos(x)$ as $x \rightarrow \infty$ does not exist. Shouldn't this be true for the second term too when $t \rightarrow \infty$? –  Mar 07 '20 at 13:04
  • $\cos(tx) \to 0$ and $\sin(tx) \to 0$ as distributions when $t \to \infty$. That is because the functions will oscillate quickly, and the "peaks" and "valleys" will cancel. This will happen for both terms away from origin. At origin the factor $\text{pv}\frac{1}{x}$ will have an important impact, though. – md2perpe Mar 08 '20 at 08:51