Can I change the limits in the Fourier transform definition of the Dirac delta function?

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The Dirac delta function is often defined as
$$delta(x)=frac12piint_-infty^infty e^i p x dp$$
Is there a way in which
$$delta(x)=int_0^infty e^ipx dp$$
is also correct? For instance if $x$ or $p$ obey some condition.

fourier-transform dirac-delta

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edited Jul 17 at 16:46

Davide Morgante

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asked Jul 17 at 16:07

Andreu

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up vote
1
down vote

favorite

The Dirac delta function is often defined as
$$delta(x)=frac12piint_-infty^infty e^i p x dp$$
Is there a way in which
$$delta(x)=int_0^infty e^ipx dp$$
is also correct? For instance if $x$ or $p$ obey some condition.

fourier-transform dirac-delta

share|cite|improve this question

edited Jul 17 at 16:46

Davide Morgante

1,875220

asked Jul 17 at 16:07

Andreu

62

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

The Dirac delta function is often defined as
$$delta(x)=frac12piint_-infty^infty e^i p x dp$$
Is there a way in which
$$delta(x)=int_0^infty e^ipx dp$$
is also correct? For instance if $x$ or $p$ obey some condition.

fourier-transform dirac-delta

share|cite|improve this question

edited Jul 17 at 16:46

Davide Morgante

1,875220

asked Jul 17 at 16:07

Andreu

62

The Dirac delta function is often defined as
$$delta(x)=frac12piint_-infty^infty e^i p x dp$$
Is there a way in which
$$delta(x)=int_0^infty e^ipx dp$$
is also correct? For instance if $x$ or $p$ obey some condition.

fourier-transform dirac-delta

share|cite|improve this question

edited Jul 17 at 16:46

Davide Morgante

1,875220

asked Jul 17 at 16:07

Andreu

62

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 16:46

Davide Morgante

1,875220

edited Jul 17 at 16:46

Davide Morgante

1,875220

edited Jul 17 at 16:46

Davide Morgante

1,875220

1,875220

asked Jul 17 at 16:07

Andreu

62

asked Jul 17 at 16:07

Andreu

62

asked Jul 17 at 16:07

Andreu

62

62

add a comment | 

add a comment | 

2 Answers
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Not exactly like that. Formally,
$$
delta(x)
= frac12piint_-infty^infty e^i p x , dp
= frac12pi left( int_-infty^0 e^i p x , dp + int_0^infty e^i p x , dp right) \
= frac12pi left( int_0^infty e^-i p x , dp + int_0^infty e^i p x , dp right) \
= frac12pi int_0^infty left( e^-i p x + e^i p x right) , dp \
= frac1pi int_0^infty cos(p x) , dp \
$$

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answered Jul 17 at 16:20

md2perpe

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No, you can't. The definition you're giving is the one arising from the Fourier transform. It can be easily proven: take $fin L^1(mathbbR)$ then $$f(x) = 1over2piint_-infty^inftye^ikxleft(int_-infty^inftye^-ikyf(y)dyright)dk$$ here you cannot interchange the integrals, but you can write $$f(x) = lim_Nrightarrowinfty1over2piint_-N^Ne^ikxleft(int_-infty^inftye^-ikyf(y)dyright)dktag1$$ By the properties that the Dirac delta should have, there is $$f(x)=int_-infty^inftydelta(y-x)f(y)dy$$ so from $(1)$ we get that $$delta_N(x) = 1over2pi int_-N^Ne^ikxdk = 1over2piint_-N^0 e^ikxdk+1over2piint_0^Ne^ikxdk = \
1over2piint_0^N e^-ikxdk+1over2piint_0^Ne^ikxdk = \
1over2piint_0^Ne^ikx+e^-ikx = 1overpiint_0^N cos(kx)dk = 1overpifracsin(Nx)x tag2$$ Does this definition of the delta, in the limit, follow the properties that should have? Yes, it does, mainly $$int_-infty^inftydelta_n(x)dx=1;;;;lim_Nrightarrowpminftydelta_n(x)=0$$ So we can define the delta function as the limit of the sequence $$delta(x) = lim_Nrightarrowinfty1overpifracsin(Nx)x = lim_Nrightarrowinfty 1over2pi int_-N^Ne^ikxdk$$ The Fourier representation of the delta function is as follows $$delta(x) = int_-infty^inftye^ikxdx$$ As you can see from equation $(2)$ the only integral of $int_0^Ne^ikxdk$ it's not enough to define a Dirac delta

share|cite|improve this answer

edited Jul 17 at 16:39

answered Jul 17 at 16:24

Davide Morgante

1,875220

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

Not exactly like that. Formally,
$$
delta(x)
= frac12piint_-infty^infty e^i p x , dp
= frac12pi left( int_-infty^0 e^i p x , dp + int_0^infty e^i p x , dp right) \
= frac12pi left( int_0^infty e^-i p x , dp + int_0^infty e^i p x , dp right) \
= frac12pi int_0^infty left( e^-i p x + e^i p x right) , dp \
= frac1pi int_0^infty cos(p x) , dp \
$$

share|cite|improve this answer

answered Jul 17 at 16:20

md2perpe

5,93011022

add a comment | 

up vote
2
down vote

Not exactly like that. Formally,
$$
delta(x)
= frac12piint_-infty^infty e^i p x , dp
= frac12pi left( int_-infty^0 e^i p x , dp + int_0^infty e^i p x , dp right) \
= frac12pi left( int_0^infty e^-i p x , dp + int_0^infty e^i p x , dp right) \
= frac12pi int_0^infty left( e^-i p x + e^i p x right) , dp \
= frac1pi int_0^infty cos(p x) , dp \
$$

share|cite|improve this answer

answered Jul 17 at 16:20

md2perpe

5,93011022

add a comment | 

up vote
2
down vote

up vote
2
down vote

Not exactly like that. Formally,
$$
delta(x)
= frac12piint_-infty^infty e^i p x , dp
= frac12pi left( int_-infty^0 e^i p x , dp + int_0^infty e^i p x , dp right) \
= frac12pi left( int_0^infty e^-i p x , dp + int_0^infty e^i p x , dp right) \
= frac12pi int_0^infty left( e^-i p x + e^i p x right) , dp \
= frac1pi int_0^infty cos(p x) , dp \
$$

share|cite|improve this answer

answered Jul 17 at 16:20

md2perpe

5,93011022

Not exactly like that. Formally,
$$
delta(x)
= frac12piint_-infty^infty e^i p x , dp
= frac12pi left( int_-infty^0 e^i p x , dp + int_0^infty e^i p x , dp right) \
= frac12pi left( int_0^infty e^-i p x , dp + int_0^infty e^i p x , dp right) \
= frac12pi int_0^infty left( e^-i p x + e^i p x right) , dp \
= frac1pi int_0^infty cos(p x) , dp \
$$

share|cite|improve this answer

answered Jul 17 at 16:20

md2perpe

5,93011022

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 16:20

md2perpe

5,93011022

answered Jul 17 at 16:20

md2perpe

5,93011022

answered Jul 17 at 16:20

md2perpe

5,93011022

5,93011022

add a comment | 

add a comment | 

up vote
0
down vote

No, you can't. The definition you're giving is the one arising from the Fourier transform. It can be easily proven: take $fin L^1(mathbbR)$ then $$f(x) = 1over2piint_-infty^inftye^ikxleft(int_-infty^inftye^-ikyf(y)dyright)dk$$ here you cannot interchange the integrals, but you can write $$f(x) = lim_Nrightarrowinfty1over2piint_-N^Ne^ikxleft(int_-infty^inftye^-ikyf(y)dyright)dktag1$$ By the properties that the Dirac delta should have, there is $$f(x)=int_-infty^inftydelta(y-x)f(y)dy$$ so from $(1)$ we get that $$delta_N(x) = 1over2pi int_-N^Ne^ikxdk = 1over2piint_-N^0 e^ikxdk+1over2piint_0^Ne^ikxdk = \
1over2piint_0^N e^-ikxdk+1over2piint_0^Ne^ikxdk = \
1over2piint_0^Ne^ikx+e^-ikx = 1overpiint_0^N cos(kx)dk = 1overpifracsin(Nx)x tag2$$ Does this definition of the delta, in the limit, follow the properties that should have? Yes, it does, mainly $$int_-infty^inftydelta_n(x)dx=1;;;;lim_Nrightarrowpminftydelta_n(x)=0$$ So we can define the delta function as the limit of the sequence $$delta(x) = lim_Nrightarrowinfty1overpifracsin(Nx)x = lim_Nrightarrowinfty 1over2pi int_-N^Ne^ikxdk$$ The Fourier representation of the delta function is as follows $$delta(x) = int_-infty^inftye^ikxdx$$ As you can see from equation $(2)$ the only integral of $int_0^Ne^ikxdk$ it's not enough to define a Dirac delta

share|cite|improve this answer

edited Jul 17 at 16:39

answered Jul 17 at 16:24

Davide Morgante

1,875220

add a comment | 

up vote
0
down vote

No, you can't. The definition you're giving is the one arising from the Fourier transform. It can be easily proven: take $fin L^1(mathbbR)$ then $$f(x) = 1over2piint_-infty^inftye^ikxleft(int_-infty^inftye^-ikyf(y)dyright)dk$$ here you cannot interchange the integrals, but you can write $$f(x) = lim_Nrightarrowinfty1over2piint_-N^Ne^ikxleft(int_-infty^inftye^-ikyf(y)dyright)dktag1$$ By the properties that the Dirac delta should have, there is $$f(x)=int_-infty^inftydelta(y-x)f(y)dy$$ so from $(1)$ we get that $$delta_N(x) = 1over2pi int_-N^Ne^ikxdk = 1over2piint_-N^0 e^ikxdk+1over2piint_0^Ne^ikxdk = \
1over2piint_0^N e^-ikxdk+1over2piint_0^Ne^ikxdk = \
1over2piint_0^Ne^ikx+e^-ikx = 1overpiint_0^N cos(kx)dk = 1overpifracsin(Nx)x tag2$$ Does this definition of the delta, in the limit, follow the properties that should have? Yes, it does, mainly $$int_-infty^inftydelta_n(x)dx=1;;;;lim_Nrightarrowpminftydelta_n(x)=0$$ So we can define the delta function as the limit of the sequence $$delta(x) = lim_Nrightarrowinfty1overpifracsin(Nx)x = lim_Nrightarrowinfty 1over2pi int_-N^Ne^ikxdk$$ The Fourier representation of the delta function is as follows $$delta(x) = int_-infty^inftye^ikxdx$$ As you can see from equation $(2)$ the only integral of $int_0^Ne^ikxdk$ it's not enough to define a Dirac delta

share|cite|improve this answer

edited Jul 17 at 16:39

answered Jul 17 at 16:24

Davide Morgante

1,875220

add a comment | 

up vote
0
down vote

up vote
0
down vote

No, you can't. The definition you're giving is the one arising from the Fourier transform. It can be easily proven: take $fin L^1(mathbbR)$ then $$f(x) = 1over2piint_-infty^inftye^ikxleft(int_-infty^inftye^-ikyf(y)dyright)dk$$ here you cannot interchange the integrals, but you can write $$f(x) = lim_Nrightarrowinfty1over2piint_-N^Ne^ikxleft(int_-infty^inftye^-ikyf(y)dyright)dktag1$$ By the properties that the Dirac delta should have, there is $$f(x)=int_-infty^inftydelta(y-x)f(y)dy$$ so from $(1)$ we get that $$delta_N(x) = 1over2pi int_-N^Ne^ikxdk = 1over2piint_-N^0 e^ikxdk+1over2piint_0^Ne^ikxdk = \
1over2piint_0^N e^-ikxdk+1over2piint_0^Ne^ikxdk = \
1over2piint_0^Ne^ikx+e^-ikx = 1overpiint_0^N cos(kx)dk = 1overpifracsin(Nx)x tag2$$ Does this definition of the delta, in the limit, follow the properties that should have? Yes, it does, mainly $$int_-infty^inftydelta_n(x)dx=1;;;;lim_Nrightarrowpminftydelta_n(x)=0$$ So we can define the delta function as the limit of the sequence $$delta(x) = lim_Nrightarrowinfty1overpifracsin(Nx)x = lim_Nrightarrowinfty 1over2pi int_-N^Ne^ikxdk$$ The Fourier representation of the delta function is as follows $$delta(x) = int_-infty^inftye^ikxdx$$ As you can see from equation $(2)$ the only integral of $int_0^Ne^ikxdk$ it's not enough to define a Dirac delta

share|cite|improve this answer

edited Jul 17 at 16:39

answered Jul 17 at 16:24

Davide Morgante

1,875220

No, you can't. The definition you're giving is the one arising from the Fourier transform. It can be easily proven: take $fin L^1(mathbbR)$ then $$f(x) = 1over2piint_-infty^inftye^ikxleft(int_-infty^inftye^-ikyf(y)dyright)dk$$ here you cannot interchange the integrals, but you can write $$f(x) = lim_Nrightarrowinfty1over2piint_-N^Ne^ikxleft(int_-infty^inftye^-ikyf(y)dyright)dktag1$$ By the properties that the Dirac delta should have, there is $$f(x)=int_-infty^inftydelta(y-x)f(y)dy$$ so from $(1)$ we get that $$delta_N(x) = 1over2pi int_-N^Ne^ikxdk = 1over2piint_-N^0 e^ikxdk+1over2piint_0^Ne^ikxdk = \
1over2piint_0^N e^-ikxdk+1over2piint_0^Ne^ikxdk = \
1over2piint_0^Ne^ikx+e^-ikx = 1overpiint_0^N cos(kx)dk = 1overpifracsin(Nx)x tag2$$ Does this definition of the delta, in the limit, follow the properties that should have? Yes, it does, mainly $$int_-infty^inftydelta_n(x)dx=1;;;;lim_Nrightarrowpminftydelta_n(x)=0$$ So we can define the delta function as the limit of the sequence $$delta(x) = lim_Nrightarrowinfty1overpifracsin(Nx)x = lim_Nrightarrowinfty 1over2pi int_-N^Ne^ikxdk$$ The Fourier representation of the delta function is as follows $$delta(x) = int_-infty^inftye^ikxdx$$ As you can see from equation $(2)$ the only integral of $int_0^Ne^ikxdk$ it's not enough to define a Dirac delta

share|cite|improve this answer

edited Jul 17 at 16:39

answered Jul 17 at 16:24

Davide Morgante

1,875220

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 16:39

edited Jul 17 at 16:39

edited Jul 17 at 16:39

answered Jul 17 at 16:24

Davide Morgante

1,875220

answered Jul 17 at 16:24

Davide Morgante

1,875220

answered Jul 17 at 16:24

Davide Morgante

1,875220

1,875220

add a comment | 

add a comment | 

 

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