[tex]\delta(x)=\frac{d}{dx}\Theta(x)[/tex] - delta function like a first derivative of Heaviside step function(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\Theta(x)=\int dx\delta(x)[/tex]

We use integral representation of delta function

[tex]\delta(x)=\frac{1}{2\pi}\int^{\infty}_{-\infty}dke^{-ikx}[/tex]

from that we get

[tex]\Theta(x)=\int dx\frac{1}{2\pi}\int^{\infty}_{-\infty}dke^{-ikx} [/tex]

And now we use one trick!

[tex]\Theta(x)=\int dx\frac{1}{2\pi}\int^{\infty}_{-\infty}dke^{-ikx}lim_{\epsilon \rightarrow 0^+}e^{\epsilon k} [/tex]

Consider now the expression without the limit

[tex]\frac{1}{2\pi}\int^{\infty}_{-\infty}dk\int dxe^{-i(k+i\epsilon)x}=\frac{i}{2\pi}\int^{\infty}_{-\infty}dk\frac{e^{-ikx}}{k+i\epsilon}[/tex]

we get that

[tex]\Theta(x)=\frac{i}{2\pi} lim_{\epsilon \rightarrow 0+}\int^{\infty}_{-\infty}dk\frac{e^{-ikx}}{k+i\epsilon}[/tex]

Using complex integration in lower half plane for [tex]x>0[/tex] and upper half plane for [tex]x<0[/tex] we get that the upper expression is correct.But why we can change the place of limit and integral? I'm not sure.

Thanks for your answer!

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# Changing place of limit and integral - problem

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