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Computing an infinite limit involving a double integral
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$$lim_Tto inftyfracint_0^Tcos^2(s) exp(-s)int_0^s cos(cos a)exp(a),mathrmda,mathrmdsT$$
I tried computing this limit in maple but got the answer: 'undefined'.
How to compute this limit, I accept special functions.
limits special-functions
edited Aug 2 at 15:41
Robert Howard
1,269620
asked Aug 2 at 13:26
MathematicalPhysicist
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up vote
3
down vote
favorite
$$lim_Tto inftyfracint_0^Tcos^2(s) exp(-s)int_0^s cos(cos a)exp(a),mathrmda,mathrmdsT$$
I tried computing this limit in maple but got the answer: 'undefined'.
How to compute this limit, I accept special functions.
limits special-functions
edited Aug 2 at 15:41
Robert Howard
1,269620
asked Aug 2 at 13:26
MathematicalPhysicist
1,1751924
1
Have you tried l'Hospitals rule?
– Rumpelstiltskin
Aug 2 at 13:30
@Adam no i haven't since I am not sure that the numerator converges to $infty$.
– MathematicalPhysicist
Aug 2 at 13:37
Have you tried computing the integral over $a$ first?
– Andrei
Aug 2 at 14:21
@Andrei I have: $int_0^s cos(cos a) exp(a)da = int_0^s (exp(-icos a+a)+exp(icos a+a))/2da$, but how to proceed from there?
– MathematicalPhysicist
Aug 2 at 14:45
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$$lim_Tto inftyfracint_0^Tcos^2(s) exp(-s)int_0^s cos(cos a)exp(a),mathrmda,mathrmdsT$$
I tried computing this limit in maple but got the answer: 'undefined'.
How to compute this limit, I accept special functions.
limits special-functions
edited Aug 2 at 15:41
Robert Howard
1,269620
asked Aug 2 at 13:26
MathematicalPhysicist
1,1751924
$$lim_Tto inftyfracint_0^Tcos^2(s) exp(-s)int_0^s cos(cos a)exp(a),mathrmda,mathrmdsT$$
I tried computing this limit in maple but got the answer: 'undefined'.
How to compute this limit, I accept special functions.
limits special-functions
edited Aug 2 at 15:41
Robert Howard
1,269620
asked Aug 2 at 13:26
MathematicalPhysicist
1,1751924
edited Aug 2 at 15:41
Robert Howard
1,269620
edited Aug 2 at 15:41
Robert Howard
1,269620
edited Aug 2 at 15:41
Robert Howard
1,269620
1,269620
asked Aug 2 at 13:26
MathematicalPhysicist
1,1751924
asked Aug 2 at 13:26
MathematicalPhysicist
1,1751924
asked Aug 2 at 13:26
MathematicalPhysicist
1,1751924
1,1751924
1
Have you tried l'Hospitals rule?
– Rumpelstiltskin
Aug 2 at 13:30
@Adam no i haven't since I am not sure that the numerator converges to $infty$.
– MathematicalPhysicist
Aug 2 at 13:37
Have you tried computing the integral over $a$ first?
– Andrei
Aug 2 at 14:21
@Andrei I have: $int_0^s cos(cos a) exp(a)da = int_0^s (exp(-icos a+a)+exp(icos a+a))/2da$, but how to proceed from there?
– MathematicalPhysicist
Aug 2 at 14:45
add a comment |Â
1
Have you tried l'Hospitals rule?
– Rumpelstiltskin
Aug 2 at 13:30
@Adam no i haven't since I am not sure that the numerator converges to $infty$.
– MathematicalPhysicist
Aug 2 at 13:37
Have you tried computing the integral over $a$ first?
– Andrei
Aug 2 at 14:21
@Andrei I have: $int_0^s cos(cos a) exp(a)da = int_0^s (exp(-icos a+a)+exp(icos a+a))/2da$, but how to proceed from there?
– MathematicalPhysicist
Aug 2 at 14:45
1
1
Have you tried l'Hospitals rule?
– Rumpelstiltskin
Aug 2 at 13:30
Have you tried l'Hospitals rule?
– Rumpelstiltskin
Aug 2 at 13:30
@Adam no i haven't since I am not sure that the numerator converges to $infty$.
– MathematicalPhysicist
Aug 2 at 13:37
@Adam no i haven't since I am not sure that the numerator converges to $infty$.
– MathematicalPhysicist
Aug 2 at 13:37
Have you tried computing the integral over $a$ first?
– Andrei
Aug 2 at 14:21
Have you tried computing the integral over $a$ first?
– Andrei
Aug 2 at 14:21
@Andrei I have: $int_0^s cos(cos a) exp(a)da = int_0^s (exp(-icos a+a)+exp(icos a+a))/2da$, but how to proceed from there?
– MathematicalPhysicist
Aug 2 at 14:45
@Andrei I have: $int_0^s cos(cos a) exp(a)da = int_0^s (exp(-icos a+a)+exp(icos a+a))/2da$, but how to proceed from there?
– MathematicalPhysicist
Aug 2 at 14:45
add a comment |Â
2 Answers
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up vote
1
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accepted
The limit exists and is finite with the value
$$lim_Ttoinfty J(T)/T = frac12pi int_0^picos(cos(x))dx = 0.38259884...$$
Here,
$$J(T):=int_0^T , ds cos^2(s)e^-sint_0^s , dx cos(cos(x))e^x. $$
The proof consists of an integration by parts to get a 'nice' expression, then bounding all the terms except for one remaining, and dealing with it specially.
The integration by parts shows that
$$ J(T) = Big( int_0^T , dx cos(cos(x))e^x Big)Big(frac-110e^-T
big(5+cos(2T)-2sin(2T)big) Big)+$$
$$+frac110int_0^T e^-sbig(5+cos(2s)-2sin(2s)big)cos(cos(s))e^s ds .$$
Let's deal with the first line of the previous question; call it $J_1.$ The second line is $J_2.$
$$J_1 = int_0^T , dx f(x,T)e^x-T , , , f(x,T)=frac-110cos(cos(x))big(5+cos(2T)-2sin(2T)big).$$ It's easy to see that $|f(x,T)|<1$ and thus $|J_1|<1-e^-T.$ In dividing by $T$ and taking the limit, it is obvious that $|J_1|/T to 0.$ Note that in $J_2$ I've written it as would come out of an integration by parts but the important thing is that the exponential growth and decay within the integrand cancel. Furthermore define
$$J_2^a:= int_0^T dxcos(cos(x)) , , , J_2^b:= int_0^T dx , cos(2x)cos(cos(x))$$
$$J_2^c:= int_0^T dx , sin(2x)cos(cos(x))
= 2 int_0^T dx ,cos(x) sin(x)cos(cos(x)).$$
$J_2^c$ is easily explicitly integrated and can be bounded by a constant. Thus $J_2^c/T to 0$ as $T to infty.$ For $J_2^b,$
$$fracJ_2^bT = int_0^1 dx , cos(2x,T)cos(cos(x ,T)) to 0$$
by the Riemann-Lebesque lemma. For $J_2^a$ notice that the integrand is periodic of period $pi.$ Assume $T=n,pi + b$ with $0<b<pi.$ All continuous $T$ is covered because of the extra $b,$ even as we let $n to infty$ through discrete values.
$$fracJ_2^bT=frac1npi+b Big( int_0^pi(*) +int_pi^2pi(*)
+ ...int_(n-1)pi^npi(*) + int_npi^npi+b(*) Big)$$
where the (*) stands for the integrand which is not shown for it obscures the argument I'm making. There are $n$ identical integrals by periodicty so
$$fracJ_2^bT=frac1npi+bBig( n, int_0^pi dxcos(cos(x)) + CBig).$$
The iterated cosine can be bounded by 1 so $|C|<pi,$ no matter what the value of $b$ is. On letting $n to infty$ and picking up the factor of 1/2 from the definition of $J_2,$ we get the first formula for this answer. As a numerical check I let T=1000 and used lots of precision to get a brute force evaluation of the J(1000)/1000 = 0.371.
answered Aug 2 at 19:54
skbmoore
96526
add a comment |Â
up vote
4
down vote
Here is some partial solution:
We want to check if the numerator goes to infinity first, so we can apply l'Hospital. We have $-1lecos ale1$, and $-pi/2<-1<1<pi/2$. We also know that the cosine function is positive in the above interval. You can therefore show that $$cos(1)le cos(cos s)le 1$$
The value for $cos(1)$ is about $0.54$. Then we have $$int_0^scos(1)e^a dale int_0^scos(cos a)e^a daleint_0^se^a da$$or $$cos(1)(e^s-1)leint_0^scos(cos a)e^a dale e^s-1$$
The integral $int_0^Tcos^se^-sds$ is finite, so in the limit when $Trightarrowinfty$ the ratio of a constant divided by $T$ goes to zero. What we have is that $$lim_Trightarrow inftycos(1)fracint_0^Tcos^2(s )dsTle lim_Trightarrow inftyfracint_0^Tcos^2(s)e^-sint_0^scos(cos a)e^a da dsTlelim_Trightarrowinftyfracint_0^Tcos^2(s )dsT$$
You can compute now the integral, to get that $$int_0^Tcos^2(s )ds=fracT2+fracsin 2T4$$
You can therefore apply l'Hospitals rule.You have $fracddxint_0^xf(t)dt=f(x)$, so:
$$lim_Trightarrow inftyfracint_0^Tcos^2(s)e^-sint_0^scos(cos a)e^a da dsT=lim_Trightarrow inftycos^2(T)e^-Tint_0^Tcos(cos a)e^a da$$
You can rewrite the last limit as
$$lim_Trightarrow inftyfracint_0^Tcos(cos a)e^a dafrace^Tcos^2(T)$$
If you apply l'Hospital again, I get that the limit does not exist.
answered Aug 2 at 15:47
Andrei
7,3502822
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The limit exists and is finite with the value
$$lim_Ttoinfty J(T)/T = frac12pi int_0^picos(cos(x))dx = 0.38259884...$$
Here,
$$J(T):=int_0^T , ds cos^2(s)e^-sint_0^s , dx cos(cos(x))e^x. $$
The proof consists of an integration by parts to get a 'nice' expression, then bounding all the terms except for one remaining, and dealing with it specially.
The integration by parts shows that
$$ J(T) = Big( int_0^T , dx cos(cos(x))e^x Big)Big(frac-110e^-T
big(5+cos(2T)-2sin(2T)big) Big)+$$
$$+frac110int_0^T e^-sbig(5+cos(2s)-2sin(2s)big)cos(cos(s))e^s ds .$$
Let's deal with the first line of the previous question; call it $J_1.$ The second line is $J_2.$
$$J_1 = int_0^T , dx f(x,T)e^x-T , , , f(x,T)=frac-110cos(cos(x))big(5+cos(2T)-2sin(2T)big).$$ It's easy to see that $|f(x,T)|<1$ and thus $|J_1|<1-e^-T.$ In dividing by $T$ and taking the limit, it is obvious that $|J_1|/T to 0.$ Note that in $J_2$ I've written it as would come out of an integration by parts but the important thing is that the exponential growth and decay within the integrand cancel. Furthermore define
$$J_2^a:= int_0^T dxcos(cos(x)) , , , J_2^b:= int_0^T dx , cos(2x)cos(cos(x))$$
$$J_2^c:= int_0^T dx , sin(2x)cos(cos(x))
= 2 int_0^T dx ,cos(x) sin(x)cos(cos(x)).$$
$J_2^c$ is easily explicitly integrated and can be bounded by a constant. Thus $J_2^c/T to 0$ as $T to infty.$ For $J_2^b,$
$$fracJ_2^bT = int_0^1 dx , cos(2x,T)cos(cos(x ,T)) to 0$$
by the Riemann-Lebesque lemma. For $J_2^a$ notice that the integrand is periodic of period $pi.$ Assume $T=n,pi + b$ with $0<b<pi.$ All continuous $T$ is covered because of the extra $b,$ even as we let $n to infty$ through discrete values.
$$fracJ_2^bT=frac1npi+b Big( int_0^pi(*) +int_pi^2pi(*)
+ ...int_(n-1)pi^npi(*) + int_npi^npi+b(*) Big)$$
where the (*) stands for the integrand which is not shown for it obscures the argument I'm making. There are $n$ identical integrals by periodicty so
$$fracJ_2^bT=frac1npi+bBig( n, int_0^pi dxcos(cos(x)) + CBig).$$
The iterated cosine can be bounded by 1 so $|C|<pi,$ no matter what the value of $b$ is. On letting $n to infty$ and picking up the factor of 1/2 from the definition of $J_2,$ we get the first formula for this answer. As a numerical check I let T=1000 and used lots of precision to get a brute force evaluation of the J(1000)/1000 = 0.371.
answered Aug 2 at 19:54
skbmoore
96526
add a comment |Â
up vote
1
down vote
accepted
The limit exists and is finite with the value
$$lim_Ttoinfty J(T)/T = frac12pi int_0^picos(cos(x))dx = 0.38259884...$$
Here,
$$J(T):=int_0^T , ds cos^2(s)e^-sint_0^s , dx cos(cos(x))e^x. $$
The proof consists of an integration by parts to get a 'nice' expression, then bounding all the terms except for one remaining, and dealing with it specially.
The integration by parts shows that
$$ J(T) = Big( int_0^T , dx cos(cos(x))e^x Big)Big(frac-110e^-T
big(5+cos(2T)-2sin(2T)big) Big)+$$
$$+frac110int_0^T e^-sbig(5+cos(2s)-2sin(2s)big)cos(cos(s))e^s ds .$$
Let's deal with the first line of the previous question; call it $J_1.$ The second line is $J_2.$
$$J_1 = int_0^T , dx f(x,T)e^x-T , , , f(x,T)=frac-110cos(cos(x))big(5+cos(2T)-2sin(2T)big).$$ It's easy to see that $|f(x,T)|<1$ and thus $|J_1|<1-e^-T.$ In dividing by $T$ and taking the limit, it is obvious that $|J_1|/T to 0.$ Note that in $J_2$ I've written it as would come out of an integration by parts but the important thing is that the exponential growth and decay within the integrand cancel. Furthermore define
$$J_2^a:= int_0^T dxcos(cos(x)) , , , J_2^b:= int_0^T dx , cos(2x)cos(cos(x))$$
$$J_2^c:= int_0^T dx , sin(2x)cos(cos(x))
= 2 int_0^T dx ,cos(x) sin(x)cos(cos(x)).$$
$J_2^c$ is easily explicitly integrated and can be bounded by a constant. Thus $J_2^c/T to 0$ as $T to infty.$ For $J_2^b,$
$$fracJ_2^bT = int_0^1 dx , cos(2x,T)cos(cos(x ,T)) to 0$$
by the Riemann-Lebesque lemma. For $J_2^a$ notice that the integrand is periodic of period $pi.$ Assume $T=n,pi + b$ with $0<b<pi.$ All continuous $T$ is covered because of the extra $b,$ even as we let $n to infty$ through discrete values.
$$fracJ_2^bT=frac1npi+b Big( int_0^pi(*) +int_pi^2pi(*)
+ ...int_(n-1)pi^npi(*) + int_npi^npi+b(*) Big)$$
where the (*) stands for the integrand which is not shown for it obscures the argument I'm making. There are $n$ identical integrals by periodicty so
$$fracJ_2^bT=frac1npi+bBig( n, int_0^pi dxcos(cos(x)) + CBig).$$
The iterated cosine can be bounded by 1 so $|C|<pi,$ no matter what the value of $b$ is. On letting $n to infty$ and picking up the factor of 1/2 from the definition of $J_2,$ we get the first formula for this answer. As a numerical check I let T=1000 and used lots of precision to get a brute force evaluation of the J(1000)/1000 = 0.371.
answered Aug 2 at 19:54
skbmoore
96526
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The limit exists and is finite with the value
$$lim_Ttoinfty J(T)/T = frac12pi int_0^picos(cos(x))dx = 0.38259884...$$
Here,
$$J(T):=int_0^T , ds cos^2(s)e^-sint_0^s , dx cos(cos(x))e^x. $$
The proof consists of an integration by parts to get a 'nice' expression, then bounding all the terms except for one remaining, and dealing with it specially.
The integration by parts shows that
$$ J(T) = Big( int_0^T , dx cos(cos(x))e^x Big)Big(frac-110e^-T
big(5+cos(2T)-2sin(2T)big) Big)+$$
$$+frac110int_0^T e^-sbig(5+cos(2s)-2sin(2s)big)cos(cos(s))e^s ds .$$
Let's deal with the first line of the previous question; call it $J_1.$ The second line is $J_2.$
$$J_1 = int_0^T , dx f(x,T)e^x-T , , , f(x,T)=frac-110cos(cos(x))big(5+cos(2T)-2sin(2T)big).$$ It's easy to see that $|f(x,T)|<1$ and thus $|J_1|<1-e^-T.$ In dividing by $T$ and taking the limit, it is obvious that $|J_1|/T to 0.$ Note that in $J_2$ I've written it as would come out of an integration by parts but the important thing is that the exponential growth and decay within the integrand cancel. Furthermore define
$$J_2^a:= int_0^T dxcos(cos(x)) , , , J_2^b:= int_0^T dx , cos(2x)cos(cos(x))$$
$$J_2^c:= int_0^T dx , sin(2x)cos(cos(x))
= 2 int_0^T dx ,cos(x) sin(x)cos(cos(x)).$$
$J_2^c$ is easily explicitly integrated and can be bounded by a constant. Thus $J_2^c/T to 0$ as $T to infty.$ For $J_2^b,$
$$fracJ_2^bT = int_0^1 dx , cos(2x,T)cos(cos(x ,T)) to 0$$
by the Riemann-Lebesque lemma. For $J_2^a$ notice that the integrand is periodic of period $pi.$ Assume $T=n,pi + b$ with $0<b<pi.$ All continuous $T$ is covered because of the extra $b,$ even as we let $n to infty$ through discrete values.
$$fracJ_2^bT=frac1npi+b Big( int_0^pi(*) +int_pi^2pi(*)
+ ...int_(n-1)pi^npi(*) + int_npi^npi+b(*) Big)$$
where the (*) stands for the integrand which is not shown for it obscures the argument I'm making. There are $n$ identical integrals by periodicty so
$$fracJ_2^bT=frac1npi+bBig( n, int_0^pi dxcos(cos(x)) + CBig).$$
The iterated cosine can be bounded by 1 so $|C|<pi,$ no matter what the value of $b$ is. On letting $n to infty$ and picking up the factor of 1/2 from the definition of $J_2,$ we get the first formula for this answer. As a numerical check I let T=1000 and used lots of precision to get a brute force evaluation of the J(1000)/1000 = 0.371.
answered Aug 2 at 19:54
skbmoore
96526
The limit exists and is finite with the value
$$lim_Ttoinfty J(T)/T = frac12pi int_0^picos(cos(x))dx = 0.38259884...$$
Here,
$$J(T):=int_0^T , ds cos^2(s)e^-sint_0^s , dx cos(cos(x))e^x. $$
The proof consists of an integration by parts to get a 'nice' expression, then bounding all the terms except for one remaining, and dealing with it specially.
The integration by parts shows that
$$ J(T) = Big( int_0^T , dx cos(cos(x))e^x Big)Big(frac-110e^-T
big(5+cos(2T)-2sin(2T)big) Big)+$$
$$+frac110int_0^T e^-sbig(5+cos(2s)-2sin(2s)big)cos(cos(s))e^s ds .$$
Let's deal with the first line of the previous question; call it $J_1.$ The second line is $J_2.$
$$J_1 = int_0^T , dx f(x,T)e^x-T , , , f(x,T)=frac-110cos(cos(x))big(5+cos(2T)-2sin(2T)big).$$ It's easy to see that $|f(x,T)|<1$ and thus $|J_1|<1-e^-T.$ In dividing by $T$ and taking the limit, it is obvious that $|J_1|/T to 0.$ Note that in $J_2$ I've written it as would come out of an integration by parts but the important thing is that the exponential growth and decay within the integrand cancel. Furthermore define
$$J_2^a:= int_0^T dxcos(cos(x)) , , , J_2^b:= int_0^T dx , cos(2x)cos(cos(x))$$
$$J_2^c:= int_0^T dx , sin(2x)cos(cos(x))
= 2 int_0^T dx ,cos(x) sin(x)cos(cos(x)).$$
$J_2^c$ is easily explicitly integrated and can be bounded by a constant. Thus $J_2^c/T to 0$ as $T to infty.$ For $J_2^b,$
$$fracJ_2^bT = int_0^1 dx , cos(2x,T)cos(cos(x ,T)) to 0$$
by the Riemann-Lebesque lemma. For $J_2^a$ notice that the integrand is periodic of period $pi.$ Assume $T=n,pi + b$ with $0<b<pi.$ All continuous $T$ is covered because of the extra $b,$ even as we let $n to infty$ through discrete values.
$$fracJ_2^bT=frac1npi+b Big( int_0^pi(*) +int_pi^2pi(*)
+ ...int_(n-1)pi^npi(*) + int_npi^npi+b(*) Big)$$
where the (*) stands for the integrand which is not shown for it obscures the argument I'm making. There are $n$ identical integrals by periodicty so
$$fracJ_2^bT=frac1npi+bBig( n, int_0^pi dxcos(cos(x)) + CBig).$$
The iterated cosine can be bounded by 1 so $|C|<pi,$ no matter what the value of $b$ is. On letting $n to infty$ and picking up the factor of 1/2 from the definition of $J_2,$ we get the first formula for this answer. As a numerical check I let T=1000 and used lots of precision to get a brute force evaluation of the J(1000)/1000 = 0.371.
answered Aug 2 at 19:54
skbmoore
96526
answered Aug 2 at 19:54
skbmoore
96526
answered Aug 2 at 19:54
skbmoore
96526
answered Aug 2 at 19:54
skbmoore
96526
96526
add a comment |Â
add a comment |Â
up vote
4
down vote
Here is some partial solution:
We want to check if the numerator goes to infinity first, so we can apply l'Hospital. We have $-1lecos ale1$, and $-pi/2<-1<1<pi/2$. We also know that the cosine function is positive in the above interval. You can therefore show that $$cos(1)le cos(cos s)le 1$$
The value for $cos(1)$ is about $0.54$. Then we have $$int_0^scos(1)e^a dale int_0^scos(cos a)e^a daleint_0^se^a da$$or $$cos(1)(e^s-1)leint_0^scos(cos a)e^a dale e^s-1$$
The integral $int_0^Tcos^se^-sds$ is finite, so in the limit when $Trightarrowinfty$ the ratio of a constant divided by $T$ goes to zero. What we have is that $$lim_Trightarrow inftycos(1)fracint_0^Tcos^2(s )dsTle lim_Trightarrow inftyfracint_0^Tcos^2(s)e^-sint_0^scos(cos a)e^a da dsTlelim_Trightarrowinftyfracint_0^Tcos^2(s )dsT$$
You can compute now the integral, to get that $$int_0^Tcos^2(s )ds=fracT2+fracsin 2T4$$
You can therefore apply l'Hospitals rule.You have $fracddxint_0^xf(t)dt=f(x)$, so:
$$lim_Trightarrow inftyfracint_0^Tcos^2(s)e^-sint_0^scos(cos a)e^a da dsT=lim_Trightarrow inftycos^2(T)e^-Tint_0^Tcos(cos a)e^a da$$
You can rewrite the last limit as
$$lim_Trightarrow inftyfracint_0^Tcos(cos a)e^a dafrace^Tcos^2(T)$$
If you apply l'Hospital again, I get that the limit does not exist.
answered Aug 2 at 15:47
Andrei
7,3502822
add a comment |Â
up vote
4
down vote
Here is some partial solution:
We want to check if the numerator goes to infinity first, so we can apply l'Hospital. We have $-1lecos ale1$, and $-pi/2<-1<1<pi/2$. We also know that the cosine function is positive in the above interval. You can therefore show that $$cos(1)le cos(cos s)le 1$$
The value for $cos(1)$ is about $0.54$. Then we have $$int_0^scos(1)e^a dale int_0^scos(cos a)e^a daleint_0^se^a da$$or $$cos(1)(e^s-1)leint_0^scos(cos a)e^a dale e^s-1$$
The integral $int_0^Tcos^se^-sds$ is finite, so in the limit when $Trightarrowinfty$ the ratio of a constant divided by $T$ goes to zero. What we have is that $$lim_Trightarrow inftycos(1)fracint_0^Tcos^2(s )dsTle lim_Trightarrow inftyfracint_0^Tcos^2(s)e^-sint_0^scos(cos a)e^a da dsTlelim_Trightarrowinftyfracint_0^Tcos^2(s )dsT$$
You can compute now the integral, to get that $$int_0^Tcos^2(s )ds=fracT2+fracsin 2T4$$
You can therefore apply l'Hospitals rule.You have $fracddxint_0^xf(t)dt=f(x)$, so:
$$lim_Trightarrow inftyfracint_0^Tcos^2(s)e^-sint_0^scos(cos a)e^a da dsT=lim_Trightarrow inftycos^2(T)e^-Tint_0^Tcos(cos a)e^a da$$
You can rewrite the last limit as
$$lim_Trightarrow inftyfracint_0^Tcos(cos a)e^a dafrace^Tcos^2(T)$$
If you apply l'Hospital again, I get that the limit does not exist.
answered Aug 2 at 15:47
Andrei
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Here is some partial solution:
We want to check if the numerator goes to infinity first, so we can apply l'Hospital. We have $-1lecos ale1$, and $-pi/2<-1<1<pi/2$. We also know that the cosine function is positive in the above interval. You can therefore show that $$cos(1)le cos(cos s)le 1$$
The value for $cos(1)$ is about $0.54$. Then we have $$int_0^scos(1)e^a dale int_0^scos(cos a)e^a daleint_0^se^a da$$or $$cos(1)(e^s-1)leint_0^scos(cos a)e^a dale e^s-1$$
The integral $int_0^Tcos^se^-sds$ is finite, so in the limit when $Trightarrowinfty$ the ratio of a constant divided by $T$ goes to zero. What we have is that $$lim_Trightarrow inftycos(1)fracint_0^Tcos^2(s )dsTle lim_Trightarrow inftyfracint_0^Tcos^2(s)e^-sint_0^scos(cos a)e^a da dsTlelim_Trightarrowinftyfracint_0^Tcos^2(s )dsT$$
You can compute now the integral, to get that $$int_0^Tcos^2(s )ds=fracT2+fracsin 2T4$$
You can therefore apply l'Hospitals rule.You have $fracddxint_0^xf(t)dt=f(x)$, so:
$$lim_Trightarrow inftyfracint_0^Tcos^2(s)e^-sint_0^scos(cos a)e^a da dsT=lim_Trightarrow inftycos^2(T)e^-Tint_0^Tcos(cos a)e^a da$$
You can rewrite the last limit as
$$lim_Trightarrow inftyfracint_0^Tcos(cos a)e^a dafrace^Tcos^2(T)$$
If you apply l'Hospital again, I get that the limit does not exist.
answered Aug 2 at 15:47
Andrei
7,3502822
Here is some partial solution:
We want to check if the numerator goes to infinity first, so we can apply l'Hospital. We have $-1lecos ale1$, and $-pi/2<-1<1<pi/2$. We also know that the cosine function is positive in the above interval. You can therefore show that $$cos(1)le cos(cos s)le 1$$
The value for $cos(1)$ is about $0.54$. Then we have $$int_0^scos(1)e^a dale int_0^scos(cos a)e^a daleint_0^se^a da$$or $$cos(1)(e^s-1)leint_0^scos(cos a)e^a dale e^s-1$$
The integral $int_0^Tcos^se^-sds$ is finite, so in the limit when $Trightarrowinfty$ the ratio of a constant divided by $T$ goes to zero. What we have is that $$lim_Trightarrow inftycos(1)fracint_0^Tcos^2(s )dsTle lim_Trightarrow inftyfracint_0^Tcos^2(s)e^-sint_0^scos(cos a)e^a da dsTlelim_Trightarrowinftyfracint_0^Tcos^2(s )dsT$$
You can compute now the integral, to get that $$int_0^Tcos^2(s )ds=fracT2+fracsin 2T4$$
You can therefore apply l'Hospitals rule.You have $fracddxint_0^xf(t)dt=f(x)$, so:
$$lim_Trightarrow inftyfracint_0^Tcos^2(s)e^-sint_0^scos(cos a)e^a da dsT=lim_Trightarrow inftycos^2(T)e^-Tint_0^Tcos(cos a)e^a da$$
You can rewrite the last limit as
$$lim_Trightarrow inftyfracint_0^Tcos(cos a)e^a dafrace^Tcos^2(T)$$
If you apply l'Hospital again, I get that the limit does not exist.
answered Aug 2 at 15:47
Andrei
7,3502822
answered Aug 2 at 15:47
Andrei
7,3502822
answered Aug 2 at 15:47
Andrei
7,3502822
answered Aug 2 at 15:47
Andrei
7,3502822
7,3502822
add a comment |Â
add a comment |Â
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1
Have you tried l'Hospitals rule?
– Rumpelstiltskin
Aug 2 at 13:30
@Adam no i haven't since I am not sure that the numerator converges to $infty$.
– MathematicalPhysicist
Aug 2 at 13:37
Have you tried computing the integral over $a$ first?
– Andrei
Aug 2 at 14:21
@Andrei I have: $int_0^s cos(cos a) exp(a)da = int_0^s (exp(-icos a+a)+exp(icos a+a))/2da$, but how to proceed from there?
– MathematicalPhysicist
Aug 2 at 14:45