Integral on Dirac delta with null argument
0
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Making some calculus I end it up with the following expression:
$$int_{mathbb{R^2}} dxdy delta(x - y)delta(x - y) = int_{mathbb{R}} dx delta(0) tag1$$
Taking into account that
$$int_{mathbb{R}}dx = int_{mathbb{R}}dx e^{ixt}Big|_{t = 0} = delta(0)(2pi) tag2$$
Then, would Eq. (1) be equal to $[delta(0)]^2(2pi)$ or there is another way to compute this? Maybe there is some property of Dirac delta that I don't know, that's why I'm asking
calculus integration dirac-delta
asked Dec 3 '18 at 0:06
VickyVicky
1407
$endgroup$
add a comment |
0
$begingroup$
Making some calculus I end it up with the following expression:
$$int_{mathbb{R^2}} dxdy delta(x - y)delta(x - y) = int_{mathbb{R}} dx delta(0) tag1$$
Taking into account that
$$int_{mathbb{R}}dx = int_{mathbb{R}}dx e^{ixt}Big|_{t = 0} = delta(0)(2pi) tag2$$
Then, would Eq. (1) be equal to $[delta(0)]^2(2pi)$ or there is another way to compute this? Maybe there is some property of Dirac delta that I don't know, that's why I'm asking
calculus integration dirac-delta
asked Dec 3 '18 at 0:06
VickyVicky
1407
$endgroup$
1
$begingroup$
What you wrote doesn't make sense under any definition of $delta$
$endgroup$
– reuns
Dec 3 '18 at 0:09
$begingroup$
Unless you know the material on Schwartz distributions, you should not try to do things like this. $delta$ is a distribution, not a function, so you cannot do things to it that work for functions. You can probably search in this forum to find that "the square of the delta function" is a nonsensical thing.
$endgroup$
– GEdgar
Dec 3 '18 at 0:15
1
$begingroup$
You want to do funny calculations involving (for example) the Dirac function squared. The only meaningful approach is then to express the Dirac function as the limit of some usable function, for example a Gaussian with unit area under the curve and with a narrow width. If you substitute this into the left hand side of (1), the integration can indeed be performed. The result (after taking the limit of the width of the Gaussian to +0) is that the integral diverges, i.e. goes to +infinity.
$endgroup$
– M. Wind
Dec 3 '18 at 6:32
add a comment |
0
0
0
$begingroup$
Making some calculus I end it up with the following expression:
$$int_{mathbb{R^2}} dxdy delta(x - y)delta(x - y) = int_{mathbb{R}} dx delta(0) tag1$$
Taking into account that
$$int_{mathbb{R}}dx = int_{mathbb{R}}dx e^{ixt}Big|_{t = 0} = delta(0)(2pi) tag2$$
Then, would Eq. (1) be equal to $[delta(0)]^2(2pi)$ or there is another way to compute this? Maybe there is some property of Dirac delta that I don't know, that's why I'm asking
calculus integration dirac-delta
asked Dec 3 '18 at 0:06
VickyVicky
1407
$endgroup$
Making some calculus I end it up with the following expression:
$$int_{mathbb{R^2}} dxdy delta(x - y)delta(x - y) = int_{mathbb{R}} dx delta(0) tag1$$
Taking into account that
$$int_{mathbb{R}}dx = int_{mathbb{R}}dx e^{ixt}Big|_{t = 0} = delta(0)(2pi) tag2$$
Then, would Eq. (1) be equal to $[delta(0)]^2(2pi)$ or there is another way to compute this? Maybe there is some property of Dirac delta that I don't know, that's why I'm asking
calculus integration dirac-delta
calculus integration dirac-delta
asked Dec 3 '18 at 0:06
VickyVicky
1407
asked Dec 3 '18 at 0:06
VickyVicky
1407
edited Dec 3 '18 at 0:20
Vicky
asked Dec 3 '18 at 0:06
VickyVicky
1407
asked Dec 3 '18 at 0:06
VickyVicky
1407
asked Dec 3 '18 at 0:06
VickyVicky
1407
1407
1
$begingroup$
What you wrote doesn't make sense under any definition of $delta$
$endgroup$
– reuns
Dec 3 '18 at 0:09
$begingroup$
Unless you know the material on Schwartz distributions, you should not try to do things like this. $delta$ is a distribution, not a function, so you cannot do things to it that work for functions. You can probably search in this forum to find that "the square of the delta function" is a nonsensical thing.
$endgroup$
– GEdgar
Dec 3 '18 at 0:15
1
$begingroup$
You want to do funny calculations involving (for example) the Dirac function squared. The only meaningful approach is then to express the Dirac function as the limit of some usable function, for example a Gaussian with unit area under the curve and with a narrow width. If you substitute this into the left hand side of (1), the integration can indeed be performed. The result (after taking the limit of the width of the Gaussian to +0) is that the integral diverges, i.e. goes to +infinity.
$endgroup$
– M. Wind
Dec 3 '18 at 6:32
add a comment |
1
$begingroup$
What you wrote doesn't make sense under any definition of $delta$
$endgroup$
– reuns
Dec 3 '18 at 0:09
$begingroup$
Unless you know the material on Schwartz distributions, you should not try to do things like this. $delta$ is a distribution, not a function, so you cannot do things to it that work for functions. You can probably search in this forum to find that "the square of the delta function" is a nonsensical thing.
$endgroup$
– GEdgar
Dec 3 '18 at 0:15
1
$begingroup$
You want to do funny calculations involving (for example) the Dirac function squared. The only meaningful approach is then to express the Dirac function as the limit of some usable function, for example a Gaussian with unit area under the curve and with a narrow width. If you substitute this into the left hand side of (1), the integration can indeed be performed. The result (after taking the limit of the width of the Gaussian to +0) is that the integral diverges, i.e. goes to +infinity.
$endgroup$
– M. Wind
Dec 3 '18 at 6:32
1
1
$begingroup$
What you wrote doesn't make sense under any definition of $delta$
$endgroup$
– reuns
Dec 3 '18 at 0:09
$begingroup$
What you wrote doesn't make sense under any definition of $delta$
$endgroup$
– reuns
Dec 3 '18 at 0:09
$begingroup$
Unless you know the material on Schwartz distributions, you should not try to do things like this. $delta$ is a distribution, not a function, so you cannot do things to it that work for functions. You can probably search in this forum to find that "the square of the delta function" is a nonsensical thing.
$endgroup$
– GEdgar
Dec 3 '18 at 0:15
$begingroup$
Unless you know the material on Schwartz distributions, you should not try to do things like this. $delta$ is a distribution, not a function, so you cannot do things to it that work for functions. You can probably search in this forum to find that "the square of the delta function" is a nonsensical thing.
$endgroup$
– GEdgar
Dec 3 '18 at 0:15
1
1
$begingroup$
You want to do funny calculations involving (for example) the Dirac function squared. The only meaningful approach is then to express the Dirac function as the limit of some usable function, for example a Gaussian with unit area under the curve and with a narrow width. If you substitute this into the left hand side of (1), the integration can indeed be performed. The result (after taking the limit of the width of the Gaussian to +0) is that the integral diverges, i.e. goes to +infinity.
$endgroup$
– M. Wind
Dec 3 '18 at 6:32
$begingroup$
You want to do funny calculations involving (for example) the Dirac function squared. The only meaningful approach is then to express the Dirac function as the limit of some usable function, for example a Gaussian with unit area under the curve and with a narrow width. If you substitute this into the left hand side of (1), the integration can indeed be performed. The result (after taking the limit of the width of the Gaussian to +0) is that the integral diverges, i.e. goes to +infinity.
$endgroup$
– M. Wind
Dec 3 '18 at 6:32
add a comment |
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1
$begingroup$
What you wrote doesn't make sense under any definition of $delta$
$endgroup$
– reuns
Dec 3 '18 at 0:09
$begingroup$
Unless you know the material on Schwartz distributions, you should not try to do things like this. $delta$ is a distribution, not a function, so you cannot do things to it that work for functions. You can probably search in this forum to find that "the square of the delta function" is a nonsensical thing.
$endgroup$
– GEdgar
Dec 3 '18 at 0:15
1
$begingroup$
You want to do funny calculations involving (for example) the Dirac function squared. The only meaningful approach is then to express the Dirac function as the limit of some usable function, for example a Gaussian with unit area under the curve and with a narrow width. If you substitute this into the left hand side of (1), the integration can indeed be performed. The result (after taking the limit of the width of the Gaussian to +0) is that the integral diverges, i.e. goes to +infinity.
$endgroup$
– M. Wind
Dec 3 '18 at 6:32