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What is principal value in delta function integral? [closed]
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The delta function may have different forms of definition. One related to Fourier transform is shown below,
$$int_-infty^infty!dt ~e^iomega t~=~2pidelta(omega).$$
then I wonder what if only the positive (or negative) frequency part be considered. We may run into singularity for the integral below,
$$int_0^infty!dt ~e^iomega t~=~?$$
I find it's related to principal value. Can anyone give a solid explanation and derivation for this integral?
fourier-transform dirac-delta-distributions
asked Aug 8 at 18:08
kinder chan
174
closed as off-topic by AccidentalFourierTransform, Emilio Pisanty, Kyle Kanos, glS, Jon Custer Aug 12 at 17:18
- This question does not appear to be about physics within the scope defined in the help center.
add a comment |Â
up vote
0
down vote
favorite
The delta function may have different forms of definition. One related to Fourier transform is shown below,
$$int_-infty^infty!dt ~e^iomega t~=~2pidelta(omega).$$
then I wonder what if only the positive (or negative) frequency part be considered. We may run into singularity for the integral below,
$$int_0^infty!dt ~e^iomega t~=~?$$
I find it's related to principal value. Can anyone give a solid explanation and derivation for this integral?
fourier-transform dirac-delta-distributions
asked Aug 8 at 18:08
kinder chan
174
closed as off-topic by AccidentalFourierTransform, Emilio Pisanty, Kyle Kanos, glS, Jon Custer Aug 12 at 17:18
- This question does not appear to be about physics within the scope defined in the help center.
1
Why should the integral on half-domain be the half of the first result? $e^iwt$ is not even on the domain. Why do you expect the second result?
– apt45
Aug 8 at 18:12
1
The delta function is even
– kinder chan
Aug 8 at 18:27
3
Would Mathematics be a better home for this question?
– Qmechanic♦
Aug 8 at 18:43
2
Your second integral is (by definition) the positive-frequency delta function $2pidelta_+(w)$. It has a well-known representation in terms of the distribution $frac1w+i0^+$. In any case, I voted to migrate to math.SE.
– AccidentalFourierTransform
Aug 8 at 18:43
The function under the integral oscillates with any $omega ne 0$, but does not oscillate for $omega=0$. The oscillations cancel themselves from $-infty$ to $+infty$, but not form $0$ to $+infty$. BTW this is not a definition of the Delta Function (or not a good one at least), but just a Fourier transform of it. You could say though that, if $int_-infty^+inftyf(x)delta(0)dx=f(0)$, then $int_0^+inftyf(x)delta(0)dx=dfrac12f(0)$
– safesphere
Aug 8 at 22:41
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The delta function may have different forms of definition. One related to Fourier transform is shown below,
$$int_-infty^infty!dt ~e^iomega t~=~2pidelta(omega).$$
then I wonder what if only the positive (or negative) frequency part be considered. We may run into singularity for the integral below,
$$int_0^infty!dt ~e^iomega t~=~?$$
I find it's related to principal value. Can anyone give a solid explanation and derivation for this integral?
fourier-transform dirac-delta-distributions
asked Aug 8 at 18:08
kinder chan
174
The delta function may have different forms of definition. One related to Fourier transform is shown below,
$$int_-infty^infty!dt ~e^iomega t~=~2pidelta(omega).$$
then I wonder what if only the positive (or negative) frequency part be considered. We may run into singularity for the integral below,
$$int_0^infty!dt ~e^iomega t~=~?$$
I find it's related to principal value. Can anyone give a solid explanation and derivation for this integral?
fourier-transform dirac-delta-distributions
asked Aug 8 at 18:08
kinder chan
174
edited Aug 8 at 23:53
asked Aug 8 at 18:08
kinder chan
174
asked Aug 8 at 18:08
kinder chan
174
asked Aug 8 at 18:08
kinder chan
174
174
closed as off-topic by AccidentalFourierTransform, Emilio Pisanty, Kyle Kanos, glS, Jon Custer Aug 12 at 17:18
- This question does not appear to be about physics within the scope defined in the help center.
closed as off-topic by AccidentalFourierTransform, Emilio Pisanty, Kyle Kanos, glS, Jon Custer Aug 12 at 17:18
- This question does not appear to be about physics within the scope defined in the help center.
1
Why should the integral on half-domain be the half of the first result? $e^iwt$ is not even on the domain. Why do you expect the second result?
– apt45
Aug 8 at 18:12
1
The delta function is even
– kinder chan
Aug 8 at 18:27
3
Would Mathematics be a better home for this question?
– Qmechanic♦
Aug 8 at 18:43
2
Your second integral is (by definition) the positive-frequency delta function $2pidelta_+(w)$. It has a well-known representation in terms of the distribution $frac1w+i0^+$. In any case, I voted to migrate to math.SE.
– AccidentalFourierTransform
Aug 8 at 18:43
The function under the integral oscillates with any $omega ne 0$, but does not oscillate for $omega=0$. The oscillations cancel themselves from $-infty$ to $+infty$, but not form $0$ to $+infty$. BTW this is not a definition of the Delta Function (or not a good one at least), but just a Fourier transform of it. You could say though that, if $int_-infty^+inftyf(x)delta(0)dx=f(0)$, then $int_0^+inftyf(x)delta(0)dx=dfrac12f(0)$
– safesphere
Aug 8 at 22:41
add a comment |Â
1
Why should the integral on half-domain be the half of the first result? $e^iwt$ is not even on the domain. Why do you expect the second result?
– apt45
Aug 8 at 18:12
1
The delta function is even
– kinder chan
Aug 8 at 18:27
3
Would Mathematics be a better home for this question?
– Qmechanic♦
Aug 8 at 18:43
2
Your second integral is (by definition) the positive-frequency delta function $2pidelta_+(w)$. It has a well-known representation in terms of the distribution $frac1w+i0^+$. In any case, I voted to migrate to math.SE.
– AccidentalFourierTransform
Aug 8 at 18:43
The function under the integral oscillates with any $omega ne 0$, but does not oscillate for $omega=0$. The oscillations cancel themselves from $-infty$ to $+infty$, but not form $0$ to $+infty$. BTW this is not a definition of the Delta Function (or not a good one at least), but just a Fourier transform of it. You could say though that, if $int_-infty^+inftyf(x)delta(0)dx=f(0)$, then $int_0^+inftyf(x)delta(0)dx=dfrac12f(0)$
– safesphere
Aug 8 at 22:41
1
1
Why should the integral on half-domain be the half of the first result? $e^iwt$ is not even on the domain. Why do you expect the second result?
– apt45
Aug 8 at 18:12
Why should the integral on half-domain be the half of the first result? $e^iwt$ is not even on the domain. Why do you expect the second result?
– apt45
Aug 8 at 18:12
1
1
The delta function is even
– kinder chan
Aug 8 at 18:27
The delta function is even
– kinder chan
Aug 8 at 18:27
3
3
Would Mathematics be a better home for this question?
– Qmechanic♦
Aug 8 at 18:43
Would Mathematics be a better home for this question?
– Qmechanic♦
Aug 8 at 18:43
2
2
Your second integral is (by definition) the positive-frequency delta function $2pidelta_+(w)$. It has a well-known representation in terms of the distribution $frac1w+i0^+$. In any case, I voted to migrate to math.SE.
– AccidentalFourierTransform
Aug 8 at 18:43
Your second integral is (by definition) the positive-frequency delta function $2pidelta_+(w)$. It has a well-known representation in terms of the distribution $frac1w+i0^+$. In any case, I voted to migrate to math.SE.
– AccidentalFourierTransform
Aug 8 at 18:43
The function under the integral oscillates with any $omega ne 0$, but does not oscillate for $omega=0$. The oscillations cancel themselves from $-infty$ to $+infty$, but not form $0$ to $+infty$. BTW this is not a definition of the Delta Function (or not a good one at least), but just a Fourier transform of it. You could say though that, if $int_-infty^+inftyf(x)delta(0)dx=f(0)$, then $int_0^+inftyf(x)delta(0)dx=dfrac12f(0)$
– safesphere
Aug 8 at 22:41
The function under the integral oscillates with any $omega ne 0$, but does not oscillate for $omega=0$. The oscillations cancel themselves from $-infty$ to $+infty$, but not form $0$ to $+infty$. BTW this is not a definition of the Delta Function (or not a good one at least), but just a Fourier transform of it. You could say though that, if $int_-infty^+inftyf(x)delta(0)dx=f(0)$, then $int_0^+inftyf(x)delta(0)dx=dfrac12f(0)$
– safesphere
Aug 8 at 22:41
add a comment |Â
3 Answers
3
active
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up vote
4
down vote
my question is how to derive the second integral in the original
question.
Try using the unit step function $u(t)$ to rewrite the integral as a Fourier transform:
$$int_0^inftymathrmdt,e^iomega t = int_-infty^inftymathrmdt,e^iomega t,u(t) = int_-infty^inftymathrmdtau,e^-iomega tau,u(-tau),quad tau = -t$$
So, the second integral is Fourier transform of the time reversed unit step function. The Fourier transform of the unit step function can be looked up in a table of Fourier transform pairs:
$$u(t) leftrightarrow left(frac1iomega + pidelta(omega) right) $$
Using the time reversal property, $mathcalFx(-t) = X(-omega)$, we have
$$int_0^inftymathrmdt,e^iomega t = fraciomega + pidelta(omega)$$
answered Aug 8 at 20:40
Hal Hollis
1,23728
add a comment |Â
up vote
3
down vote
So what does it mean to be a delta function? It means that something is a mathematical object which satisfies for all other nice $f$,
$$int_-infty^infty domega~f(omega)~delta(omega) = f(0).$$
More formally, it turns out that the $delta(omega)$ can never truly be separated from the $int_-infty^infty domega$ sign: our writing of $delta(omega)$ is really a failure of notation where we would want to actually describe something more like, $$int_mathbb R dt~left[f mapsto int_mathbb R domega ~e^iomega t f(omega)right] = fmapsto 2pi~f(0),$$ where $mapsto$ denotes a function which accepts a smooth function as input and integration (really, addition and multiplication) of such functions is done "pointwise" (so that $f + g = xmapsto f(x) + g(x)$ for example).
Setting up the integral
Anyway the general approach to this problem is to shift the subproblems off the real axis by an infinitesimal amount, so that $$int_-infty^infty dt~e^iomega t =
lim_epsilonto 0^+left(int_-infty^0 dt~e^i(omega - i epsilon) t + int_0^infty dt~e^i(omega + iepsilon)t right),$$hence your question is actually one of the "stepping stones" we need to address in order to handle the above integrals. One finds of course,$$int_0^infty dt~ e^(iomega - epsilon) t = -frac 1iomega - epsilon = fraciomega + i epsilon, ~~epsilon > 0,$$ or substituting in the fuller expression using $mapsto$ we would find that this half-integral is, $$int_0^infty dt~left[f mapsto int_mathbb R domega~e^iomega t f(omega)right] = fmapsto lim_epsilonto 0^+int_mathbb R domega~fraciomega + i epsilon~f(omega). $$ It is when we try to evaluate this with complex analysis that we need the Cauchy principal value.
Contour deformation
Cauchy's integral theorem allows us to deform contours off the real line into the complex plane without changing the value. If we do this before we take the limit of $epsilon to 0$ we can make taking that limit very easy. So we deform the contour a little into the positive imaginary half of the plane around $omega = 0$, a half-circle with radius $r$, which we will later shrink to zero size. So on this circle $omega = r~e^itheta$ with $theta: pi to 0$ and thus $domega = i~r~e^itheta~dtheta.$ To save some space let $k_epsilon(omega) = frac iomega + iepsilon,$ the resulting expression for the right-hand-side is,
$$f mapsto lim_rto 0^+lim_epsilonto 0^+ left(int_-infty^-rdomega ~k_epsilon(omega)~f(omega) + int_r^inftydomega ~k_epsilon(omega)~f(omega) + int_pi^0 i~r~e^itheta dtheta~k_epsilon(re^itheta)~f(r~e^itheta) right).$$
We can now take the limit $epsilon to 0^+$ by simply substituting $epsilon = 0$ in here; this contour does not hit any singularities when we do so. We find that $k_0 = i/omega$ and the first term in the limit as $rto 0^+$ is just the Cauchy principal value of an integrand, while the rightmost term is also interesting due to cancellation of the a term $r e^itheta$ from $domega$ with the term $1/(r e^itheta)$ from $k_0$:
$$
f mapsto operatornamePVint_-infty^infty domega ~frac iomega ~f(omega) - lim_rto 0^+ int_pi^0 dtheta~f(r~e^itheta).$$
So we can also take this limit as $rto 0^+$ by direct substitution and we find,
$$int_0^infty dt~left[f mapsto int_mathbb R domega~e^iomega t f(omega)right] =
f mapsto operatornamePVint_-infty^infty domega ~frac iomega ~f(omega) + pi~f(0).$$
If the principal value is understood as implied, one might instead write, $$ int_0^infty domega~ e^iomega t = fraciomega + pi~delta(omega).$$
What about the other side of the original integral?
As one might imagine, the integral for negative $t$ is largely the reverse of this. Indeed one has $$int_-infty^0 dt~e^(iomega + epsilon)t = frac1iomega + epsilon = frac -iomega - i epsilon.$$ The principal value analysis is absolutely identical except that the numerator contains $-i$ rather than $+i$, but the semicircle must dip into the negative half-plane rather than the positive half plane, yielding:
$$f mapsto operatornamePVint_-infty^infty domega ~frac -iomega ~f(omega) +
lim_rto 0^+ int_-pi^0 i~r~e^itheta dtheta~frac-ir e^itheta ~f(r~e^itheta).$$
Thus we have, perhaps surprisingly,
$$int_-infty^0 dt~e^iomega t = - fraciomega + pi~delta(omega).$$The sum of these then proves our original statement: $$
int_-infty^infty dt~e^iomega t = 2pi~delta(omega).
$$
Where were your intuitions failing you?
You perhaps thought that since $delta(omega)$ was even in $omega$, it meant that $e^iomega t$ was even in $t$. That is somewhat dubious reasoning in general: usually the (indefinite) integral of an even function is an odd function. However that reasoning is not 100% bad, it is just targeted at the wrong domain: it needs to be directed at a function of $t$. The real part of $e^iomega t$ is indeed even and hence we can expect that the real part of the integral $0toinfty$ is $pi~delta(omega)$ just as we eventually derived. But the imaginary part is odd in $t$ and hence, we would think, it's doomed to cancel on the domain $-infty to infty.$ Accordingly we could have expected that the integral $0 to infty$ contains some imaginary term which is the precise negative of the integral $-infty to 0$ such that they both cancel in the end.
answered Aug 8 at 21:55
CR Drost
20.2k11757
+1, this answer clearly took some time to compose and format.
– Alfred Centauri
Aug 8 at 23:33
@AlfredCentauri Most answers by Chris tend to have a nice and clean formatting, which add value to his clear explanations. Not sure why I'm telling you this. I guess this time is as good as any other to acknowledge and appreciate his great contributions to this site. Cheers!
– AccidentalFourierTransform
Aug 9 at 0:02
Well thanks for saying so! In retrospect I suppose that you're right, it's probably a dupe of something on maths.se.
– CR Drost
Aug 9 at 2:36
add a comment |Â
up vote
1
down vote
why the half of the integral is not half of delta function?
Because the function is not even.
To illustrate why, consider:
$$
int_-infty^infty dt e^iomega t
=
int_-infty^0 dt e^iomega t+int_0^infty dt e^iomega t
$$
$$
=
int_0^infty dse^-iomega s + int_0^infty dt e^iomega t;,
$$
where I made the change of variable $s=-t$ in the first integral of the lower line.
But since $s$ is just a dummy variable, I can also write:
$$
=int_0^infty dte^-iomega t + int_0^infty dt e^iomega t
$$
$$
neq 2int_0^infty dt e^iomega t
$$
answered Aug 8 at 18:29
hft
3,8931821
Thanks, my point is not to discuss the odd and even, my question is how to derive the second integral in the original question.
– kinder chan
Aug 8 at 18:31
what do you mean "how to derive the second integral?" Starting from what? Ending with what? You seem to have edited your question to make it even more vague and hard to understand. Could you edit to provide some context?
– hft
Aug 8 at 19:03
My bad for the confusion. I mean $int_0^infty!dt ~e^iomega t$, I also re-edit the question.
– kinder chan
Aug 8 at 19:12
I advise you to edit. What you wrote is right, but this explains why the second integral is not a delta whose argument $omega $ is defined on the whole real axis. The integral, infact, is a delta defined on the positive frequencies, as already mentioned in a comment above. Explain that in definition of the delta function in its integrale form you have to take into account of the definition domain of the argument.
– MRT
Aug 8 at 20:26
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
my question is how to derive the second integral in the original
question.
Try using the unit step function $u(t)$ to rewrite the integral as a Fourier transform:
$$int_0^inftymathrmdt,e^iomega t = int_-infty^inftymathrmdt,e^iomega t,u(t) = int_-infty^inftymathrmdtau,e^-iomega tau,u(-tau),quad tau = -t$$
So, the second integral is Fourier transform of the time reversed unit step function. The Fourier transform of the unit step function can be looked up in a table of Fourier transform pairs:
$$u(t) leftrightarrow left(frac1iomega + pidelta(omega) right) $$
Using the time reversal property, $mathcalFx(-t) = X(-omega)$, we have
$$int_0^inftymathrmdt,e^iomega t = fraciomega + pidelta(omega)$$
answered Aug 8 at 20:40
Hal Hollis
1,23728
add a comment |Â
up vote
4
down vote
my question is how to derive the second integral in the original
question.
Try using the unit step function $u(t)$ to rewrite the integral as a Fourier transform:
$$int_0^inftymathrmdt,e^iomega t = int_-infty^inftymathrmdt,e^iomega t,u(t) = int_-infty^inftymathrmdtau,e^-iomega tau,u(-tau),quad tau = -t$$
So, the second integral is Fourier transform of the time reversed unit step function. The Fourier transform of the unit step function can be looked up in a table of Fourier transform pairs:
$$u(t) leftrightarrow left(frac1iomega + pidelta(omega) right) $$
Using the time reversal property, $mathcalFx(-t) = X(-omega)$, we have
$$int_0^inftymathrmdt,e^iomega t = fraciomega + pidelta(omega)$$
answered Aug 8 at 20:40
Hal Hollis
1,23728
add a comment |Â
up vote
4
down vote
up vote
4
down vote
my question is how to derive the second integral in the original
question.
Try using the unit step function $u(t)$ to rewrite the integral as a Fourier transform:
$$int_0^inftymathrmdt,e^iomega t = int_-infty^inftymathrmdt,e^iomega t,u(t) = int_-infty^inftymathrmdtau,e^-iomega tau,u(-tau),quad tau = -t$$
So, the second integral is Fourier transform of the time reversed unit step function. The Fourier transform of the unit step function can be looked up in a table of Fourier transform pairs:
$$u(t) leftrightarrow left(frac1iomega + pidelta(omega) right) $$
Using the time reversal property, $mathcalFx(-t) = X(-omega)$, we have
$$int_0^inftymathrmdt,e^iomega t = fraciomega + pidelta(omega)$$
answered Aug 8 at 20:40
Hal Hollis
1,23728
my question is how to derive the second integral in the original
question.
Try using the unit step function $u(t)$ to rewrite the integral as a Fourier transform:
$$int_0^inftymathrmdt,e^iomega t = int_-infty^inftymathrmdt,e^iomega t,u(t) = int_-infty^inftymathrmdtau,e^-iomega tau,u(-tau),quad tau = -t$$
So, the second integral is Fourier transform of the time reversed unit step function. The Fourier transform of the unit step function can be looked up in a table of Fourier transform pairs:
$$u(t) leftrightarrow left(frac1iomega + pidelta(omega) right) $$
Using the time reversal property, $mathcalFx(-t) = X(-omega)$, we have
$$int_0^inftymathrmdt,e^iomega t = fraciomega + pidelta(omega)$$
answered Aug 8 at 20:40
Hal Hollis
1,23728
answered Aug 8 at 20:40
Hal Hollis
1,23728
answered Aug 8 at 20:40
Hal Hollis
1,23728
answered Aug 8 at 20:40
Hal Hollis
1,23728
1,23728
add a comment |Â
add a comment |Â
up vote
3
down vote
So what does it mean to be a delta function? It means that something is a mathematical object which satisfies for all other nice $f$,
$$int_-infty^infty domega~f(omega)~delta(omega) = f(0).$$
More formally, it turns out that the $delta(omega)$ can never truly be separated from the $int_-infty^infty domega$ sign: our writing of $delta(omega)$ is really a failure of notation where we would want to actually describe something more like, $$int_mathbb R dt~left[f mapsto int_mathbb R domega ~e^iomega t f(omega)right] = fmapsto 2pi~f(0),$$ where $mapsto$ denotes a function which accepts a smooth function as input and integration (really, addition and multiplication) of such functions is done "pointwise" (so that $f + g = xmapsto f(x) + g(x)$ for example).
Setting up the integral
Anyway the general approach to this problem is to shift the subproblems off the real axis by an infinitesimal amount, so that $$int_-infty^infty dt~e^iomega t =
lim_epsilonto 0^+left(int_-infty^0 dt~e^i(omega - i epsilon) t + int_0^infty dt~e^i(omega + iepsilon)t right),$$hence your question is actually one of the "stepping stones" we need to address in order to handle the above integrals. One finds of course,$$int_0^infty dt~ e^(iomega - epsilon) t = -frac 1iomega - epsilon = fraciomega + i epsilon, ~~epsilon > 0,$$ or substituting in the fuller expression using $mapsto$ we would find that this half-integral is, $$int_0^infty dt~left[f mapsto int_mathbb R domega~e^iomega t f(omega)right] = fmapsto lim_epsilonto 0^+int_mathbb R domega~fraciomega + i epsilon~f(omega). $$ It is when we try to evaluate this with complex analysis that we need the Cauchy principal value.
Contour deformation
Cauchy's integral theorem allows us to deform contours off the real line into the complex plane without changing the value. If we do this before we take the limit of $epsilon to 0$ we can make taking that limit very easy. So we deform the contour a little into the positive imaginary half of the plane around $omega = 0$, a half-circle with radius $r$, which we will later shrink to zero size. So on this circle $omega = r~e^itheta$ with $theta: pi to 0$ and thus $domega = i~r~e^itheta~dtheta.$ To save some space let $k_epsilon(omega) = frac iomega + iepsilon,$ the resulting expression for the right-hand-side is,
$$f mapsto lim_rto 0^+lim_epsilonto 0^+ left(int_-infty^-rdomega ~k_epsilon(omega)~f(omega) + int_r^inftydomega ~k_epsilon(omega)~f(omega) + int_pi^0 i~r~e^itheta dtheta~k_epsilon(re^itheta)~f(r~e^itheta) right).$$
We can now take the limit $epsilon to 0^+$ by simply substituting $epsilon = 0$ in here; this contour does not hit any singularities when we do so. We find that $k_0 = i/omega$ and the first term in the limit as $rto 0^+$ is just the Cauchy principal value of an integrand, while the rightmost term is also interesting due to cancellation of the a term $r e^itheta$ from $domega$ with the term $1/(r e^itheta)$ from $k_0$:
$$
f mapsto operatornamePVint_-infty^infty domega ~frac iomega ~f(omega) - lim_rto 0^+ int_pi^0 dtheta~f(r~e^itheta).$$
So we can also take this limit as $rto 0^+$ by direct substitution and we find,
$$int_0^infty dt~left[f mapsto int_mathbb R domega~e^iomega t f(omega)right] =
f mapsto operatornamePVint_-infty^infty domega ~frac iomega ~f(omega) + pi~f(0).$$
If the principal value is understood as implied, one might instead write, $$ int_0^infty domega~ e^iomega t = fraciomega + pi~delta(omega).$$
What about the other side of the original integral?
As one might imagine, the integral for negative $t$ is largely the reverse of this. Indeed one has $$int_-infty^0 dt~e^(iomega + epsilon)t = frac1iomega + epsilon = frac -iomega - i epsilon.$$ The principal value analysis is absolutely identical except that the numerator contains $-i$ rather than $+i$, but the semicircle must dip into the negative half-plane rather than the positive half plane, yielding:
$$f mapsto operatornamePVint_-infty^infty domega ~frac -iomega ~f(omega) +
lim_rto 0^+ int_-pi^0 i~r~e^itheta dtheta~frac-ir e^itheta ~f(r~e^itheta).$$
Thus we have, perhaps surprisingly,
$$int_-infty^0 dt~e^iomega t = - fraciomega + pi~delta(omega).$$The sum of these then proves our original statement: $$
int_-infty^infty dt~e^iomega t = 2pi~delta(omega).
$$
Where were your intuitions failing you?
You perhaps thought that since $delta(omega)$ was even in $omega$, it meant that $e^iomega t$ was even in $t$. That is somewhat dubious reasoning in general: usually the (indefinite) integral of an even function is an odd function. However that reasoning is not 100% bad, it is just targeted at the wrong domain: it needs to be directed at a function of $t$. The real part of $e^iomega t$ is indeed even and hence we can expect that the real part of the integral $0toinfty$ is $pi~delta(omega)$ just as we eventually derived. But the imaginary part is odd in $t$ and hence, we would think, it's doomed to cancel on the domain $-infty to infty.$ Accordingly we could have expected that the integral $0 to infty$ contains some imaginary term which is the precise negative of the integral $-infty to 0$ such that they both cancel in the end.
answered Aug 8 at 21:55
CR Drost
20.2k11757
+1, this answer clearly took some time to compose and format.
– Alfred Centauri
Aug 8 at 23:33
@AlfredCentauri Most answers by Chris tend to have a nice and clean formatting, which add value to his clear explanations. Not sure why I'm telling you this. I guess this time is as good as any other to acknowledge and appreciate his great contributions to this site. Cheers!
– AccidentalFourierTransform
Aug 9 at 0:02
Well thanks for saying so! In retrospect I suppose that you're right, it's probably a dupe of something on maths.se.
– CR Drost
Aug 9 at 2:36
add a comment |Â
up vote
3
down vote
So what does it mean to be a delta function? It means that something is a mathematical object which satisfies for all other nice $f$,
$$int_-infty^infty domega~f(omega)~delta(omega) = f(0).$$
More formally, it turns out that the $delta(omega)$ can never truly be separated from the $int_-infty^infty domega$ sign: our writing of $delta(omega)$ is really a failure of notation where we would want to actually describe something more like, $$int_mathbb R dt~left[f mapsto int_mathbb R domega ~e^iomega t f(omega)right] = fmapsto 2pi~f(0),$$ where $mapsto$ denotes a function which accepts a smooth function as input and integration (really, addition and multiplication) of such functions is done "pointwise" (so that $f + g = xmapsto f(x) + g(x)$ for example).
Setting up the integral
Anyway the general approach to this problem is to shift the subproblems off the real axis by an infinitesimal amount, so that $$int_-infty^infty dt~e^iomega t =
lim_epsilonto 0^+left(int_-infty^0 dt~e^i(omega - i epsilon) t + int_0^infty dt~e^i(omega + iepsilon)t right),$$hence your question is actually one of the "stepping stones" we need to address in order to handle the above integrals. One finds of course,$$int_0^infty dt~ e^(iomega - epsilon) t = -frac 1iomega - epsilon = fraciomega + i epsilon, ~~epsilon > 0,$$ or substituting in the fuller expression using $mapsto$ we would find that this half-integral is, $$int_0^infty dt~left[f mapsto int_mathbb R domega~e^iomega t f(omega)right] = fmapsto lim_epsilonto 0^+int_mathbb R domega~fraciomega + i epsilon~f(omega). $$ It is when we try to evaluate this with complex analysis that we need the Cauchy principal value.
Contour deformation
Cauchy's integral theorem allows us to deform contours off the real line into the complex plane without changing the value. If we do this before we take the limit of $epsilon to 0$ we can make taking that limit very easy. So we deform the contour a little into the positive imaginary half of the plane around $omega = 0$, a half-circle with radius $r$, which we will later shrink to zero size. So on this circle $omega = r~e^itheta$ with $theta: pi to 0$ and thus $domega = i~r~e^itheta~dtheta.$ To save some space let $k_epsilon(omega) = frac iomega + iepsilon,$ the resulting expression for the right-hand-side is,
$$f mapsto lim_rto 0^+lim_epsilonto 0^+ left(int_-infty^-rdomega ~k_epsilon(omega)~f(omega) + int_r^inftydomega ~k_epsilon(omega)~f(omega) + int_pi^0 i~r~e^itheta dtheta~k_epsilon(re^itheta)~f(r~e^itheta) right).$$
We can now take the limit $epsilon to 0^+$ by simply substituting $epsilon = 0$ in here; this contour does not hit any singularities when we do so. We find that $k_0 = i/omega$ and the first term in the limit as $rto 0^+$ is just the Cauchy principal value of an integrand, while the rightmost term is also interesting due to cancellation of the a term $r e^itheta$ from $domega$ with the term $1/(r e^itheta)$ from $k_0$:
$$
f mapsto operatornamePVint_-infty^infty domega ~frac iomega ~f(omega) - lim_rto 0^+ int_pi^0 dtheta~f(r~e^itheta).$$
So we can also take this limit as $rto 0^+$ by direct substitution and we find,
$$int_0^infty dt~left[f mapsto int_mathbb R domega~e^iomega t f(omega)right] =
f mapsto operatornamePVint_-infty^infty domega ~frac iomega ~f(omega) + pi~f(0).$$
If the principal value is understood as implied, one might instead write, $$ int_0^infty domega~ e^iomega t = fraciomega + pi~delta(omega).$$
What about the other side of the original integral?
As one might imagine, the integral for negative $t$ is largely the reverse of this. Indeed one has $$int_-infty^0 dt~e^(iomega + epsilon)t = frac1iomega + epsilon = frac -iomega - i epsilon.$$ The principal value analysis is absolutely identical except that the numerator contains $-i$ rather than $+i$, but the semicircle must dip into the negative half-plane rather than the positive half plane, yielding:
$$f mapsto operatornamePVint_-infty^infty domega ~frac -iomega ~f(omega) +
lim_rto 0^+ int_-pi^0 i~r~e^itheta dtheta~frac-ir e^itheta ~f(r~e^itheta).$$
Thus we have, perhaps surprisingly,
$$int_-infty^0 dt~e^iomega t = - fraciomega + pi~delta(omega).$$The sum of these then proves our original statement: $$
int_-infty^infty dt~e^iomega t = 2pi~delta(omega).
$$
Where were your intuitions failing you?
You perhaps thought that since $delta(omega)$ was even in $omega$, it meant that $e^iomega t$ was even in $t$. That is somewhat dubious reasoning in general: usually the (indefinite) integral of an even function is an odd function. However that reasoning is not 100% bad, it is just targeted at the wrong domain: it needs to be directed at a function of $t$. The real part of $e^iomega t$ is indeed even and hence we can expect that the real part of the integral $0toinfty$ is $pi~delta(omega)$ just as we eventually derived. But the imaginary part is odd in $t$ and hence, we would think, it's doomed to cancel on the domain $-infty to infty.$ Accordingly we could have expected that the integral $0 to infty$ contains some imaginary term which is the precise negative of the integral $-infty to 0$ such that they both cancel in the end.
answered Aug 8 at 21:55
CR Drost
20.2k11757
+1, this answer clearly took some time to compose and format.
– Alfred Centauri
Aug 8 at 23:33
@AlfredCentauri Most answers by Chris tend to have a nice and clean formatting, which add value to his clear explanations. Not sure why I'm telling you this. I guess this time is as good as any other to acknowledge and appreciate his great contributions to this site. Cheers!
– AccidentalFourierTransform
Aug 9 at 0:02
Well thanks for saying so! In retrospect I suppose that you're right, it's probably a dupe of something on maths.se.
– CR Drost
Aug 9 at 2:36
add a comment |Â
up vote
3
down vote
up vote
3
down vote
So what does it mean to be a delta function? It means that something is a mathematical object which satisfies for all other nice $f$,
$$int_-infty^infty domega~f(omega)~delta(omega) = f(0).$$
More formally, it turns out that the $delta(omega)$ can never truly be separated from the $int_-infty^infty domega$ sign: our writing of $delta(omega)$ is really a failure of notation where we would want to actually describe something more like, $$int_mathbb R dt~left[f mapsto int_mathbb R domega ~e^iomega t f(omega)right] = fmapsto 2pi~f(0),$$ where $mapsto$ denotes a function which accepts a smooth function as input and integration (really, addition and multiplication) of such functions is done "pointwise" (so that $f + g = xmapsto f(x) + g(x)$ for example).
Setting up the integral
Anyway the general approach to this problem is to shift the subproblems off the real axis by an infinitesimal amount, so that $$int_-infty^infty dt~e^iomega t =
lim_epsilonto 0^+left(int_-infty^0 dt~e^i(omega - i epsilon) t + int_0^infty dt~e^i(omega + iepsilon)t right),$$hence your question is actually one of the "stepping stones" we need to address in order to handle the above integrals. One finds of course,$$int_0^infty dt~ e^(iomega - epsilon) t = -frac 1iomega - epsilon = fraciomega + i epsilon, ~~epsilon > 0,$$ or substituting in the fuller expression using $mapsto$ we would find that this half-integral is, $$int_0^infty dt~left[f mapsto int_mathbb R domega~e^iomega t f(omega)right] = fmapsto lim_epsilonto 0^+int_mathbb R domega~fraciomega + i epsilon~f(omega). $$ It is when we try to evaluate this with complex analysis that we need the Cauchy principal value.
Contour deformation
Cauchy's integral theorem allows us to deform contours off the real line into the complex plane without changing the value. If we do this before we take the limit of $epsilon to 0$ we can make taking that limit very easy. So we deform the contour a little into the positive imaginary half of the plane around $omega = 0$, a half-circle with radius $r$, which we will later shrink to zero size. So on this circle $omega = r~e^itheta$ with $theta: pi to 0$ and thus $domega = i~r~e^itheta~dtheta.$ To save some space let $k_epsilon(omega) = frac iomega + iepsilon,$ the resulting expression for the right-hand-side is,
$$f mapsto lim_rto 0^+lim_epsilonto 0^+ left(int_-infty^-rdomega ~k_epsilon(omega)~f(omega) + int_r^inftydomega ~k_epsilon(omega)~f(omega) + int_pi^0 i~r~e^itheta dtheta~k_epsilon(re^itheta)~f(r~e^itheta) right).$$
We can now take the limit $epsilon to 0^+$ by simply substituting $epsilon = 0$ in here; this contour does not hit any singularities when we do so. We find that $k_0 = i/omega$ and the first term in the limit as $rto 0^+$ is just the Cauchy principal value of an integrand, while the rightmost term is also interesting due to cancellation of the a term $r e^itheta$ from $domega$ with the term $1/(r e^itheta)$ from $k_0$:
$$
f mapsto operatornamePVint_-infty^infty domega ~frac iomega ~f(omega) - lim_rto 0^+ int_pi^0 dtheta~f(r~e^itheta).$$
So we can also take this limit as $rto 0^+$ by direct substitution and we find,
$$int_0^infty dt~left[f mapsto int_mathbb R domega~e^iomega t f(omega)right] =
f mapsto operatornamePVint_-infty^infty domega ~frac iomega ~f(omega) + pi~f(0).$$
If the principal value is understood as implied, one might instead write, $$ int_0^infty domega~ e^iomega t = fraciomega + pi~delta(omega).$$
What about the other side of the original integral?
As one might imagine, the integral for negative $t$ is largely the reverse of this. Indeed one has $$int_-infty^0 dt~e^(iomega + epsilon)t = frac1iomega + epsilon = frac -iomega - i epsilon.$$ The principal value analysis is absolutely identical except that the numerator contains $-i$ rather than $+i$, but the semicircle must dip into the negative half-plane rather than the positive half plane, yielding:
$$f mapsto operatornamePVint_-infty^infty domega ~frac -iomega ~f(omega) +
lim_rto 0^+ int_-pi^0 i~r~e^itheta dtheta~frac-ir e^itheta ~f(r~e^itheta).$$
Thus we have, perhaps surprisingly,
$$int_-infty^0 dt~e^iomega t = - fraciomega + pi~delta(omega).$$The sum of these then proves our original statement: $$
int_-infty^infty dt~e^iomega t = 2pi~delta(omega).
$$
Where were your intuitions failing you?
You perhaps thought that since $delta(omega)$ was even in $omega$, it meant that $e^iomega t$ was even in $t$. That is somewhat dubious reasoning in general: usually the (indefinite) integral of an even function is an odd function. However that reasoning is not 100% bad, it is just targeted at the wrong domain: it needs to be directed at a function of $t$. The real part of $e^iomega t$ is indeed even and hence we can expect that the real part of the integral $0toinfty$ is $pi~delta(omega)$ just as we eventually derived. But the imaginary part is odd in $t$ and hence, we would think, it's doomed to cancel on the domain $-infty to infty.$ Accordingly we could have expected that the integral $0 to infty$ contains some imaginary term which is the precise negative of the integral $-infty to 0$ such that they both cancel in the end.
answered Aug 8 at 21:55
CR Drost
20.2k11757
So what does it mean to be a delta function? It means that something is a mathematical object which satisfies for all other nice $f$,
$$int_-infty^infty domega~f(omega)~delta(omega) = f(0).$$
More formally, it turns out that the $delta(omega)$ can never truly be separated from the $int_-infty^infty domega$ sign: our writing of $delta(omega)$ is really a failure of notation where we would want to actually describe something more like, $$int_mathbb R dt~left[f mapsto int_mathbb R domega ~e^iomega t f(omega)right] = fmapsto 2pi~f(0),$$ where $mapsto$ denotes a function which accepts a smooth function as input and integration (really, addition and multiplication) of such functions is done "pointwise" (so that $f + g = xmapsto f(x) + g(x)$ for example).
Setting up the integral
Anyway the general approach to this problem is to shift the subproblems off the real axis by an infinitesimal amount, so that $$int_-infty^infty dt~e^iomega t =
lim_epsilonto 0^+left(int_-infty^0 dt~e^i(omega - i epsilon) t + int_0^infty dt~e^i(omega + iepsilon)t right),$$hence your question is actually one of the "stepping stones" we need to address in order to handle the above integrals. One finds of course,$$int_0^infty dt~ e^(iomega - epsilon) t = -frac 1iomega - epsilon = fraciomega + i epsilon, ~~epsilon > 0,$$ or substituting in the fuller expression using $mapsto$ we would find that this half-integral is, $$int_0^infty dt~left[f mapsto int_mathbb R domega~e^iomega t f(omega)right] = fmapsto lim_epsilonto 0^+int_mathbb R domega~fraciomega + i epsilon~f(omega). $$ It is when we try to evaluate this with complex analysis that we need the Cauchy principal value.
Contour deformation
Cauchy's integral theorem allows us to deform contours off the real line into the complex plane without changing the value. If we do this before we take the limit of $epsilon to 0$ we can make taking that limit very easy. So we deform the contour a little into the positive imaginary half of the plane around $omega = 0$, a half-circle with radius $r$, which we will later shrink to zero size. So on this circle $omega = r~e^itheta$ with $theta: pi to 0$ and thus $domega = i~r~e^itheta~dtheta.$ To save some space let $k_epsilon(omega) = frac iomega + iepsilon,$ the resulting expression for the right-hand-side is,
$$f mapsto lim_rto 0^+lim_epsilonto 0^+ left(int_-infty^-rdomega ~k_epsilon(omega)~f(omega) + int_r^inftydomega ~k_epsilon(omega)~f(omega) + int_pi^0 i~r~e^itheta dtheta~k_epsilon(re^itheta)~f(r~e^itheta) right).$$
We can now take the limit $epsilon to 0^+$ by simply substituting $epsilon = 0$ in here; this contour does not hit any singularities when we do so. We find that $k_0 = i/omega$ and the first term in the limit as $rto 0^+$ is just the Cauchy principal value of an integrand, while the rightmost term is also interesting due to cancellation of the a term $r e^itheta$ from $domega$ with the term $1/(r e^itheta)$ from $k_0$:
$$
f mapsto operatornamePVint_-infty^infty domega ~frac iomega ~f(omega) - lim_rto 0^+ int_pi^0 dtheta~f(r~e^itheta).$$
So we can also take this limit as $rto 0^+$ by direct substitution and we find,
$$int_0^infty dt~left[f mapsto int_mathbb R domega~e^iomega t f(omega)right] =
f mapsto operatornamePVint_-infty^infty domega ~frac iomega ~f(omega) + pi~f(0).$$
If the principal value is understood as implied, one might instead write, $$ int_0^infty domega~ e^iomega t = fraciomega + pi~delta(omega).$$
What about the other side of the original integral?
As one might imagine, the integral for negative $t$ is largely the reverse of this. Indeed one has $$int_-infty^0 dt~e^(iomega + epsilon)t = frac1iomega + epsilon = frac -iomega - i epsilon.$$ The principal value analysis is absolutely identical except that the numerator contains $-i$ rather than $+i$, but the semicircle must dip into the negative half-plane rather than the positive half plane, yielding:
$$f mapsto operatornamePVint_-infty^infty domega ~frac -iomega ~f(omega) +
lim_rto 0^+ int_-pi^0 i~r~e^itheta dtheta~frac-ir e^itheta ~f(r~e^itheta).$$
Thus we have, perhaps surprisingly,
$$int_-infty^0 dt~e^iomega t = - fraciomega + pi~delta(omega).$$The sum of these then proves our original statement: $$
int_-infty^infty dt~e^iomega t = 2pi~delta(omega).
$$
Where were your intuitions failing you?
You perhaps thought that since $delta(omega)$ was even in $omega$, it meant that $e^iomega t$ was even in $t$. That is somewhat dubious reasoning in general: usually the (indefinite) integral of an even function is an odd function. However that reasoning is not 100% bad, it is just targeted at the wrong domain: it needs to be directed at a function of $t$. The real part of $e^iomega t$ is indeed even and hence we can expect that the real part of the integral $0toinfty$ is $pi~delta(omega)$ just as we eventually derived. But the imaginary part is odd in $t$ and hence, we would think, it's doomed to cancel on the domain $-infty to infty.$ Accordingly we could have expected that the integral $0 to infty$ contains some imaginary term which is the precise negative of the integral $-infty to 0$ such that they both cancel in the end.
answered Aug 8 at 21:55
CR Drost
20.2k11757
answered Aug 8 at 21:55
CR Drost
20.2k11757
answered Aug 8 at 21:55
CR Drost
20.2k11757
answered Aug 8 at 21:55
CR Drost
20.2k11757
20.2k11757
+1, this answer clearly took some time to compose and format.
– Alfred Centauri
Aug 8 at 23:33
@AlfredCentauri Most answers by Chris tend to have a nice and clean formatting, which add value to his clear explanations. Not sure why I'm telling you this. I guess this time is as good as any other to acknowledge and appreciate his great contributions to this site. Cheers!
– AccidentalFourierTransform
Aug 9 at 0:02
Well thanks for saying so! In retrospect I suppose that you're right, it's probably a dupe of something on maths.se.
– CR Drost
Aug 9 at 2:36
add a comment |Â
+1, this answer clearly took some time to compose and format.
– Alfred Centauri
Aug 8 at 23:33
@AlfredCentauri Most answers by Chris tend to have a nice and clean formatting, which add value to his clear explanations. Not sure why I'm telling you this. I guess this time is as good as any other to acknowledge and appreciate his great contributions to this site. Cheers!
– AccidentalFourierTransform
Aug 9 at 0:02
Well thanks for saying so! In retrospect I suppose that you're right, it's probably a dupe of something on maths.se.
– CR Drost
Aug 9 at 2:36
+1, this answer clearly took some time to compose and format.
– Alfred Centauri
Aug 8 at 23:33
+1, this answer clearly took some time to compose and format.
– Alfred Centauri
Aug 8 at 23:33
@AlfredCentauri Most answers by Chris tend to have a nice and clean formatting, which add value to his clear explanations. Not sure why I'm telling you this. I guess this time is as good as any other to acknowledge and appreciate his great contributions to this site. Cheers!
– AccidentalFourierTransform
Aug 9 at 0:02
@AlfredCentauri Most answers by Chris tend to have a nice and clean formatting, which add value to his clear explanations. Not sure why I'm telling you this. I guess this time is as good as any other to acknowledge and appreciate his great contributions to this site. Cheers!
– AccidentalFourierTransform
Aug 9 at 0:02
Well thanks for saying so! In retrospect I suppose that you're right, it's probably a dupe of something on maths.se.
– CR Drost
Aug 9 at 2:36
Well thanks for saying so! In retrospect I suppose that you're right, it's probably a dupe of something on maths.se.
– CR Drost
Aug 9 at 2:36
add a comment |Â
up vote
1
down vote
why the half of the integral is not half of delta function?
Because the function is not even.
To illustrate why, consider:
$$
int_-infty^infty dt e^iomega t
=
int_-infty^0 dt e^iomega t+int_0^infty dt e^iomega t
$$
$$
=
int_0^infty dse^-iomega s + int_0^infty dt e^iomega t;,
$$
where I made the change of variable $s=-t$ in the first integral of the lower line.
But since $s$ is just a dummy variable, I can also write:
$$
=int_0^infty dte^-iomega t + int_0^infty dt e^iomega t
$$
$$
neq 2int_0^infty dt e^iomega t
$$
answered Aug 8 at 18:29
hft
3,8931821
Thanks, my point is not to discuss the odd and even, my question is how to derive the second integral in the original question.
– kinder chan
Aug 8 at 18:31
what do you mean "how to derive the second integral?" Starting from what? Ending with what? You seem to have edited your question to make it even more vague and hard to understand. Could you edit to provide some context?
– hft
Aug 8 at 19:03
My bad for the confusion. I mean $int_0^infty!dt ~e^iomega t$, I also re-edit the question.
– kinder chan
Aug 8 at 19:12
I advise you to edit. What you wrote is right, but this explains why the second integral is not a delta whose argument $omega $ is defined on the whole real axis. The integral, infact, is a delta defined on the positive frequencies, as already mentioned in a comment above. Explain that in definition of the delta function in its integrale form you have to take into account of the definition domain of the argument.
– MRT
Aug 8 at 20:26
add a comment |Â
up vote
1
down vote
why the half of the integral is not half of delta function?
Because the function is not even.
To illustrate why, consider:
$$
int_-infty^infty dt e^iomega t
=
int_-infty^0 dt e^iomega t+int_0^infty dt e^iomega t
$$
$$
=
int_0^infty dse^-iomega s + int_0^infty dt e^iomega t;,
$$
where I made the change of variable $s=-t$ in the first integral of the lower line.
But since $s$ is just a dummy variable, I can also write:
$$
=int_0^infty dte^-iomega t + int_0^infty dt e^iomega t
$$
$$
neq 2int_0^infty dt e^iomega t
$$
answered Aug 8 at 18:29
hft
3,8931821
Thanks, my point is not to discuss the odd and even, my question is how to derive the second integral in the original question.
– kinder chan
Aug 8 at 18:31
what do you mean "how to derive the second integral?" Starting from what? Ending with what? You seem to have edited your question to make it even more vague and hard to understand. Could you edit to provide some context?
– hft
Aug 8 at 19:03
My bad for the confusion. I mean $int_0^infty!dt ~e^iomega t$, I also re-edit the question.
– kinder chan
Aug 8 at 19:12
I advise you to edit. What you wrote is right, but this explains why the second integral is not a delta whose argument $omega $ is defined on the whole real axis. The integral, infact, is a delta defined on the positive frequencies, as already mentioned in a comment above. Explain that in definition of the delta function in its integrale form you have to take into account of the definition domain of the argument.
– MRT
Aug 8 at 20:26
add a comment |Â
up vote
1
down vote
up vote
1
down vote
why the half of the integral is not half of delta function?
Because the function is not even.
To illustrate why, consider:
$$
int_-infty^infty dt e^iomega t
=
int_-infty^0 dt e^iomega t+int_0^infty dt e^iomega t
$$
$$
=
int_0^infty dse^-iomega s + int_0^infty dt e^iomega t;,
$$
where I made the change of variable $s=-t$ in the first integral of the lower line.
But since $s$ is just a dummy variable, I can also write:
$$
=int_0^infty dte^-iomega t + int_0^infty dt e^iomega t
$$
$$
neq 2int_0^infty dt e^iomega t
$$
answered Aug 8 at 18:29
hft
3,8931821
why the half of the integral is not half of delta function?
Because the function is not even.
To illustrate why, consider:
$$
int_-infty^infty dt e^iomega t
=
int_-infty^0 dt e^iomega t+int_0^infty dt e^iomega t
$$
$$
=
int_0^infty dse^-iomega s + int_0^infty dt e^iomega t;,
$$
where I made the change of variable $s=-t$ in the first integral of the lower line.
But since $s$ is just a dummy variable, I can also write:
$$
=int_0^infty dte^-iomega t + int_0^infty dt e^iomega t
$$
$$
neq 2int_0^infty dt e^iomega t
$$
answered Aug 8 at 18:29
hft
3,8931821
answered Aug 8 at 18:29
hft
3,8931821
answered Aug 8 at 18:29
hft
3,8931821
answered Aug 8 at 18:29
hft
3,8931821
3,8931821
Thanks, my point is not to discuss the odd and even, my question is how to derive the second integral in the original question.
– kinder chan
Aug 8 at 18:31
what do you mean "how to derive the second integral?" Starting from what? Ending with what? You seem to have edited your question to make it even more vague and hard to understand. Could you edit to provide some context?
– hft
Aug 8 at 19:03
My bad for the confusion. I mean $int_0^infty!dt ~e^iomega t$, I also re-edit the question.
– kinder chan
Aug 8 at 19:12
I advise you to edit. What you wrote is right, but this explains why the second integral is not a delta whose argument $omega $ is defined on the whole real axis. The integral, infact, is a delta defined on the positive frequencies, as already mentioned in a comment above. Explain that in definition of the delta function in its integrale form you have to take into account of the definition domain of the argument.
– MRT
Aug 8 at 20:26
add a comment |Â
Thanks, my point is not to discuss the odd and even, my question is how to derive the second integral in the original question.
– kinder chan
Aug 8 at 18:31
what do you mean "how to derive the second integral?" Starting from what? Ending with what? You seem to have edited your question to make it even more vague and hard to understand. Could you edit to provide some context?
– hft
Aug 8 at 19:03
My bad for the confusion. I mean $int_0^infty!dt ~e^iomega t$, I also re-edit the question.
– kinder chan
Aug 8 at 19:12
I advise you to edit. What you wrote is right, but this explains why the second integral is not a delta whose argument $omega $ is defined on the whole real axis. The integral, infact, is a delta defined on the positive frequencies, as already mentioned in a comment above. Explain that in definition of the delta function in its integrale form you have to take into account of the definition domain of the argument.
– MRT
Aug 8 at 20:26
Thanks, my point is not to discuss the odd and even, my question is how to derive the second integral in the original question.
– kinder chan
Aug 8 at 18:31
Thanks, my point is not to discuss the odd and even, my question is how to derive the second integral in the original question.
– kinder chan
Aug 8 at 18:31
what do you mean "how to derive the second integral?" Starting from what? Ending with what? You seem to have edited your question to make it even more vague and hard to understand. Could you edit to provide some context?
– hft
Aug 8 at 19:03
what do you mean "how to derive the second integral?" Starting from what? Ending with what? You seem to have edited your question to make it even more vague and hard to understand. Could you edit to provide some context?
– hft
Aug 8 at 19:03
My bad for the confusion. I mean $int_0^infty!dt ~e^iomega t$, I also re-edit the question.
– kinder chan
Aug 8 at 19:12
My bad for the confusion. I mean $int_0^infty!dt ~e^iomega t$, I also re-edit the question.
– kinder chan
Aug 8 at 19:12
I advise you to edit. What you wrote is right, but this explains why the second integral is not a delta whose argument $omega $ is defined on the whole real axis. The integral, infact, is a delta defined on the positive frequencies, as already mentioned in a comment above. Explain that in definition of the delta function in its integrale form you have to take into account of the definition domain of the argument.
– MRT
Aug 8 at 20:26
I advise you to edit. What you wrote is right, but this explains why the second integral is not a delta whose argument $omega $ is defined on the whole real axis. The integral, infact, is a delta defined on the positive frequencies, as already mentioned in a comment above. Explain that in definition of the delta function in its integrale form you have to take into account of the definition domain of the argument.
– MRT
Aug 8 at 20:26
add a comment |Â
1
Why should the integral on half-domain be the half of the first result? $e^iwt$ is not even on the domain. Why do you expect the second result?
– apt45
Aug 8 at 18:12
1
The delta function is even
– kinder chan
Aug 8 at 18:27
3
Would Mathematics be a better home for this question?
– Qmechanic♦
Aug 8 at 18:43
2
Your second integral is (by definition) the positive-frequency delta function $2pidelta_+(w)$. It has a well-known representation in terms of the distribution $frac1w+i0^+$. In any case, I voted to migrate to math.SE.
– AccidentalFourierTransform
Aug 8 at 18:43
The function under the integral oscillates with any $omega ne 0$, but does not oscillate for $omega=0$. The oscillations cancel themselves from $-infty$ to $+infty$, but not form $0$ to $+infty$. BTW this is not a definition of the Delta Function (or not a good one at least), but just a Fourier transform of it. You could say though that, if $int_-infty^+inftyf(x)delta(0)dx=f(0)$, then $int_0^+inftyf(x)delta(0)dx=dfrac12f(0)$
– safesphere
Aug 8 at 22:41