Finding $limlimits_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4,mathrm dx$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Evaluating $lim limits_nto infty,,, n!! intlimits_0^pi/2!! left(1-sqrt [n]sin x right),mathrm dx$Evaluate $int_0^1 frac1 x^2+2x+3,mathrm dx$Show how $fracpartialpartial x left[int_0^x (x-t)g(t),mathrmdtright] = int_0^x g(t),mathrmdt$Evaluate $large int_0^1left(frac1ln x + frac11-xright)^2 mathrm dx $ using elementary, high school techniquesFind the limit $lim_limitsxto 0^+left( e^frac1sin x-e^frac1xright)$L'Hopital rule to solve $ limlimits_xto 0 left(fracsin xxright)^frac1x^2 $Compute the limit $lim_ntoinfty nleft [int_0^frac pi4tan^n left ( fracxn right )mathrm dxright]^frac1n$How to find $limlimits_xto 0^+ frac1x int_0^x sin(frac1t),mathrm dt$$n^2 int_0^1 (1-x)^n sin(pi x) mathrmdx$If $I_n=int_0^1 fracx^nx^2+2019,mathrm dx$, evaluate $limlimits_nto infty nI_n$
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Finding $limlimits_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4,mathrm dx$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Evaluating $lim limits_nto infty,,, n!! intlimits_0^pi/2!! left(1-sqrt [n]sin x right),mathrm dx$Evaluate $int_0^1 frac1 x^2+2x+3,mathrm dx$Show how $fracpartialpartial x left[int_0^x (x-t)g(t),mathrmdtright] = int_0^x g(t),mathrmdt$Evaluate $large int_0^1left(frac1ln x + frac11-xright)^2 mathrm dx $ using elementary, high school techniquesFind the limit $lim_limitsxto 0^+left( e^frac1sin x-e^frac1xright)$L'Hopital rule to solve $ limlimits_xto 0 left(fracsin xxright)^frac1x^2 $Compute the limit $lim_ntoinfty nleft [int_0^frac pi4tan^n left ( fracxn right )mathrm dxright]^frac1n$How to find $limlimits_xto 0^+ frac1x int_0^x sin(frac1t),mathrm dt$$n^2 int_0^1 (1-x)^n sin(pi x) mathrmdx$If $I_n=int_0^1 fracx^nx^2+2019,mathrm dx$, evaluate $limlimits_nto infty nI_n$
$begingroup$
$$lim_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4,mathrm dx$$I don't know how to solve this limit problem. I'm roughly clear about using the pinch theorem, but I haven't done it yet. Could somebody please tell me? Thanks!
calculus integration limits
edited Mar 27 at 11:17
rash
568216
asked Mar 27 at 6:45
赵家艺赵家艺
382
$endgroup$
|
show 1 more comment
$begingroup$
$$lim_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4,mathrm dx$$I don't know how to solve this limit problem. I'm roughly clear about using the pinch theorem, but I haven't done it yet. Could somebody please tell me? Thanks!
calculus integration limits
edited Mar 27 at 11:17
rash
568216
asked Mar 27 at 6:45
赵家艺赵家艺
382
$endgroup$
$begingroup$
(pinch theorem)
$endgroup$
– mathworker21
Mar 27 at 7:08
$begingroup$
there's some formula like $sum_n e(ntheta) = fracsin(Ntheta)sin(theta)$. Something like that. I think that should be useful
$endgroup$
– mathworker21
Mar 27 at 7:26
$begingroup$
I checked through W|A for values of $n$ as high as $nsim 20000$ and it seems that the limit does exist and is equal to $ln 2$. But at present I don't have a rigorous proof for the same.
$endgroup$
– Darkrai
Mar 27 at 8:00
$begingroup$
After a lengthy semi-heuristic study I suspect the simple result $lim = log(2) simeq 0.6931471805599453..$
$endgroup$
– Dr. Wolfgang Hintze
Mar 27 at 8:09
2
$begingroup$
Another way to reach the integral in Tianlalu's answer is as follows: Substituting $nx = u$ and writing $operatornamesinc(x) = fracsin xx$, we have $$ frac1n^2 int_0^fracpi2 x fracsin^4(nx)sin^4 x , mathrmdx = int_0^fracn pi2 fracsin^4 uu^3 operatornamesinc^4(u/n) , mathrmd u xrightarrow[ntoinfty]textDCT int_0^infty fracsin^4 uu^3 , mathrmd u, $$ where the convergence follows from the dominated convergence together with the integrable dominating function $Cu^-3sin^4 u$ for some constant $C > 0$.
$endgroup$
– Sangchul Lee
Mar 27 at 11:48
|
show 1 more comment
$begingroup$
$$lim_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4,mathrm dx$$I don't know how to solve this limit problem. I'm roughly clear about using the pinch theorem, but I haven't done it yet. Could somebody please tell me? Thanks!
calculus integration limits
edited Mar 27 at 11:17
rash
568216
asked Mar 27 at 6:45
赵家艺赵家艺
382
$endgroup$
$$lim_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4,mathrm dx$$I don't know how to solve this limit problem. I'm roughly clear about using the pinch theorem, but I haven't done it yet. Could somebody please tell me? Thanks!
calculus integration limits
calculus integration limits
edited Mar 27 at 11:17
rash
568216
asked Mar 27 at 6:45
赵家艺赵家艺
382
edited Mar 27 at 11:17
rash
568216
asked Mar 27 at 6:45
赵家艺赵家艺
382
edited Mar 27 at 11:17
rash
568216
edited Mar 27 at 11:17
rash
568216
edited Mar 27 at 11:17
rash
568216
568216
asked Mar 27 at 6:45
赵家艺赵家艺
382
asked Mar 27 at 6:45
赵家艺赵家艺
382
asked Mar 27 at 6:45
赵家艺赵家艺
382
382
$begingroup$
(pinch theorem)
$endgroup$
– mathworker21
Mar 27 at 7:08
$begingroup$
there's some formula like $sum_n e(ntheta) = fracsin(Ntheta)sin(theta)$. Something like that. I think that should be useful
$endgroup$
– mathworker21
Mar 27 at 7:26
$begingroup$
I checked through W|A for values of $n$ as high as $nsim 20000$ and it seems that the limit does exist and is equal to $ln 2$. But at present I don't have a rigorous proof for the same.
$endgroup$
– Darkrai
Mar 27 at 8:00
$begingroup$
After a lengthy semi-heuristic study I suspect the simple result $lim = log(2) simeq 0.6931471805599453..$
$endgroup$
– Dr. Wolfgang Hintze
Mar 27 at 8:09
2
$begingroup$
Another way to reach the integral in Tianlalu's answer is as follows: Substituting $nx = u$ and writing $operatornamesinc(x) = fracsin xx$, we have $$ frac1n^2 int_0^fracpi2 x fracsin^4(nx)sin^4 x , mathrmdx = int_0^fracn pi2 fracsin^4 uu^3 operatornamesinc^4(u/n) , mathrmd u xrightarrow[ntoinfty]textDCT int_0^infty fracsin^4 uu^3 , mathrmd u, $$ where the convergence follows from the dominated convergence together with the integrable dominating function $Cu^-3sin^4 u$ for some constant $C > 0$.
$endgroup$
– Sangchul Lee
Mar 27 at 11:48
|
show 1 more comment
$begingroup$
(pinch theorem)
$endgroup$
– mathworker21
Mar 27 at 7:08
$begingroup$
there's some formula like $sum_n e(ntheta) = fracsin(Ntheta)sin(theta)$. Something like that. I think that should be useful
$endgroup$
– mathworker21
Mar 27 at 7:26
$begingroup$
I checked through W|A for values of $n$ as high as $nsim 20000$ and it seems that the limit does exist and is equal to $ln 2$. But at present I don't have a rigorous proof for the same.
$endgroup$
– Darkrai
Mar 27 at 8:00
$begingroup$
After a lengthy semi-heuristic study I suspect the simple result $lim = log(2) simeq 0.6931471805599453..$
$endgroup$
– Dr. Wolfgang Hintze
Mar 27 at 8:09
2
$begingroup$
Another way to reach the integral in Tianlalu's answer is as follows: Substituting $nx = u$ and writing $operatornamesinc(x) = fracsin xx$, we have $$ frac1n^2 int_0^fracpi2 x fracsin^4(nx)sin^4 x , mathrmdx = int_0^fracn pi2 fracsin^4 uu^3 operatornamesinc^4(u/n) , mathrmd u xrightarrow[ntoinfty]textDCT int_0^infty fracsin^4 uu^3 , mathrmd u, $$ where the convergence follows from the dominated convergence together with the integrable dominating function $Cu^-3sin^4 u$ for some constant $C > 0$.
$endgroup$
– Sangchul Lee
Mar 27 at 11:48
$begingroup$
(pinch theorem)
$endgroup$
– mathworker21
Mar 27 at 7:08
$begingroup$
(pinch theorem)
$endgroup$
– mathworker21
Mar 27 at 7:08
$begingroup$
there's some formula like $sum_n e(ntheta) = fracsin(Ntheta)sin(theta)$. Something like that. I think that should be useful
$endgroup$
– mathworker21
Mar 27 at 7:26
$begingroup$
there's some formula like $sum_n e(ntheta) = fracsin(Ntheta)sin(theta)$. Something like that. I think that should be useful
$endgroup$
– mathworker21
Mar 27 at 7:26
$begingroup$
I checked through W|A for values of $n$ as high as $nsim 20000$ and it seems that the limit does exist and is equal to $ln 2$. But at present I don't have a rigorous proof for the same.
$endgroup$
– Darkrai
Mar 27 at 8:00
$begingroup$
I checked through W|A for values of $n$ as high as $nsim 20000$ and it seems that the limit does exist and is equal to $ln 2$. But at present I don't have a rigorous proof for the same.
$endgroup$
– Darkrai
Mar 27 at 8:00
$begingroup$
After a lengthy semi-heuristic study I suspect the simple result $lim = log(2) simeq 0.6931471805599453..$
$endgroup$
– Dr. Wolfgang Hintze
Mar 27 at 8:09
$begingroup$
After a lengthy semi-heuristic study I suspect the simple result $lim = log(2) simeq 0.6931471805599453..$
$endgroup$
– Dr. Wolfgang Hintze
Mar 27 at 8:09
2
2
$begingroup$
Another way to reach the integral in Tianlalu's answer is as follows: Substituting $nx = u$ and writing $operatornamesinc(x) = fracsin xx$, we have $$ frac1n^2 int_0^fracpi2 x fracsin^4(nx)sin^4 x , mathrmdx = int_0^fracn pi2 fracsin^4 uu^3 operatornamesinc^4(u/n) , mathrmd u xrightarrow[ntoinfty]textDCT int_0^infty fracsin^4 uu^3 , mathrmd u, $$ where the convergence follows from the dominated convergence together with the integrable dominating function $Cu^-3sin^4 u$ for some constant $C > 0$.
$endgroup$
– Sangchul Lee
Mar 27 at 11:48
$begingroup$
Another way to reach the integral in Tianlalu's answer is as follows: Substituting $nx = u$ and writing $operatornamesinc(x) = fracsin xx$, we have $$ frac1n^2 int_0^fracpi2 x fracsin^4(nx)sin^4 x , mathrmdx = int_0^fracn pi2 fracsin^4 uu^3 operatornamesinc^4(u/n) , mathrmd u xrightarrow[ntoinfty]textDCT int_0^infty fracsin^4 uu^3 , mathrmd u, $$ where the convergence follows from the dominated convergence together with the integrable dominating function $Cu^-3sin^4 u$ for some constant $C > 0$.
$endgroup$
– Sangchul Lee
Mar 27 at 11:48
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
From
$$lim_xto 0x^2left(frac1sin^4 x-frac1x^4right)=frac23,$$
for $xin(0,fracpi 2)$, there is $C$ such that
$$left|fracxsin^4 nxsin^4 x-fracsin^4 nxx^3right|le fracCsin^4 nxxle Cn.$$
Thus,
beginalign*
lim_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4, mathrm dx&=lim_ntoinftyfrac1n^2int_0^fracpi2fracsin^4 nxx^3, mathrm dx\
&= lim_ntoinftyint_0^fracnpi2fracsin^4 xx^3, mathrm dx\
&= int_0^inftyfracsin^4 xx^3, mathrm dx\
text(IBP)quad &= int_0^inftyfrac2sin^3 xcos xx^2, mathrm dx\
text(IBP)quad &= int_0^inftyfrac6sin^2 xcos^2 x-2sin^4 xx, mathrm dx\
&= int_0^inftyfraccos 2x-cos 4xx, mathrm dx\
text(Frullani)quad &=ln 2.
endalign*
answered Mar 27 at 10:11
TianlaluTianlalu
3,29021238
$endgroup$
add a comment |
$begingroup$
Result
Let
$$f(n) = int_0^fracpi 2 x left(fracsin (n x)sin (x)right)^4 , dx$$
then
$$lim_nto infty , fracf(n)n^2=log (2)simeq 0.6931471805599453 ...$$
This result is in good agreement with the numerical estimate of the integral.
Derivation
This is more heuristic and not really strict. Maybe others can fill in the gaps.
We split the integral into equally spaced intervals, and observe $sin(pi x/n) simeq pi x/n$ for $nto infty$ and get
$$lim_nto infty f(n)/n^2 =lim_nto infty sum_k=1^infty a(k)tag1$$
where
$$a(k) = int_k-1^k pi ^2 s left(fracsin (pi s)pi sright)^4 , ds$$
The first two values are
$$a(1) = textCi(2 pi )-textCi(4 pi )+log (2)$$
$$a(2) = -textCi(-8 pi )+textCi(-4 pi )-textCi(2 pi )+textCi(4 pi )$$
and for $k ge 3$
$$a(k) = -textCi(2 (k-1) pi )+textCi(4 (k-1) pi )-textCi(-4 k pi )+textCi(-2 k pi )$$
Here $textCi(z) = -int_-infty ^z fraccos (t)t , dt$ is the cosine integral.
The sum in $(1)$ telescopes and we are left with the announced result.
Discussion
My statement "the sum telescopes" is fairly bold.
In fact, the cancellation is like this
$$textCi(-4 pi )-textCi(4 pi )=i pi$$
$$textCi(8 pi )-textCi(-8 pi )=-i pi$$
see symmetry relations in http://people.math.sfu.ca/~cbm/aands/abramowitz_and_stegun.pdf, 5.2.20, and the sum of these two lines is $0$.
answered Mar 27 at 8:34
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,960621
$endgroup$
add a comment |
Your Answer
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2 Answers
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
From
$$lim_xto 0x^2left(frac1sin^4 x-frac1x^4right)=frac23,$$
for $xin(0,fracpi 2)$, there is $C$ such that
$$left|fracxsin^4 nxsin^4 x-fracsin^4 nxx^3right|le fracCsin^4 nxxle Cn.$$
Thus,
beginalign*
lim_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4, mathrm dx&=lim_ntoinftyfrac1n^2int_0^fracpi2fracsin^4 nxx^3, mathrm dx\
&= lim_ntoinftyint_0^fracnpi2fracsin^4 xx^3, mathrm dx\
&= int_0^inftyfracsin^4 xx^3, mathrm dx\
text(IBP)quad &= int_0^inftyfrac2sin^3 xcos xx^2, mathrm dx\
text(IBP)quad &= int_0^inftyfrac6sin^2 xcos^2 x-2sin^4 xx, mathrm dx\
&= int_0^inftyfraccos 2x-cos 4xx, mathrm dx\
text(Frullani)quad &=ln 2.
endalign*
answered Mar 27 at 10:11
TianlaluTianlalu
3,29021238
$endgroup$
add a comment |
$begingroup$
From
$$lim_xto 0x^2left(frac1sin^4 x-frac1x^4right)=frac23,$$
for $xin(0,fracpi 2)$, there is $C$ such that
$$left|fracxsin^4 nxsin^4 x-fracsin^4 nxx^3right|le fracCsin^4 nxxle Cn.$$
Thus,
beginalign*
lim_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4, mathrm dx&=lim_ntoinftyfrac1n^2int_0^fracpi2fracsin^4 nxx^3, mathrm dx\
&= lim_ntoinftyint_0^fracnpi2fracsin^4 xx^3, mathrm dx\
&= int_0^inftyfracsin^4 xx^3, mathrm dx\
text(IBP)quad &= int_0^inftyfrac2sin^3 xcos xx^2, mathrm dx\
text(IBP)quad &= int_0^inftyfrac6sin^2 xcos^2 x-2sin^4 xx, mathrm dx\
&= int_0^inftyfraccos 2x-cos 4xx, mathrm dx\
text(Frullani)quad &=ln 2.
endalign*
answered Mar 27 at 10:11
TianlaluTianlalu
3,29021238
$endgroup$
add a comment |
$begingroup$
From
$$lim_xto 0x^2left(frac1sin^4 x-frac1x^4right)=frac23,$$
for $xin(0,fracpi 2)$, there is $C$ such that
$$left|fracxsin^4 nxsin^4 x-fracsin^4 nxx^3right|le fracCsin^4 nxxle Cn.$$
Thus,
beginalign*
lim_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4, mathrm dx&=lim_ntoinftyfrac1n^2int_0^fracpi2fracsin^4 nxx^3, mathrm dx\
&= lim_ntoinftyint_0^fracnpi2fracsin^4 xx^3, mathrm dx\
&= int_0^inftyfracsin^4 xx^3, mathrm dx\
text(IBP)quad &= int_0^inftyfrac2sin^3 xcos xx^2, mathrm dx\
text(IBP)quad &= int_0^inftyfrac6sin^2 xcos^2 x-2sin^4 xx, mathrm dx\
&= int_0^inftyfraccos 2x-cos 4xx, mathrm dx\
text(Frullani)quad &=ln 2.
endalign*
answered Mar 27 at 10:11
TianlaluTianlalu
3,29021238
$endgroup$
From
$$lim_xto 0x^2left(frac1sin^4 x-frac1x^4right)=frac23,$$
for $xin(0,fracpi 2)$, there is $C$ such that
$$left|fracxsin^4 nxsin^4 x-fracsin^4 nxx^3right|le fracCsin^4 nxxle Cn.$$
Thus,
beginalign*
lim_ntoinftyfrac1n^2int_0^fracpi2xleft(fracsin nxsin xright)^4, mathrm dx&=lim_ntoinftyfrac1n^2int_0^fracpi2fracsin^4 nxx^3, mathrm dx\
&= lim_ntoinftyint_0^fracnpi2fracsin^4 xx^3, mathrm dx\
&= int_0^inftyfracsin^4 xx^3, mathrm dx\
text(IBP)quad &= int_0^inftyfrac2sin^3 xcos xx^2, mathrm dx\
text(IBP)quad &= int_0^inftyfrac6sin^2 xcos^2 x-2sin^4 xx, mathrm dx\
&= int_0^inftyfraccos 2x-cos 4xx, mathrm dx\
text(Frullani)quad &=ln 2.
endalign*
answered Mar 27 at 10:11
TianlaluTianlalu
3,29021238
answered Mar 27 at 10:11
TianlaluTianlalu
3,29021238
answered Mar 27 at 10:11
TianlaluTianlalu
3,29021238
answered Mar 27 at 10:11
TianlaluTianlalu
3,29021238
3,29021238
add a comment |
add a comment |
$begingroup$
Result
Let
$$f(n) = int_0^fracpi 2 x left(fracsin (n x)sin (x)right)^4 , dx$$
then
$$lim_nto infty , fracf(n)n^2=log (2)simeq 0.6931471805599453 ...$$
This result is in good agreement with the numerical estimate of the integral.
Derivation
This is more heuristic and not really strict. Maybe others can fill in the gaps.
We split the integral into equally spaced intervals, and observe $sin(pi x/n) simeq pi x/n$ for $nto infty$ and get
$$lim_nto infty f(n)/n^2 =lim_nto infty sum_k=1^infty a(k)tag1$$
where
$$a(k) = int_k-1^k pi ^2 s left(fracsin (pi s)pi sright)^4 , ds$$
The first two values are
$$a(1) = textCi(2 pi )-textCi(4 pi )+log (2)$$
$$a(2) = -textCi(-8 pi )+textCi(-4 pi )-textCi(2 pi )+textCi(4 pi )$$
and for $k ge 3$
$$a(k) = -textCi(2 (k-1) pi )+textCi(4 (k-1) pi )-textCi(-4 k pi )+textCi(-2 k pi )$$
Here $textCi(z) = -int_-infty ^z fraccos (t)t , dt$ is the cosine integral.
The sum in $(1)$ telescopes and we are left with the announced result.
Discussion
My statement "the sum telescopes" is fairly bold.
In fact, the cancellation is like this
$$textCi(-4 pi )-textCi(4 pi )=i pi$$
$$textCi(8 pi )-textCi(-8 pi )=-i pi$$
see symmetry relations in http://people.math.sfu.ca/~cbm/aands/abramowitz_and_stegun.pdf, 5.2.20, and the sum of these two lines is $0$.
answered Mar 27 at 8:34
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,960621
$endgroup$
add a comment |
$begingroup$
Result
Let
$$f(n) = int_0^fracpi 2 x left(fracsin (n x)sin (x)right)^4 , dx$$
then
$$lim_nto infty , fracf(n)n^2=log (2)simeq 0.6931471805599453 ...$$
This result is in good agreement with the numerical estimate of the integral.
Derivation
This is more heuristic and not really strict. Maybe others can fill in the gaps.
We split the integral into equally spaced intervals, and observe $sin(pi x/n) simeq pi x/n$ for $nto infty$ and get
$$lim_nto infty f(n)/n^2 =lim_nto infty sum_k=1^infty a(k)tag1$$
where
$$a(k) = int_k-1^k pi ^2 s left(fracsin (pi s)pi sright)^4 , ds$$
The first two values are
$$a(1) = textCi(2 pi )-textCi(4 pi )+log (2)$$
$$a(2) = -textCi(-8 pi )+textCi(-4 pi )-textCi(2 pi )+textCi(4 pi )$$
and for $k ge 3$
$$a(k) = -textCi(2 (k-1) pi )+textCi(4 (k-1) pi )-textCi(-4 k pi )+textCi(-2 k pi )$$
Here $textCi(z) = -int_-infty ^z fraccos (t)t , dt$ is the cosine integral.
The sum in $(1)$ telescopes and we are left with the announced result.
Discussion
My statement "the sum telescopes" is fairly bold.
In fact, the cancellation is like this
$$textCi(-4 pi )-textCi(4 pi )=i pi$$
$$textCi(8 pi )-textCi(-8 pi )=-i pi$$
see symmetry relations in http://people.math.sfu.ca/~cbm/aands/abramowitz_and_stegun.pdf, 5.2.20, and the sum of these two lines is $0$.
answered Mar 27 at 8:34
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,960621
$endgroup$
add a comment |
$begingroup$
Result
Let
$$f(n) = int_0^fracpi 2 x left(fracsin (n x)sin (x)right)^4 , dx$$
then
$$lim_nto infty , fracf(n)n^2=log (2)simeq 0.6931471805599453 ...$$
This result is in good agreement with the numerical estimate of the integral.
Derivation
This is more heuristic and not really strict. Maybe others can fill in the gaps.
We split the integral into equally spaced intervals, and observe $sin(pi x/n) simeq pi x/n$ for $nto infty$ and get
$$lim_nto infty f(n)/n^2 =lim_nto infty sum_k=1^infty a(k)tag1$$
where
$$a(k) = int_k-1^k pi ^2 s left(fracsin (pi s)pi sright)^4 , ds$$
The first two values are
$$a(1) = textCi(2 pi )-textCi(4 pi )+log (2)$$
$$a(2) = -textCi(-8 pi )+textCi(-4 pi )-textCi(2 pi )+textCi(4 pi )$$
and for $k ge 3$
$$a(k) = -textCi(2 (k-1) pi )+textCi(4 (k-1) pi )-textCi(-4 k pi )+textCi(-2 k pi )$$
Here $textCi(z) = -int_-infty ^z fraccos (t)t , dt$ is the cosine integral.
The sum in $(1)$ telescopes and we are left with the announced result.
Discussion
My statement "the sum telescopes" is fairly bold.
In fact, the cancellation is like this
$$textCi(-4 pi )-textCi(4 pi )=i pi$$
$$textCi(8 pi )-textCi(-8 pi )=-i pi$$
see symmetry relations in http://people.math.sfu.ca/~cbm/aands/abramowitz_and_stegun.pdf, 5.2.20, and the sum of these two lines is $0$.
answered Mar 27 at 8:34
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,960621
$endgroup$
Result
Let
$$f(n) = int_0^fracpi 2 x left(fracsin (n x)sin (x)right)^4 , dx$$
then
$$lim_nto infty , fracf(n)n^2=log (2)simeq 0.6931471805599453 ...$$
This result is in good agreement with the numerical estimate of the integral.
Derivation
This is more heuristic and not really strict. Maybe others can fill in the gaps.
We split the integral into equally spaced intervals, and observe $sin(pi x/n) simeq pi x/n$ for $nto infty$ and get
$$lim_nto infty f(n)/n^2 =lim_nto infty sum_k=1^infty a(k)tag1$$
where
$$a(k) = int_k-1^k pi ^2 s left(fracsin (pi s)pi sright)^4 , ds$$
The first two values are
$$a(1) = textCi(2 pi )-textCi(4 pi )+log (2)$$
$$a(2) = -textCi(-8 pi )+textCi(-4 pi )-textCi(2 pi )+textCi(4 pi )$$
and for $k ge 3$
$$a(k) = -textCi(2 (k-1) pi )+textCi(4 (k-1) pi )-textCi(-4 k pi )+textCi(-2 k pi )$$
Here $textCi(z) = -int_-infty ^z fraccos (t)t , dt$ is the cosine integral.
The sum in $(1)$ telescopes and we are left with the announced result.
Discussion
My statement "the sum telescopes" is fairly bold.
In fact, the cancellation is like this
$$textCi(-4 pi )-textCi(4 pi )=i pi$$
$$textCi(8 pi )-textCi(-8 pi )=-i pi$$
see symmetry relations in http://people.math.sfu.ca/~cbm/aands/abramowitz_and_stegun.pdf, 5.2.20, and the sum of these two lines is $0$.
answered Mar 27 at 8:34
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,960621
edited Mar 27 at 11:09
answered Mar 27 at 8:34
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,960621
answered Mar 27 at 8:34
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,960621
answered Mar 27 at 8:34
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,960621
3,960621
add a comment |
add a comment |
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$begingroup$
(pinch theorem)
$endgroup$
– mathworker21
Mar 27 at 7:08
$begingroup$
there's some formula like $sum_n e(ntheta) = fracsin(Ntheta)sin(theta)$. Something like that. I think that should be useful
$endgroup$
– mathworker21
Mar 27 at 7:26
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I checked through W|A for values of $n$ as high as $nsim 20000$ and it seems that the limit does exist and is equal to $ln 2$. But at present I don't have a rigorous proof for the same.
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– Darkrai
Mar 27 at 8:00
$begingroup$
After a lengthy semi-heuristic study I suspect the simple result $lim = log(2) simeq 0.6931471805599453..$
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– Dr. Wolfgang Hintze
Mar 27 at 8:09
2
$begingroup$
Another way to reach the integral in Tianlalu's answer is as follows: Substituting $nx = u$ and writing $operatornamesinc(x) = fracsin xx$, we have $$ frac1n^2 int_0^fracpi2 x fracsin^4(nx)sin^4 x , mathrmdx = int_0^fracn pi2 fracsin^4 uu^3 operatornamesinc^4(u/n) , mathrmd u xrightarrow[ntoinfty]textDCT int_0^infty fracsin^4 uu^3 , mathrmd u, $$ where the convergence follows from the dominated convergence together with the integrable dominating function $Cu^-3sin^4 u$ for some constant $C > 0$.
$endgroup$
– Sangchul Lee
Mar 27 at 11:48