pellman
Apr16-03, 05:54 PM
Supposedly,
∫ ez*(z - z0)f(z) dz*dz
is proportional to f(z0) much in the same way that
(1/2π)∫ eiy(x - x0)f(x) dxdy
= ∫ δ(x - x0)f(x) dx
= f(x0)
Is this true? Could someone help convince me of it, or point me to a text?
I would say that even if true, it would be incorrect to say that
∫ ez*(z - z0)dz* = δ(z - z0)
because the integration over dz and dz* cannot be done independently in the same way that a surface integral over dxdy in the plane can (sometimes) be separated into independent integrations over x and y. Or can it?
∫ ez*(z - z0)f(z) dz*dz
is proportional to f(z0) much in the same way that
(1/2π)∫ eiy(x - x0)f(x) dxdy
= ∫ δ(x - x0)f(x) dx
= f(x0)
Is this true? Could someone help convince me of it, or point me to a text?
I would say that even if true, it would be incorrect to say that
∫ ez*(z - z0)dz* = δ(z - z0)
because the integration over dz and dz* cannot be done independently in the same way that a surface integral over dxdy in the plane can (sometimes) be separated into independent integrations over x and y. Or can it?