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arildno

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"How will this look like when I integrate the expression?"

I haven't seen the result you post before, but do remember the two following results:

[tex]\int_{-\infty}^{\infty}f(x)\delta{(x-y)}dx=f(y)[/tex]

[tex]\int_{-\infty}^{\infty}f(y)\delta{(x-y)}dy=f(x)[/tex]

Most likely, your result follows from some clever manipulation of these two basic results.

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Thanks for the replies.

Have found the answer in Parr and Yang's book on Density Functional Theory:

For a functional F=F[f] have

[tex]\delta F = \int \frac {\delta F} {\delta f(y)} \delta f(y) \,dy[/tex]

In a special case that F=F(f), i.e. F is just some function of f it is required that:

[tex]\frac {\delta F(f(x))} {\delta f(y)} = \frac {dF}{df} \delta(x-y) [/tex] in order to have:

[tex] \delta F = \int \frac {\delta F} {\delta f(y)} \delta f(y) \,dx = \frac {dF} {df} \delta f(x) [/tex]

So that taking F = f, get:

[tex] \frac {\delta f(x)} {\delta f(y)} = \delta (x-y) [/tex]

Have found the answer in Parr and Yang's book on Density Functional Theory:

For a functional F=F[f] have

[tex]\delta F = \int \frac {\delta F} {\delta f(y)} \delta f(y) \,dy[/tex]

In a special case that F=F(f), i.e. F is just some function of f it is required that:

[tex]\frac {\delta F(f(x))} {\delta f(y)} = \frac {dF}{df} \delta(x-y) [/tex] in order to have:

[tex] \delta F = \int \frac {\delta F} {\delta f(y)} \delta f(y) \,dx = \frac {dF} {df} \delta f(x) [/tex]

So that taking F = f, get:

[tex] \frac {\delta f(x)} {\delta f(y)} = \delta (x-y) [/tex]

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