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TIA.

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- Thread starter JayFsd
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TIA.

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HallsofIvy

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[tex]\int_C f(x)\delta (x)dx[/tex]

is equal to 1 if 0 is in set C, equal to 0 if 0 is not in set C.

More generally

[tex]\int_C f(x)\delta (x-a) dx[/itex]

is equal to 1 if a is in set C, equal to 0 if 0 is not in set C.

There is no reason C cannot be a subset of the complex numbers and a a complex number.

It can be thought of, roughly, as the limit of a sequence of functions, f

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Suppose that [tex]a[/tex] is complex and you have the integral

[tex]\int_C f\left(x\right) \delta\left(x - a\right)dx[/tex]

such that [tex]a[/tex] lies on the countour [tex]C[/tex], you would expect the result to be:

[tex]\int_C f\left(x\right) \delta\left(x - a\right)dx = f\left(a\right)[/tex].

However, to evaluate this integral for a specific contour, let's say, we do the substitution [tex] x = x\left(t\right),\, 0 <t < 1 [/tex] and get

[tex]\int_0^1 f\left(x\left(t\right)\right)\delta\left(x\left(t\right) - x\left(t_a\right)\right)\frac{dx}{dt}dt[/tex]

where [tex] x\left(t_a\right) = a[/tex]

Ok, so the question I have:

According to Wikipedia, the dirac delta scales as

[tex]\delta\left(g\left(x\right)\right) = \sum_i\frac{\delta\left(x - x_i\right)}{|g'\left(x_i\right)}[/tex]

where the [tex]x_i[/tex] are the roots of [tex]g[/tex].

So if I use that scaling property in the above integral, then I get

[tex]\int_0^1 f\left(x\left(t\right)\right)\delta\left(x\left(t\right) - x\left(t_a\right)\right)\frac{dx}{dt}dt = \frac{dx}{dt_a}/\left|\frac{dx}{dt_a}\right|f\left(x\left(t_a\right)\right)[/tex]

which doesn't agree with the expected result. It seems as if I don't understand the derivation of the scaling property well enough -- that is with real numbers it seems fine, but when you throw complex numbers in the mix I get hopelessly lost.

Anyone care to chime in and help me?

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