Integrating a Kronecker Delta with F(u) and G(v)

[natski]

Hi alll,

I have an integral which includes a Kronecker delta:

<br /> I = \int_{u=0}^{a} \int_{v=0}^{a} F(u) G(v) \delta_{u,v} \, \mathrm{d}u \, \mathrm{d}v<br />

I know that for a 1D integral there exists the special property: \int F(u) DiracDelta(u-a) = F(a)

However, is there something equivalent for the problem I have stated in the first equation? I was thinking perhaps that:

I = \int F(u) G(u) du dv or something similar...

Cheers,

Natski

 

[Cyosis]

Fixing your latex to make it readable:

<br /> I = \int_{u=0}^{a} \int_{v=0}^{a} F(u) G(v) \delta_{u,v} du dv<br />

 

[Cyosis]

Are you sure that it is a Kronecker Delta and not a Dirac Delta?

 

[natski]

Yes I believe so, but I did derive the equation myself. I thought there would be a way to convert between the two types of delta functions...

 

[natski]

I have just obtained the following simplification, which may make the next step easier:

<br /> I = \int_{u=0}^{a} \int_{v=0}^{a} F(u) F(v) \delta_{u,v} \, \mathrm{d}u \, \mathrm{d}v<br />

i.e. the two functions are the same, just with a different variable.

Natski

 

[GRFrones]

I think it should be zero... the only thing that makes the integral with a Dirac Delta produce something different of zero is that it assumes infinite value at a certain point. For any finite value, which is the case for any function by definition (the Dirac Delta is NOT a function by definition), the Riemann integral should be zero, as it's not altered by a single point.

 

[Mute]

You're really sure it's a Kronecker delta? Were you always dealing with integrals, or did you have sums that you approximated as integrals?

How did you arrive at this equation? Kronecker deltas do not take continuous variables as their arguments, so I would suspect that you either converted sums to integrals but didn't convert the Kronecker delta or you made some other error. Please tell us how you got the Kronecker so that we can try to figure out what the problem is.

 

[natski]

Yes it is definitely Kronecker.

I also think that this integral is zero. When I plot the function, it looks like a 2D function in a 3D environment and thus the "area" under the curve is infinitesimal.

Thanks for your advise on this!

Natski

 

[HallsofIvy]

Hi alll,

I have an integral which includes a Kronecker delta:

<br /> I = \int_{u=0}^{a} \int_{v=0}^{a} F(u) G(v) \delta_{u,v} \, \mathrm{d}u \, \mathrm{d}v<br />

I know that for a 1D integral there exists the special property: \int F(u) DiracDelta(u-a) = F(a)

However, is there something equivalent for the problem I have stated in the first equation? I was thinking perhaps that:

I = \int F(u) G(u) du dv or something similar...

Cheers,

Natski

If this is, in fact, a continuous Kronecker \delta, \delta_{xy}= 1 if x= y, 0 other wise, then F(u)G(u)\delta_{uv} is 0 except on the line u= v. Since that is one dimensional it has two dimensional measure 0 and the double integral of any function on that set is 0.