When considering tempered distributions, I am only aware of the definition of test functions of a real variable. However, is it okay to use test functions of a complex variable z that are analytic in a strip that includes the real axis? (of course they still must fall off fast enough as the real part of z goes to +/- infinity). I know of at least one test function of a real variable, [itex]e^{-x^2}[/itex] for whom the analytic continuation to the strip is trivial.
I am asking because (just for fun) I am looking at the Fourier transform of the Heaviside step function, [itex]u(t)[/itex] that is zero for t<0 and one for t>0. If we let [itex]\hat{f} [/itex] denote the Fourier transform of an arbitrary [itex]f[/itex] (test function or distribution), and let [itex]\psi[/itex] be a test function, then by definition of the Fourier transform for distributions
[tex]
\langle \hat{u}, \psi \rangle = \langle u, \hat{\psi} \rangle = \int_0^{\infty} dx\, \int_{-\infty}^\infty dt\, e^{-i x t} \psi(t)
[/tex]
I want to swap the order of integration but cannot since the reverse order integral is not defined. However, the above integral is well behaved so it should be equal to
[tex]
\lim_{\epsilon \rightarrow 0} \int_0^{\infty} dx\, e^{-\epsilon x} \int_{-\infty}^\infty dt\, e^{-i x t} \psi(t).
[/tex]
Now I can swap the order of integration and perform the first integration to obtain,
[tex]
\lim_{\epsilon \rightarrow 0} \frac{1}{i} \int_{-\infty}^\infty dt\, \frac{\psi(t)}{t - i \epsilon}.
[/tex]
This is where I would like [itex]\psi[/itex] to be analytic in a strip including the real axis and extending into the lower half plane (even by just a little bit). In that case the above integral is simple by contour integral techniques - the integral is along the real axis and as [itex]\epsilon \rightarrow 0[/itex] we indent the contour into the lower half plane and I get
[tex]
\langle \hat{u}, \psi \rangle = \pi \psi(0) + PV\int_{-\infty}^\infty dt\, \frac{\psi(t)}{it}
[/tex]
where [itex]PV[/itex] indicates the Cauchy principle value (symmetric limits about t=0). Hence,
[tex]
\hat{u}(x) = \pi \delta(x) + PV \frac{1}{ix}.
[/tex]
It works out so nice it seems like it should be fine to do this, but lots of things can be done formally that make no sense! Everything up to where I want to use contour integration is easy to find in textbooks (right now I am looking at "waves and distributions" by Jonsson and Yngvason), but I don't see authors doing the last step with contour integration and I am wondering if there is a reason - I am guessing there is but I just don't see it.
Thanks,
Jason
I am asking because (just for fun) I am looking at the Fourier transform of the Heaviside step function, [itex]u(t)[/itex] that is zero for t<0 and one for t>0. If we let [itex]\hat{f} [/itex] denote the Fourier transform of an arbitrary [itex]f[/itex] (test function or distribution), and let [itex]\psi[/itex] be a test function, then by definition of the Fourier transform for distributions
[tex]
\langle \hat{u}, \psi \rangle = \langle u, \hat{\psi} \rangle = \int_0^{\infty} dx\, \int_{-\infty}^\infty dt\, e^{-i x t} \psi(t)
[/tex]
I want to swap the order of integration but cannot since the reverse order integral is not defined. However, the above integral is well behaved so it should be equal to
[tex]
\lim_{\epsilon \rightarrow 0} \int_0^{\infty} dx\, e^{-\epsilon x} \int_{-\infty}^\infty dt\, e^{-i x t} \psi(t).
[/tex]
Now I can swap the order of integration and perform the first integration to obtain,
[tex]
\lim_{\epsilon \rightarrow 0} \frac{1}{i} \int_{-\infty}^\infty dt\, \frac{\psi(t)}{t - i \epsilon}.
[/tex]
This is where I would like [itex]\psi[/itex] to be analytic in a strip including the real axis and extending into the lower half plane (even by just a little bit). In that case the above integral is simple by contour integral techniques - the integral is along the real axis and as [itex]\epsilon \rightarrow 0[/itex] we indent the contour into the lower half plane and I get
[tex]
\langle \hat{u}, \psi \rangle = \pi \psi(0) + PV\int_{-\infty}^\infty dt\, \frac{\psi(t)}{it}
[/tex]
where [itex]PV[/itex] indicates the Cauchy principle value (symmetric limits about t=0). Hence,
[tex]
\hat{u}(x) = \pi \delta(x) + PV \frac{1}{ix}.
[/tex]
It works out so nice it seems like it should be fine to do this, but lots of things can be done formally that make no sense! Everything up to where I want to use contour integration is easy to find in textbooks (right now I am looking at "waves and distributions" by Jonsson and Yngvason), but I don't see authors doing the last step with contour integration and I am wondering if there is a reason - I am guessing there is but I just don't see it.
Thanks,
Jason
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