Prove that $ int _0^inftyfracsin xxdx=fracpi2$$sum_n=-infty^inftyfrac1(n+alpha)^2=fracpi^2(sin pi alpha)^2$?Show $lim_Ntoinftyint_0^pileft(frac1sinfracx2-frac2xright)sinleft((N+frac12)xright)dx=0$Show that Fourier transformation is differentiable if $int|xf(x)|,dlambda<infty$Riemann Lebesgue Lemma Clarification$lim_lambda to infty int^b_0 f(t) fracsin(lambda t)t $The improper integrals of $fracsin tt$ and $left|fracsin ttright|$Calculate limits of integralsAlternate proof of Dirichlet integral $fracsin(x)x$.Riemann-Lebesgue Lemma And Integral Having $sin(cdot)$Abel means of function f when having a jump discontinuity
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Prove that $ int _0^inftyfracsin xxdx=fracpi2$
$sum_n=-infty^inftyfrac1(n+alpha)^2=fracpi^2(sin pi alpha)^2$?Show $lim_Ntoinftyint_0^pileft(frac1sinfracx2-frac2xright)sinleft((N+frac12)xright)dx=0$Show that Fourier transformation is differentiable if $int|xf(x)|,dlambda<infty$Riemann Lebesgue Lemma Clarification$lim_lambda to infty int^b_0 f(t) fracsin(lambda t)t $The improper integrals of $fracsin tt$ and $left|fracsin ttright|$Calculate limits of integralsAlternate proof of Dirichlet integral $fracsin(x)x$.Riemann-Lebesgue Lemma And Integral Having $sin(cdot)$Abel means of function f when having a jump discontinuity
$begingroup$
Prove
$$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$$But I don't know the steps to prove this.
fourier-analysis
asked 15 hours ago
John HeJohn He
415
$endgroup$
|
show 2 more comments
$begingroup$
Prove
$$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$$But I don't know the steps to prove this.
fourier-analysis
asked 15 hours ago
John HeJohn He
415
$endgroup$
$begingroup$
It appears you're trying to follow a specific argument for the integral here. In order for the rest of us to understand what you're doing, we need more context about that argument. For example, what is $D_N$?
$endgroup$
– jmerry
15 hours ago
$begingroup$
you have a limit as $n to infty$ on the LHS and an $n$ on the RHS
$endgroup$
– mathworker21
15 hours ago
$begingroup$
On the last line, both integrals are badly non-convergent.
$endgroup$
– Lord Shark the Unknown
15 hours ago
$begingroup$
I changed it according to the form of Riemann-Lebesgue lemma.
$endgroup$
– John He
15 hours ago
$begingroup$
@jmerry - If I had to guess at a glance, it probably is tied to the Dirichlet kernel somehow or another
$endgroup$
– Eevee Trainer
15 hours ago
|
show 2 more comments
$begingroup$
Prove
$$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$$But I don't know the steps to prove this.
fourier-analysis
asked 15 hours ago
John HeJohn He
415
$endgroup$
Prove
$$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 15 hours ago
John HeJohn He
415
asked 15 hours ago
John HeJohn He
415
edited 15 hours ago
John He
asked 15 hours ago
John HeJohn He
415
asked 15 hours ago
John HeJohn He
415
asked 15 hours ago
John HeJohn He
415
415
$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
15 hours ago
$begingroup$
you have a limit as $n to infty$ on the LHS and an $n$ on the RHS
$endgroup$
– mathworker21
15 hours ago
$begingroup$
On the last line, both integrals are badly non-convergent.
$endgroup$
– Lord Shark the Unknown
15 hours ago
$begingroup$
I changed it according to the form of Riemann-Lebesgue lemma.
$endgroup$
– John He
15 hours ago
$begingroup$
@jmerry - If I had to guess at a glance, it probably is tied to the Dirichlet kernel somehow or another
$endgroup$
– Eevee Trainer
15 hours ago
|
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
15 hours ago
$begingroup$
you have a limit as $n to infty$ on the LHS and an $n$ on the RHS
$endgroup$
– mathworker21
15 hours ago
$begingroup$
On the last line, both integrals are badly non-convergent.
$endgroup$
– Lord Shark the Unknown
15 hours ago
$begingroup$
I changed it according to the form of Riemann-Lebesgue lemma.
$endgroup$
– John He
15 hours ago
$begingroup$
@jmerry - If I had to guess at a glance, it probably is tied to the Dirichlet kernel somehow or another
$endgroup$
– Eevee Trainer
15 hours ago
$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
15 hours ago
$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
15 hours ago
$begingroup$
you have a limit as $n to infty$ on the LHS and an $n$ on the RHS
$endgroup$
– mathworker21
15 hours ago
$begingroup$
you have a limit as $n to infty$ on the LHS and an $n$ on the RHS
$endgroup$
– mathworker21
15 hours ago
$begingroup$
On the last line, both integrals are badly non-convergent.
$endgroup$
– Lord Shark the Unknown
15 hours ago
$begingroup$
On the last line, both integrals are badly non-convergent.
$endgroup$
– Lord Shark the Unknown
15 hours ago
$begingroup$
I changed it according to the form of Riemann-Lebesgue lemma.
$endgroup$
– John He
15 hours ago
$begingroup$
I changed it according to the form of Riemann-Lebesgue lemma.
$endgroup$
– John He
15 hours ago
$begingroup$
@jmerry - If I had to guess at a glance, it probably is tied to the Dirichlet kernel somehow or another
$endgroup$
– Eevee Trainer
15 hours ago
$begingroup$
@jmerry - If I had to guess at a glance, it probably is tied to the Dirichlet kernel somehow or another
$endgroup$
– Eevee Trainer
15 hours ago
|
show 2 more comments
1 Answer
1
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.$$
answered 15 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
Thank you. I did It again with My original path, and I find It's essentially the same thing.
$endgroup$
– John He
15 hours ago
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Note that
$$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.$$
answered 15 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
Thank you. I did It again with My original path, and I find It's essentially the same thing.
$endgroup$
– John He
15 hours ago
add a comment |
$begingroup$
Note that
$$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.$$
answered 15 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
$begingroup$
Thank you. I did It again with My original path, and I find It's essentially the same thing.
$endgroup$
– John He
15 hours ago
add a comment |
$begingroup$
Note that
$$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.$$
answered 15 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
Note that
$$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 15 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered 15 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered 15 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered 15 hours ago
Lord Shark the UnknownLord Shark the Unknown
107k1162135
107k1162135
$begingroup$
Thank you. I did It again with My original path, and I find It's essentially the same thing.
$endgroup$
– John He
15 hours ago
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
15 hours ago
$begingroup$
Thank you. I did It again with My original path, and I find It's essentially the same thing.
$endgroup$
– John He
15 hours ago
$begingroup$
Thank you. I did It again with My original path, and I find It's essentially the same thing.
$endgroup$
– John He
15 hours ago
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
15 hours ago
$begingroup$
you have a limit as $n to infty$ on the LHS and an $n$ on the RHS
$endgroup$
– mathworker21
15 hours ago
$begingroup$
On the last line, both integrals are badly non-convergent.
$endgroup$
– Lord Shark the Unknown
15 hours ago
$begingroup$
I changed it according to the form of Riemann-Lebesgue lemma.
$endgroup$
– John He
15 hours ago
$begingroup$
@jmerry - If I had to guess at a glance, it probably is tied to the Dirichlet kernel somehow or another
$endgroup$
– Eevee Trainer
15 hours ago