Prove that $ int _0^inftyfracsin xxdx=fracpi2$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)$sum_n=-infty^inftyfrac1(n+alpha)^2=fracpi^2(sin pi alpha)^2$?Show $lim_Ntoinftyint_0^pileft(frac1sinfracx2-frac2xright)sinleft((N+frac12)xright)dx=0$Show that Fourier transformation is differentiable if $int|xf(x)|,dlambda<infty$Riemann Lebesgue Lemma Clarification$lim_lambda to infty int^b_0 f(t) fracsin(lambda t)t $The improper integrals of $fracsin tt$ and $left|fracsin ttright|$Calculate limits of integralsAlternate proof of Dirichlet integral $fracsin(x)x$.Riemann-Lebesgue Lemma And Integral Having $sin(cdot)$Abel means of function f when having a jump discontinuity
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Prove that $ int _0^inftyfracsin xxdx=fracpi2$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)$sum_n=-infty^inftyfrac1(n+alpha)^2=fracpi^2(sin pi alpha)^2$?Show $lim_Ntoinftyint_0^pileft(frac1sinfracx2-frac2xright)sinleft((N+frac12)xright)dx=0$Show that Fourier transformation is differentiable if $int|xf(x)|,dlambda<infty$Riemann Lebesgue Lemma Clarification$lim_lambda to infty int^b_0 f(t) fracsin(lambda t)t $The improper integrals of $fracsin tt$ and $left|fracsin ttright|$Calculate limits of integralsAlternate proof of Dirichlet integral $fracsin(x)x$.Riemann-Lebesgue Lemma And Integral Having $sin(cdot)$Abel means of function f when having a jump discontinuity
$begingroup$
Prove
$$int_0^inftyfracsin xxdx=fracpi2$$
Start with the fact that the integral of $D_N(theta)$ equals $2pi$, and note that the difference $$frac1sin(theta/2)-frac2theta$$
is continuous on $[-pi,pi]$. Apply the Riemann-Lebesgue lemma.
$$D_N(x)=sum_-N^Ne^inx=fracsin((N+1/2)x)sin(x/2)$$
Now I can only think to apply the Riemann-Lebesgue lemma to the function
$$frac1sin(theta/2)-frac2theta$$
To get :
$$lim_nto inftyfrac12piint_-pi^pi(frace^-inthetasin(theta/2)-frac2e^-inthetatheta)dthetato0$$But I don't know the steps to prove this.
fourier-analysis
asked Mar 27 at 4:52
John HeJohn He
668
$endgroup$
|
show 2 more comments
$begingroup$
Prove
$$int_0^inftyfracsin xxdx=fracpi2$$
Start with the fact that the integral of $D_N(theta)$ equals $2pi$, and note that the difference $$frac1sin(theta/2)-frac2theta$$
is continuous on $[-pi,pi]$. Apply the Riemann-Lebesgue lemma.
$$D_N(x)=sum_-N^Ne^inx=fracsin((N+1/2)x)sin(x/2)$$
Now I can only think to apply the Riemann-Lebesgue lemma to the function
$$frac1sin(theta/2)-frac2theta$$
To get :
$$lim_nto inftyfrac12piint_-pi^pi(frace^-inthetasin(theta/2)-frac2e^-inthetatheta)dthetato0$$But I don't know the steps to prove this.
fourier-analysis
asked Mar 27 at 4:52
John HeJohn He
668
$endgroup$
$begingroup$
It appears you're trying to follow a specific argument for the integral here. In order for the rest of us to understand what you're doing, we need more context about that argument. For example, what is $D_N$?
$endgroup$
– jmerry
Mar 27 at 4:57
$begingroup$
you have a limit as $n to infty$ on the LHS and an $n$ on the RHS
$endgroup$
– mathworker21
Mar 27 at 4:58
$begingroup$
On the last line, both integrals are badly non-convergent.
$endgroup$
– Lord Shark the Unknown
Mar 27 at 5:07
$begingroup$
I changed it according to the form of Riemann-Lebesgue lemma.
$endgroup$
– John He
Mar 27 at 5:13
$begingroup$
@jmerry - If I had to guess at a glance, it probably is tied to the Dirichlet kernel somehow or another
$endgroup$
– Eevee Trainer
Mar 27 at 5:14
|
show 2 more comments
$begingroup$
Prove
$$int_0^inftyfracsin xxdx=fracpi2$$
Start with the fact that the integral of $D_N(theta)$ equals $2pi$, and note that the difference $$frac1sin(theta/2)-frac2theta$$
is continuous on $[-pi,pi]$. Apply the Riemann-Lebesgue lemma.
$$D_N(x)=sum_-N^Ne^inx=fracsin((N+1/2)x)sin(x/2)$$
Now I can only think to apply the Riemann-Lebesgue lemma to the function
$$frac1sin(theta/2)-frac2theta$$
To get :
$$lim_nto inftyfrac12piint_-pi^pi(frace^-inthetasin(theta/2)-frac2e^-inthetatheta)dthetato0$$But I don't know the steps to prove this.
fourier-analysis
asked Mar 27 at 4:52
John HeJohn He
668
$endgroup$
Prove
$$int_0^inftyfracsin xxdx=fracpi2$$
Start with the fact that the integral of $D_N(theta)$ equals $2pi$, and note that the difference $$frac1sin(theta/2)-frac2theta$$
is continuous on $[-pi,pi]$. Apply the Riemann-Lebesgue lemma.
$$D_N(x)=sum_-N^Ne^inx=fracsin((N+1/2)x)sin(x/2)$$
Now I can only think to apply the Riemann-Lebesgue lemma to the function
$$frac1sin(theta/2)-frac2theta$$
To get :
$$lim_nto inftyfrac12piint_-pi^pi(frace^-inthetasin(theta/2)-frac2e^-inthetatheta)dthetato0$$But I don't know the steps to prove this.
fourier-analysis
fourier-analysis
asked Mar 27 at 4:52
John HeJohn He
668
asked Mar 27 at 4:52
John HeJohn He
668
edited Mar 27 at 5:11
John He
asked Mar 27 at 4:52
John HeJohn He
668
asked Mar 27 at 4:52
John HeJohn He
668
asked Mar 27 at 4:52
John HeJohn He
668
668
$begingroup$
It appears you're trying to follow a specific argument for the integral here. In order for the rest of us to understand what you're doing, we need more context about that argument. For example, what is $D_N$?
$endgroup$
– jmerry
Mar 27 at 4:57
$begingroup$
you have a limit as $n to infty$ on the LHS and an $n$ on the RHS
$endgroup$
– mathworker21
Mar 27 at 4:58
$begingroup$
On the last line, both integrals are badly non-convergent.
$endgroup$
– Lord Shark the Unknown
Mar 27 at 5:07
$begingroup$
I changed it according to the form of Riemann-Lebesgue lemma.
$endgroup$
– John He
Mar 27 at 5:13
$begingroup$
@jmerry - If I had to guess at a glance, it probably is tied to the Dirichlet kernel somehow or another
$endgroup$
– Eevee Trainer
Mar 27 at 5:14
|
show 2 more comments
$begingroup$
It appears you're trying to follow a specific argument for the integral here. In order for the rest of us to understand what you're doing, we need more context about that argument. For example, what is $D_N$?
$endgroup$
– jmerry
Mar 27 at 4:57
$begingroup$
you have a limit as $n to infty$ on the LHS and an $n$ on the RHS
$endgroup$
– mathworker21
Mar 27 at 4:58
$begingroup$
On the last line, both integrals are badly non-convergent.
$endgroup$
– Lord Shark the Unknown
Mar 27 at 5:07
$begingroup$
I changed it according to the form of Riemann-Lebesgue lemma.
$endgroup$
– John He
Mar 27 at 5:13
$begingroup$
@jmerry - If I had to guess at a glance, it probably is tied to the Dirichlet kernel somehow or another
$endgroup$
– Eevee Trainer
Mar 27 at 5:14
$begingroup$
It appears you're trying to follow a specific argument for the integral here. In order for the rest of us to understand what you're doing, we need more context about that argument. For example, what is $D_N$?
$endgroup$
– jmerry
Mar 27 at 4:57
$begingroup$
It appears you're trying to follow a specific argument for the integral here. In order for the rest of us to understand what you're doing, we need more context about that argument. For example, what is $D_N$?
$endgroup$
– jmerry
Mar 27 at 4:57
$begingroup$
you have a limit as $n to infty$ on the LHS and an $n$ on the RHS
$endgroup$
– mathworker21
Mar 27 at 4:58
$begingroup$
you have a limit as $n to infty$ on the LHS and an $n$ on the RHS
$endgroup$
– mathworker21
Mar 27 at 4:58
$begingroup$
On the last line, both integrals are badly non-convergent.
$endgroup$
– Lord Shark the Unknown
Mar 27 at 5:07
$begingroup$
On the last line, both integrals are badly non-convergent.
$endgroup$
– Lord Shark the Unknown
Mar 27 at 5:07
$begingroup$
I changed it according to the form of Riemann-Lebesgue lemma.
$endgroup$
– John He
Mar 27 at 5:13
$begingroup$
I changed it according to the form of Riemann-Lebesgue lemma.
$endgroup$
– John He
Mar 27 at 5:13
$begingroup$
@jmerry - If I had to guess at a glance, it probably is tied to the Dirichlet kernel somehow or another
$endgroup$
– Eevee Trainer
Mar 27 at 5:14
$begingroup$
@jmerry - If I had to guess at a glance, it probably is tied to the Dirichlet kernel somehow or another
$endgroup$
– Eevee Trainer
Mar 27 at 5:14
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Note that
$$int_-pi^pi D_N(t),dt=2pi.$$
Use that
$$int_-pi^pisin((N+1/2)t)left(frac1sin t/2-frac 2tright),dtto0$$
as $Ntoinfty$ (Riemann-Lebesgue) to deduce that
$$lim_Ntoinftyint_-pi^pifracsin((N+1/2)t)t,dx=pi.$$
answered Mar 27 at 5:16
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
$begingroup$
Thank you. I did It again with My original path, and I find It's essentially the same thing.
$endgroup$
– John He
Mar 27 at 5:30
add a comment |
$begingroup$
Consider the Laplace transform
$$
F(s)=int_0^inftye^-sxfracsin xxdx.
$$
Your integral is $lim_sdownarrow 0F(s)$. For $s > 0$,
$$
F'(s)=int_0^inftye^-sxsin xdx \
= Im int_0^inftye^-sxe^ixdx \
= Im int_0^inftye^-x(s+i)dx \
= Im frac1s+i=Imfracs-i1+s^2 \=-frac11+s^2
$$
Therefore $F(s)=-tan^-1(s)+C$ for some constant $C$. Because $F(infty)=0$, then $C=fracpi2$. So
$$
int_0^inftye^-sxfracsin xxdx = fracpi2-tan^-1(s).
$$
Letting $sdownarrow 0$ gives
$$
int_0^inftyfracsin xxdx=fracpi2.
$$
answered Mar 29 at 2:40
DisintegratingByPartsDisintegratingByParts
60.5k42681
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that
$$int_-pi^pi D_N(t),dt=2pi.$$
Use that
$$int_-pi^pisin((N+1/2)t)left(frac1sin t/2-frac 2tright),dtto0$$
as $Ntoinfty$ (Riemann-Lebesgue) to deduce that
$$lim_Ntoinftyint_-pi^pifracsin((N+1/2)t)t,dx=pi.$$
answered Mar 27 at 5:16
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
$begingroup$
Thank you. I did It again with My original path, and I find It's essentially the same thing.
$endgroup$
– John He
Mar 27 at 5:30
add a comment |
$begingroup$
Note that
$$int_-pi^pi D_N(t),dt=2pi.$$
Use that
$$int_-pi^pisin((N+1/2)t)left(frac1sin t/2-frac 2tright),dtto0$$
as $Ntoinfty$ (Riemann-Lebesgue) to deduce that
$$lim_Ntoinftyint_-pi^pifracsin((N+1/2)t)t,dx=pi.$$
answered Mar 27 at 5:16
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
$begingroup$
Thank you. I did It again with My original path, and I find It's essentially the same thing.
$endgroup$
– John He
Mar 27 at 5:30
add a comment |
$begingroup$
Note that
$$int_-pi^pi D_N(t),dt=2pi.$$
Use that
$$int_-pi^pisin((N+1/2)t)left(frac1sin t/2-frac 2tright),dtto0$$
as $Ntoinfty$ (Riemann-Lebesgue) to deduce that
$$lim_Ntoinftyint_-pi^pifracsin((N+1/2)t)t,dx=pi.$$
answered Mar 27 at 5:16
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
Note that
$$int_-pi^pi D_N(t),dt=2pi.$$
Use that
$$int_-pi^pisin((N+1/2)t)left(frac1sin t/2-frac 2tright),dtto0$$
as $Ntoinfty$ (Riemann-Lebesgue) to deduce that
$$lim_Ntoinftyint_-pi^pifracsin((N+1/2)t)t,dx=pi.$$
answered Mar 27 at 5:16
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered Mar 27 at 5:16
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered Mar 27 at 5:16
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered Mar 27 at 5:16
Lord Shark the UnknownLord Shark the Unknown
109k1163136
109k1163136
$begingroup$
Thank you. I did It again with My original path, and I find It's essentially the same thing.
$endgroup$
– John He
Mar 27 at 5:30
add a comment |
$begingroup$
Thank you. I did It again with My original path, and I find It's essentially the same thing.
$endgroup$
– John He
Mar 27 at 5:30
$begingroup$
Thank you. I did It again with My original path, and I find It's essentially the same thing.
$endgroup$
– John He
Mar 27 at 5:30
$begingroup$
Thank you. I did It again with My original path, and I find It's essentially the same thing.
$endgroup$
– John He
Mar 27 at 5:30
add a comment |
$begingroup$
Consider the Laplace transform
$$
F(s)=int_0^inftye^-sxfracsin xxdx.
$$
Your integral is $lim_sdownarrow 0F(s)$. For $s > 0$,
$$
F'(s)=int_0^inftye^-sxsin xdx \
= Im int_0^inftye^-sxe^ixdx \
= Im int_0^inftye^-x(s+i)dx \
= Im frac1s+i=Imfracs-i1+s^2 \=-frac11+s^2
$$
Therefore $F(s)=-tan^-1(s)+C$ for some constant $C$. Because $F(infty)=0$, then $C=fracpi2$. So
$$
int_0^inftye^-sxfracsin xxdx = fracpi2-tan^-1(s).
$$
Letting $sdownarrow 0$ gives
$$
int_0^inftyfracsin xxdx=fracpi2.
$$
answered Mar 29 at 2:40
DisintegratingByPartsDisintegratingByParts
60.5k42681
$endgroup$
add a comment |
$begingroup$
Consider the Laplace transform
$$
F(s)=int_0^inftye^-sxfracsin xxdx.
$$
Your integral is $lim_sdownarrow 0F(s)$. For $s > 0$,
$$
F'(s)=int_0^inftye^-sxsin xdx \
= Im int_0^inftye^-sxe^ixdx \
= Im int_0^inftye^-x(s+i)dx \
= Im frac1s+i=Imfracs-i1+s^2 \=-frac11+s^2
$$
Therefore $F(s)=-tan^-1(s)+C$ for some constant $C$. Because $F(infty)=0$, then $C=fracpi2$. So
$$
int_0^inftye^-sxfracsin xxdx = fracpi2-tan^-1(s).
$$
Letting $sdownarrow 0$ gives
$$
int_0^inftyfracsin xxdx=fracpi2.
$$
answered Mar 29 at 2:40
DisintegratingByPartsDisintegratingByParts
60.5k42681
$endgroup$
add a comment |
$begingroup$
Consider the Laplace transform
$$
F(s)=int_0^inftye^-sxfracsin xxdx.
$$
Your integral is $lim_sdownarrow 0F(s)$. For $s > 0$,
$$
F'(s)=int_0^inftye^-sxsin xdx \
= Im int_0^inftye^-sxe^ixdx \
= Im int_0^inftye^-x(s+i)dx \
= Im frac1s+i=Imfracs-i1+s^2 \=-frac11+s^2
$$
Therefore $F(s)=-tan^-1(s)+C$ for some constant $C$. Because $F(infty)=0$, then $C=fracpi2$. So
$$
int_0^inftye^-sxfracsin xxdx = fracpi2-tan^-1(s).
$$
Letting $sdownarrow 0$ gives
$$
int_0^inftyfracsin xxdx=fracpi2.
$$
answered Mar 29 at 2:40
DisintegratingByPartsDisintegratingByParts
60.5k42681
$endgroup$
Consider the Laplace transform
$$
F(s)=int_0^inftye^-sxfracsin xxdx.
$$
Your integral is $lim_sdownarrow 0F(s)$. For $s > 0$,
$$
F'(s)=int_0^inftye^-sxsin xdx \
= Im int_0^inftye^-sxe^ixdx \
= Im int_0^inftye^-x(s+i)dx \
= Im frac1s+i=Imfracs-i1+s^2 \=-frac11+s^2
$$
Therefore $F(s)=-tan^-1(s)+C$ for some constant $C$. Because $F(infty)=0$, then $C=fracpi2$. So
$$
int_0^inftye^-sxfracsin xxdx = fracpi2-tan^-1(s).
$$
Letting $sdownarrow 0$ gives
$$
int_0^inftyfracsin xxdx=fracpi2.
$$
answered Mar 29 at 2:40
DisintegratingByPartsDisintegratingByParts
60.5k42681
answered Mar 29 at 2:40
DisintegratingByPartsDisintegratingByParts
60.5k42681
answered Mar 29 at 2:40
DisintegratingByPartsDisintegratingByParts
60.5k42681
answered Mar 29 at 2:40
DisintegratingByPartsDisintegratingByParts
60.5k42681
60.5k42681
add a comment |
add a comment |
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$begingroup$
It appears you're trying to follow a specific argument for the integral here. In order for the rest of us to understand what you're doing, we need more context about that argument. For example, what is $D_N$?
$endgroup$
– jmerry
Mar 27 at 4:57
$begingroup$
you have a limit as $n to infty$ on the LHS and an $n$ on the RHS
$endgroup$
– mathworker21
Mar 27 at 4:58
$begingroup$
On the last line, both integrals are badly non-convergent.
$endgroup$
– Lord Shark the Unknown
Mar 27 at 5:07
$begingroup$
I changed it according to the form of Riemann-Lebesgue lemma.
$endgroup$
– John He
Mar 27 at 5:13
$begingroup$
@jmerry - If I had to guess at a glance, it probably is tied to the Dirichlet kernel somehow or another
$endgroup$
– Eevee Trainer
Mar 27 at 5:14