- #1

- 675

- 4

*∫ e*

^{z*(z - z0)}f(z) dz*dzis proportional to

*f(z*much in the same way that

_{0})*(1/2π)∫ e*

= ∫ δ(x - x

= f(x

^{iy(x - x0)}f(x) dxdy= ∫ δ(x - x

_{0})f(x) dx= f(x

_{0})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

*∫ e*

^{z*(z - z0)}dz* = δ(z - z_{0})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?