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Prove $lim_ttoinfty sin(tx) textP.V.frac1x = pi delta$ in the distributional sense
Clash Royale CLAN TAG#URR8PPP
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Prove $lim_ttoinfty sin(tx) textP.V.frac1x = pi delta$ in the distributional sense.
This was the work I had put into the question:
For $phi in mathcalD(mathbbR)$,
beginalign*
langle sin(tx) textP.V.frac1x, phi rangle &= langle textP.V.frac1x, sin(tx) phi rangle\
&= lim_epsilon to 0^+ int_ geq epsilon fracsin(tx)x phi(x) dx.
endalign*
We know that for any interval $[-R,R]$, the DCT applies and
beginalign*
lim_tto infty int_-R^R fracsin(y)yphi(y/t) dy &= int_-R^R lim_ttoinfty fracsin(y)y phi(y/t) dy \
&= phi(0) int_-R^R fracsin(y)y dy.
endalign*
Thus, we need only consider the tails, i.e. we need to show that $lim_tto infty int_R^infty fracsin(y)y[phi(y/t) - phi(0)] dy$ and $lim_ttoinfty int_-infty^-R fracsin(y)y[phi(y/t)-phi(0)]dy$ go to $0$ as $R$ increases. So
beginalign*
lim_tto infty int_R^infty fracsin(y)y [phi(y/t)-phi(0)] dy = lim_tto infty int_R/t^infty fracsin(tx)x [phi(x) - phi(0)] dx.
endalign*
The problem with this though is the fact that with u-substitution the limits of integration become impossible to get a grip on, hence we cannot utilize the first result. You can approach this problem in various ways using these same techniques and ideas, but in the end you land back with the same dilemma of being unable to properly control the limits of integration. Any help would be appreciated.
functional-analysis distribution-theory
asked Jul 29 at 20:08
mathishard.butweloveit
1069
add a comment |Â
up vote
1
down vote
favorite
Prove $lim_ttoinfty sin(tx) textP.V.frac1x = pi delta$ in the distributional sense.
This was the work I had put into the question:
For $phi in mathcalD(mathbbR)$,
beginalign*
langle sin(tx) textP.V.frac1x, phi rangle &= langle textP.V.frac1x, sin(tx) phi rangle\
&= lim_epsilon to 0^+ int_ geq epsilon fracsin(tx)x phi(x) dx.
endalign*
We know that for any interval $[-R,R]$, the DCT applies and
beginalign*
lim_tto infty int_-R^R fracsin(y)yphi(y/t) dy &= int_-R^R lim_ttoinfty fracsin(y)y phi(y/t) dy \
&= phi(0) int_-R^R fracsin(y)y dy.
endalign*
Thus, we need only consider the tails, i.e. we need to show that $lim_tto infty int_R^infty fracsin(y)y[phi(y/t) - phi(0)] dy$ and $lim_ttoinfty int_-infty^-R fracsin(y)y[phi(y/t)-phi(0)]dy$ go to $0$ as $R$ increases. So
beginalign*
lim_tto infty int_R^infty fracsin(y)y [phi(y/t)-phi(0)] dy = lim_tto infty int_R/t^infty fracsin(tx)x [phi(x) - phi(0)] dx.
endalign*
The problem with this though is the fact that with u-substitution the limits of integration become impossible to get a grip on, hence we cannot utilize the first result. You can approach this problem in various ways using these same techniques and ideas, but in the end you land back with the same dilemma of being unable to properly control the limits of integration. Any help would be appreciated.
functional-analysis distribution-theory
asked Jul 29 at 20:08
mathishard.butweloveit
1069
I think at some point my professor found the solution in a book, but I don't have the book (thought I did) and I don't remember what it was. So if you have a book reference + solution, that would be great.
– mathishard.butweloveit
Jul 29 at 21:11
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Prove $lim_ttoinfty sin(tx) textP.V.frac1x = pi delta$ in the distributional sense.
This was the work I had put into the question:
For $phi in mathcalD(mathbbR)$,
beginalign*
langle sin(tx) textP.V.frac1x, phi rangle &= langle textP.V.frac1x, sin(tx) phi rangle\
&= lim_epsilon to 0^+ int_ geq epsilon fracsin(tx)x phi(x) dx.
endalign*
We know that for any interval $[-R,R]$, the DCT applies and
beginalign*
lim_tto infty int_-R^R fracsin(y)yphi(y/t) dy &= int_-R^R lim_ttoinfty fracsin(y)y phi(y/t) dy \
&= phi(0) int_-R^R fracsin(y)y dy.
endalign*
Thus, we need only consider the tails, i.e. we need to show that $lim_tto infty int_R^infty fracsin(y)y[phi(y/t) - phi(0)] dy$ and $lim_ttoinfty int_-infty^-R fracsin(y)y[phi(y/t)-phi(0)]dy$ go to $0$ as $R$ increases. So
beginalign*
lim_tto infty int_R^infty fracsin(y)y [phi(y/t)-phi(0)] dy = lim_tto infty int_R/t^infty fracsin(tx)x [phi(x) - phi(0)] dx.
endalign*
The problem with this though is the fact that with u-substitution the limits of integration become impossible to get a grip on, hence we cannot utilize the first result. You can approach this problem in various ways using these same techniques and ideas, but in the end you land back with the same dilemma of being unable to properly control the limits of integration. Any help would be appreciated.
functional-analysis distribution-theory
asked Jul 29 at 20:08
mathishard.butweloveit
1069
Prove $lim_ttoinfty sin(tx) textP.V.frac1x = pi delta$ in the distributional sense.
This was the work I had put into the question:
For $phi in mathcalD(mathbbR)$,
beginalign*
langle sin(tx) textP.V.frac1x, phi rangle &= langle textP.V.frac1x, sin(tx) phi rangle\
&= lim_epsilon to 0^+ int_ geq epsilon fracsin(tx)x phi(x) dx.
endalign*
We know that for any interval $[-R,R]$, the DCT applies and
beginalign*
lim_tto infty int_-R^R fracsin(y)yphi(y/t) dy &= int_-R^R lim_ttoinfty fracsin(y)y phi(y/t) dy \
&= phi(0) int_-R^R fracsin(y)y dy.
endalign*
Thus, we need only consider the tails, i.e. we need to show that $lim_tto infty int_R^infty fracsin(y)y[phi(y/t) - phi(0)] dy$ and $lim_ttoinfty int_-infty^-R fracsin(y)y[phi(y/t)-phi(0)]dy$ go to $0$ as $R$ increases. So
beginalign*
lim_tto infty int_R^infty fracsin(y)y [phi(y/t)-phi(0)] dy = lim_tto infty int_R/t^infty fracsin(tx)x [phi(x) - phi(0)] dx.
endalign*
The problem with this though is the fact that with u-substitution the limits of integration become impossible to get a grip on, hence we cannot utilize the first result. You can approach this problem in various ways using these same techniques and ideas, but in the end you land back with the same dilemma of being unable to properly control the limits of integration. Any help would be appreciated.
functional-analysis distribution-theory
asked Jul 29 at 20:08
mathishard.butweloveit
1069
asked Jul 29 at 20:08
mathishard.butweloveit
1069
asked Jul 29 at 20:08
mathishard.butweloveit
1069
asked Jul 29 at 20:08
mathishard.butweloveit
1069
1069
I think at some point my professor found the solution in a book, but I don't have the book (thought I did) and I don't remember what it was. So if you have a book reference + solution, that would be great.
– mathishard.butweloveit
Jul 29 at 21:11
add a comment |Â
I think at some point my professor found the solution in a book, but I don't have the book (thought I did) and I don't remember what it was. So if you have a book reference + solution, that would be great.
– mathishard.butweloveit
Jul 29 at 21:11
I think at some point my professor found the solution in a book, but I don't have the book (thought I did) and I don't remember what it was. So if you have a book reference + solution, that would be great.
– mathishard.butweloveit
Jul 29 at 21:11
I think at some point my professor found the solution in a book, but I don't have the book (thought I did) and I don't remember what it was. So if you have a book reference + solution, that would be great.
– mathishard.butweloveit
Jul 29 at 21:11
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Let $phi_e(x)=frac12(phi(x)+phi(-x))$ denote the even part of $phi(x)$. Then, recognizing that $phi(0)=phi_e(0)$ we can write
$$beginalign
textPVint_-infty^infty fracsin(tx)x,phi(x),dx&=piphi(0)+2int_0^infty fracsin(tx)x(phi_e(x)-phi_e(0)),dx\\
tag1
endalign$$
Inasmuch as $phi(x)$ is a suitable test function, then for $xsim0$, $phi_e(x)-phi_e(0)=O(x^2)$ and $phi_e'(x)=O(x)$. Moreover, $phi(x)$ has compact support.
Hence, upon integrating by parts the integral on the right-hand side of $(1)$ with $u=displaystyle fracphi_e(x)-phi_e(0)x$ and $displaystyle v=-fraccos(tx)t$, we find that
$$beginalign
int_0^infty left(fracphi_e(x)-phi_e(0)xright),sin(tx),dx&=frac1t int_0^infty left(fracphi_e(x)-phi_e(0)x^2right),cos(tx),dx\\
&-frac1t int_0^infty left(fracphi'_e(x)xright),cos(tx),dxtag2
endalign$$
Both integrals on the right-hand side of $(2)$ converge, owing to the aforementioned small argument behavior and compact support of $phi_e(x)$. Therefore, letting $tto infty$ reveals
$$lim_ttoinftyint_0^infty fracsin(tx)x(phi_e(x)-phi_e(0)),dx=0tag3$$
Finally, substituting $(3)$ into $(1)$ we find that
$$lim_ttoinftytextPVint_-infty^infty fracsin(tx)x,phi(x),dx=piphi(0)$$
as was to be shown!
answered Jul 29 at 21:33
Mark Viola
126k1172167
That's a crazy neat argument, very clever trick identifying the even portion. Is there anything in particular that made you think of this particular trick? Also out of curiosity, do you think there might be another way to approach this problem without defining $phi_e$?
– mathishard.butweloveit
Jul 29 at 21:54
I'd speculate that there is a "slicker" way to go that streamlines the analysis. I was motivated by the fact that the kernel $fracsin(tx)x$ is even in $x$ and therefore the PV renders the integral over the odd part of $phi$ to be $0$. That left only the even part, which has the small argument properties I exploited upon integrating by parts.
– Mark Viola
Jul 29 at 21:59
Interesting, yes I am not as familiar with thinking about things like this. Thank you.
– mathishard.butweloveit
Jul 29 at 22:01
You're quite welcome. It was my pleasure.
– Mark Viola
Jul 29 at 22:03
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $phi_e(x)=frac12(phi(x)+phi(-x))$ denote the even part of $phi(x)$. Then, recognizing that $phi(0)=phi_e(0)$ we can write
$$beginalign
textPVint_-infty^infty fracsin(tx)x,phi(x),dx&=piphi(0)+2int_0^infty fracsin(tx)x(phi_e(x)-phi_e(0)),dx\\
tag1
endalign$$
Inasmuch as $phi(x)$ is a suitable test function, then for $xsim0$, $phi_e(x)-phi_e(0)=O(x^2)$ and $phi_e'(x)=O(x)$. Moreover, $phi(x)$ has compact support.
Hence, upon integrating by parts the integral on the right-hand side of $(1)$ with $u=displaystyle fracphi_e(x)-phi_e(0)x$ and $displaystyle v=-fraccos(tx)t$, we find that
$$beginalign
int_0^infty left(fracphi_e(x)-phi_e(0)xright),sin(tx),dx&=frac1t int_0^infty left(fracphi_e(x)-phi_e(0)x^2right),cos(tx),dx\\
&-frac1t int_0^infty left(fracphi'_e(x)xright),cos(tx),dxtag2
endalign$$
Both integrals on the right-hand side of $(2)$ converge, owing to the aforementioned small argument behavior and compact support of $phi_e(x)$. Therefore, letting $tto infty$ reveals
$$lim_ttoinftyint_0^infty fracsin(tx)x(phi_e(x)-phi_e(0)),dx=0tag3$$
Finally, substituting $(3)$ into $(1)$ we find that
$$lim_ttoinftytextPVint_-infty^infty fracsin(tx)x,phi(x),dx=piphi(0)$$
as was to be shown!
answered Jul 29 at 21:33
Mark Viola
126k1172167
That's a crazy neat argument, very clever trick identifying the even portion. Is there anything in particular that made you think of this particular trick? Also out of curiosity, do you think there might be another way to approach this problem without defining $phi_e$?
– mathishard.butweloveit
Jul 29 at 21:54
I'd speculate that there is a "slicker" way to go that streamlines the analysis. I was motivated by the fact that the kernel $fracsin(tx)x$ is even in $x$ and therefore the PV renders the integral over the odd part of $phi$ to be $0$. That left only the even part, which has the small argument properties I exploited upon integrating by parts.
– Mark Viola
Jul 29 at 21:59
Interesting, yes I am not as familiar with thinking about things like this. Thank you.
– mathishard.butweloveit
Jul 29 at 22:01
You're quite welcome. It was my pleasure.
– Mark Viola
Jul 29 at 22:03
add a comment |Â
up vote
2
down vote
accepted
Let $phi_e(x)=frac12(phi(x)+phi(-x))$ denote the even part of $phi(x)$. Then, recognizing that $phi(0)=phi_e(0)$ we can write
$$beginalign
textPVint_-infty^infty fracsin(tx)x,phi(x),dx&=piphi(0)+2int_0^infty fracsin(tx)x(phi_e(x)-phi_e(0)),dx\\
tag1
endalign$$
Inasmuch as $phi(x)$ is a suitable test function, then for $xsim0$, $phi_e(x)-phi_e(0)=O(x^2)$ and $phi_e'(x)=O(x)$. Moreover, $phi(x)$ has compact support.
Hence, upon integrating by parts the integral on the right-hand side of $(1)$ with $u=displaystyle fracphi_e(x)-phi_e(0)x$ and $displaystyle v=-fraccos(tx)t$, we find that
$$beginalign
int_0^infty left(fracphi_e(x)-phi_e(0)xright),sin(tx),dx&=frac1t int_0^infty left(fracphi_e(x)-phi_e(0)x^2right),cos(tx),dx\\
&-frac1t int_0^infty left(fracphi'_e(x)xright),cos(tx),dxtag2
endalign$$
Both integrals on the right-hand side of $(2)$ converge, owing to the aforementioned small argument behavior and compact support of $phi_e(x)$. Therefore, letting $tto infty$ reveals
$$lim_ttoinftyint_0^infty fracsin(tx)x(phi_e(x)-phi_e(0)),dx=0tag3$$
Finally, substituting $(3)$ into $(1)$ we find that
$$lim_ttoinftytextPVint_-infty^infty fracsin(tx)x,phi(x),dx=piphi(0)$$
as was to be shown!
answered Jul 29 at 21:33
Mark Viola
126k1172167
That's a crazy neat argument, very clever trick identifying the even portion. Is there anything in particular that made you think of this particular trick? Also out of curiosity, do you think there might be another way to approach this problem without defining $phi_e$?
– mathishard.butweloveit
Jul 29 at 21:54
I'd speculate that there is a "slicker" way to go that streamlines the analysis. I was motivated by the fact that the kernel $fracsin(tx)x$ is even in $x$ and therefore the PV renders the integral over the odd part of $phi$ to be $0$. That left only the even part, which has the small argument properties I exploited upon integrating by parts.
– Mark Viola
Jul 29 at 21:59
Interesting, yes I am not as familiar with thinking about things like this. Thank you.
– mathishard.butweloveit
Jul 29 at 22:01
You're quite welcome. It was my pleasure.
– Mark Viola
Jul 29 at 22:03
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $phi_e(x)=frac12(phi(x)+phi(-x))$ denote the even part of $phi(x)$. Then, recognizing that $phi(0)=phi_e(0)$ we can write
$$beginalign
textPVint_-infty^infty fracsin(tx)x,phi(x),dx&=piphi(0)+2int_0^infty fracsin(tx)x(phi_e(x)-phi_e(0)),dx\\
tag1
endalign$$
Inasmuch as $phi(x)$ is a suitable test function, then for $xsim0$, $phi_e(x)-phi_e(0)=O(x^2)$ and $phi_e'(x)=O(x)$. Moreover, $phi(x)$ has compact support.
Hence, upon integrating by parts the integral on the right-hand side of $(1)$ with $u=displaystyle fracphi_e(x)-phi_e(0)x$ and $displaystyle v=-fraccos(tx)t$, we find that
$$beginalign
int_0^infty left(fracphi_e(x)-phi_e(0)xright),sin(tx),dx&=frac1t int_0^infty left(fracphi_e(x)-phi_e(0)x^2right),cos(tx),dx\\
&-frac1t int_0^infty left(fracphi'_e(x)xright),cos(tx),dxtag2
endalign$$
Both integrals on the right-hand side of $(2)$ converge, owing to the aforementioned small argument behavior and compact support of $phi_e(x)$. Therefore, letting $tto infty$ reveals
$$lim_ttoinftyint_0^infty fracsin(tx)x(phi_e(x)-phi_e(0)),dx=0tag3$$
Finally, substituting $(3)$ into $(1)$ we find that
$$lim_ttoinftytextPVint_-infty^infty fracsin(tx)x,phi(x),dx=piphi(0)$$
as was to be shown!
answered Jul 29 at 21:33
Mark Viola
126k1172167
Let $phi_e(x)=frac12(phi(x)+phi(-x))$ denote the even part of $phi(x)$. Then, recognizing that $phi(0)=phi_e(0)$ we can write
$$beginalign
textPVint_-infty^infty fracsin(tx)x,phi(x),dx&=piphi(0)+2int_0^infty fracsin(tx)x(phi_e(x)-phi_e(0)),dx\\
tag1
endalign$$
Inasmuch as $phi(x)$ is a suitable test function, then for $xsim0$, $phi_e(x)-phi_e(0)=O(x^2)$ and $phi_e'(x)=O(x)$. Moreover, $phi(x)$ has compact support.
Hence, upon integrating by parts the integral on the right-hand side of $(1)$ with $u=displaystyle fracphi_e(x)-phi_e(0)x$ and $displaystyle v=-fraccos(tx)t$, we find that
$$beginalign
int_0^infty left(fracphi_e(x)-phi_e(0)xright),sin(tx),dx&=frac1t int_0^infty left(fracphi_e(x)-phi_e(0)x^2right),cos(tx),dx\\
&-frac1t int_0^infty left(fracphi'_e(x)xright),cos(tx),dxtag2
endalign$$
Both integrals on the right-hand side of $(2)$ converge, owing to the aforementioned small argument behavior and compact support of $phi_e(x)$. Therefore, letting $tto infty$ reveals
$$lim_ttoinftyint_0^infty fracsin(tx)x(phi_e(x)-phi_e(0)),dx=0tag3$$
Finally, substituting $(3)$ into $(1)$ we find that
$$lim_ttoinftytextPVint_-infty^infty fracsin(tx)x,phi(x),dx=piphi(0)$$
as was to be shown!
answered Jul 29 at 21:33
Mark Viola
126k1172167
answered Jul 29 at 21:33
Mark Viola
126k1172167
answered Jul 29 at 21:33
Mark Viola
126k1172167
answered Jul 29 at 21:33
Mark Viola
126k1172167
126k1172167
That's a crazy neat argument, very clever trick identifying the even portion. Is there anything in particular that made you think of this particular trick? Also out of curiosity, do you think there might be another way to approach this problem without defining $phi_e$?
– mathishard.butweloveit
Jul 29 at 21:54
I'd speculate that there is a "slicker" way to go that streamlines the analysis. I was motivated by the fact that the kernel $fracsin(tx)x$ is even in $x$ and therefore the PV renders the integral over the odd part of $phi$ to be $0$. That left only the even part, which has the small argument properties I exploited upon integrating by parts.
– Mark Viola
Jul 29 at 21:59
Interesting, yes I am not as familiar with thinking about things like this. Thank you.
– mathishard.butweloveit
Jul 29 at 22:01
You're quite welcome. It was my pleasure.
– Mark Viola
Jul 29 at 22:03
add a comment |Â
That's a crazy neat argument, very clever trick identifying the even portion. Is there anything in particular that made you think of this particular trick? Also out of curiosity, do you think there might be another way to approach this problem without defining $phi_e$?
– mathishard.butweloveit
Jul 29 at 21:54
I'd speculate that there is a "slicker" way to go that streamlines the analysis. I was motivated by the fact that the kernel $fracsin(tx)x$ is even in $x$ and therefore the PV renders the integral over the odd part of $phi$ to be $0$. That left only the even part, which has the small argument properties I exploited upon integrating by parts.
– Mark Viola
Jul 29 at 21:59
Interesting, yes I am not as familiar with thinking about things like this. Thank you.
– mathishard.butweloveit
Jul 29 at 22:01
You're quite welcome. It was my pleasure.
– Mark Viola
Jul 29 at 22:03
That's a crazy neat argument, very clever trick identifying the even portion. Is there anything in particular that made you think of this particular trick? Also out of curiosity, do you think there might be another way to approach this problem without defining $phi_e$?
– mathishard.butweloveit
Jul 29 at 21:54
That's a crazy neat argument, very clever trick identifying the even portion. Is there anything in particular that made you think of this particular trick? Also out of curiosity, do you think there might be another way to approach this problem without defining $phi_e$?
– mathishard.butweloveit
Jul 29 at 21:54
I'd speculate that there is a "slicker" way to go that streamlines the analysis. I was motivated by the fact that the kernel $fracsin(tx)x$ is even in $x$ and therefore the PV renders the integral over the odd part of $phi$ to be $0$. That left only the even part, which has the small argument properties I exploited upon integrating by parts.
– Mark Viola
Jul 29 at 21:59
I'd speculate that there is a "slicker" way to go that streamlines the analysis. I was motivated by the fact that the kernel $fracsin(tx)x$ is even in $x$ and therefore the PV renders the integral over the odd part of $phi$ to be $0$. That left only the even part, which has the small argument properties I exploited upon integrating by parts.
– Mark Viola
Jul 29 at 21:59
Interesting, yes I am not as familiar with thinking about things like this. Thank you.
– mathishard.butweloveit
Jul 29 at 22:01
Interesting, yes I am not as familiar with thinking about things like this. Thank you.
– mathishard.butweloveit
Jul 29 at 22:01
You're quite welcome. It was my pleasure.
– Mark Viola
Jul 29 at 22:03
You're quite welcome. It was my pleasure.
– Mark Viola
Jul 29 at 22:03
add a comment |Â
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I think at some point my professor found the solution in a book, but I don't have the book (thought I did) and I don't remember what it was. So if you have a book reference + solution, that would be great.
– mathishard.butweloveit
Jul 29 at 21:11