Deriving asymptotic expansion of $int_-1^1 e^ilambda(fracx^33+x) dx$ The Next CEO of Stack OverflowComputing the sum $sum frac1n (2n-1)$The Asymptotic Expansion of The Exponential IntegralEvaluate $int _0^inftydlambda left(lambda ^2 + 2blambda + cright)^-fracepsilon2$Asymptotic Expansion for a Function involving a Weird IntegralAsymptotic expansion of integrals and solving using integration by parts.Asymptotic expansion of $int_2^x fractlog tdt$Integrating ExpressionUniform Convergence of a Asymptotic Series (Asymptotic Expansion of Integrals)Using asymptotic expansion of integralWhy doesn't this work for integrating $x^2e^-ax^2$ by parts?
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Deriving asymptotic expansion of $int_-1^1 e^ilambda(fracx^33+x) dx$
The Next CEO of Stack OverflowComputing the sum $sum frac1n (2n-1)$The Asymptotic Expansion of The Exponential IntegralEvaluate $int _0^inftydlambda left(lambda ^2 + 2blambda + cright)^-fracepsilon2$Asymptotic Expansion for a Function involving a Weird IntegralAsymptotic expansion of integrals and solving using integration by parts.Asymptotic expansion of $int_2^x fractlog tdt$Integrating ExpressionUniform Convergence of a Asymptotic Series (Asymptotic Expansion of Integrals)Using asymptotic expansion of integralWhy doesn't this work for integrating $x^2e^-ax^2$ by parts?
$begingroup$
I'm attempting to use integration by parts to find the expansion for the above integral. I'm able to derive the correct sequence, however I'm not sure how to demonstrate that the error bound is $O(lambda ^-n)$. My working is as follows:
beginalign*int frac1(x^2+1) cdot ilambda cdot fracddxe^ilambda(fracx^33+x) dx &= frace^ilambda(fracx^33+x)(x^2+1) cdot ilambdabiggr_-1^1 +frac1ilambda int frac2x(x^2+1)^2 cdot e^ilambda(fracx^33+x) dx\
&=frac1lambdasinleft(frac4lambda3right)+R(lambda)endalign*
I'd like to show that $R(lambda)$ is of order $O(lambda^-2)$. I thought to just take the supremum of both functions over the range, but noting that $|frac2x(x^2+1)^2|<1$ on the domain, this gives the less useful
$$|R(lambda)|leq frac2lambda $$
integration analysis numerical-methods asymptotics
edited Mar 9 at 17:12
Delta-u
5,6302720
asked Mar 9 at 16:26
D.DogD.Dog
229
$endgroup$
add a comment |
$begingroup$
I'm attempting to use integration by parts to find the expansion for the above integral. I'm able to derive the correct sequence, however I'm not sure how to demonstrate that the error bound is $O(lambda ^-n)$. My working is as follows:
beginalign*int frac1(x^2+1) cdot ilambda cdot fracddxe^ilambda(fracx^33+x) dx &= frace^ilambda(fracx^33+x)(x^2+1) cdot ilambdabiggr_-1^1 +frac1ilambda int frac2x(x^2+1)^2 cdot e^ilambda(fracx^33+x) dx\
&=frac1lambdasinleft(frac4lambda3right)+R(lambda)endalign*
I'd like to show that $R(lambda)$ is of order $O(lambda^-2)$. I thought to just take the supremum of both functions over the range, but noting that $|frac2x(x^2+1)^2|<1$ on the domain, this gives the less useful
$$|R(lambda)|leq frac2lambda $$
integration analysis numerical-methods asymptotics
edited Mar 9 at 17:12
Delta-u
5,6302720
asked Mar 9 at 16:26
D.DogD.Dog
229
$endgroup$
add a comment |
$begingroup$
I'm attempting to use integration by parts to find the expansion for the above integral. I'm able to derive the correct sequence, however I'm not sure how to demonstrate that the error bound is $O(lambda ^-n)$. My working is as follows:
beginalign*int frac1(x^2+1) cdot ilambda cdot fracddxe^ilambda(fracx^33+x) dx &= frace^ilambda(fracx^33+x)(x^2+1) cdot ilambdabiggr_-1^1 +frac1ilambda int frac2x(x^2+1)^2 cdot e^ilambda(fracx^33+x) dx\
&=frac1lambdasinleft(frac4lambda3right)+R(lambda)endalign*
I'd like to show that $R(lambda)$ is of order $O(lambda^-2)$. I thought to just take the supremum of both functions over the range, but noting that $|frac2x(x^2+1)^2|<1$ on the domain, this gives the less useful
$$|R(lambda)|leq frac2lambda $$
integration analysis numerical-methods asymptotics
edited Mar 9 at 17:12
Delta-u
5,6302720
asked Mar 9 at 16:26
D.DogD.Dog
229
$endgroup$
I'm attempting to use integration by parts to find the expansion for the above integral. I'm able to derive the correct sequence, however I'm not sure how to demonstrate that the error bound is $O(lambda ^-n)$. My working is as follows:
beginalign*int frac1(x^2+1) cdot ilambda cdot fracddxe^ilambda(fracx^33+x) dx &= frace^ilambda(fracx^33+x)(x^2+1) cdot ilambdabiggr_-1^1 +frac1ilambda int frac2x(x^2+1)^2 cdot e^ilambda(fracx^33+x) dx\
&=frac1lambdasinleft(frac4lambda3right)+R(lambda)endalign*
I'd like to show that $R(lambda)$ is of order $O(lambda^-2)$. I thought to just take the supremum of both functions over the range, but noting that $|frac2x(x^2+1)^2|<1$ on the domain, this gives the less useful
$$|R(lambda)|leq frac2lambda $$
integration analysis numerical-methods asymptotics
integration analysis numerical-methods asymptotics
edited Mar 9 at 17:12
Delta-u
5,6302720
asked Mar 9 at 16:26
D.DogD.Dog
229
edited Mar 9 at 17:12
Delta-u
5,6302720
asked Mar 9 at 16:26
D.DogD.Dog
229
edited Mar 9 at 17:12
Delta-u
5,6302720
edited Mar 9 at 17:12
Delta-u
5,6302720
edited Mar 9 at 17:12
Delta-u
5,6302720
5,6302720
asked Mar 9 at 16:26
D.DogD.Dog
229
asked Mar 9 at 16:26
D.DogD.Dog
229
asked Mar 9 at 16:26
D.DogD.Dog
229
229
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
A trick is to integrate by parts once more
$$R(lambda)=frac1i lambda int_-1^1 frac2x(x^2+1)^2 e^i lambda (x^3/3+x)dx=frac1i lambda int_-1^1 frac1i lambdafrac2x(x^2+1)^3 fracddxe^i lambda (x^3/3+x)dx$$
so
$$R(lambda)= -frac1lambda^2 left.frac2x(x^2+1)^3 e^i lambda (x^3/3+x) right|_-1^1+frac1lambda^2 int_-1^1 fracddx frac2x(x^2+1)^3 e^i lambda (x^3/3+x)dx=O(lambda^-2).$$
answered Mar 9 at 17:10
Delta-uDelta-u
5,6302720
$endgroup$
$begingroup$
Thanks for the explanation. I had previously put it in this form as I was looking for the second term in the expansion and didn't think to do it again to get the bound of O(lambda^-3). Very helpful.
$endgroup$
– D.Dog
Mar 9 at 17:28
add a comment |
$begingroup$
Hint: This integral is equal to
$$int_-1^1 cos(frackx^33+kx) dx$$
answered Mar 9 at 16:54
Peter ForemanPeter Foreman
5,1941216
$endgroup$
add a comment |
$begingroup$
rewrite your integral to $int_-1^1[coslambdaint(x^2+1)dx+isinlambdaint(x^2+1)dx]dxsimsinlambdaint(x^2+1)dxvert_-1^1-isinlambdaint(x^2+1)dxvert_-1^1$
therefore, the error's order is $2lambda$, then use the condition of asymptotic exapnasion to assume $O(f(x)/g(x))=1/x^2sim1$. now you can multiple $O(2lambda)$ with $O(lambda^-2)$ to get $|R(lambda)|leq frac2lambda $. thank you!
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A trick is to integrate by parts once more
$$R(lambda)=frac1i lambda int_-1^1 frac2x(x^2+1)^2 e^i lambda (x^3/3+x)dx=frac1i lambda int_-1^1 frac1i lambdafrac2x(x^2+1)^3 fracddxe^i lambda (x^3/3+x)dx$$
so
$$R(lambda)= -frac1lambda^2 left.frac2x(x^2+1)^3 e^i lambda (x^3/3+x) right|_-1^1+frac1lambda^2 int_-1^1 fracddx frac2x(x^2+1)^3 e^i lambda (x^3/3+x)dx=O(lambda^-2).$$
answered Mar 9 at 17:10
Delta-uDelta-u
5,6302720
$endgroup$
$begingroup$
Thanks for the explanation. I had previously put it in this form as I was looking for the second term in the expansion and didn't think to do it again to get the bound of O(lambda^-3). Very helpful.
$endgroup$
– D.Dog
Mar 9 at 17:28
add a comment |
$begingroup$
A trick is to integrate by parts once more
$$R(lambda)=frac1i lambda int_-1^1 frac2x(x^2+1)^2 e^i lambda (x^3/3+x)dx=frac1i lambda int_-1^1 frac1i lambdafrac2x(x^2+1)^3 fracddxe^i lambda (x^3/3+x)dx$$
so
$$R(lambda)= -frac1lambda^2 left.frac2x(x^2+1)^3 e^i lambda (x^3/3+x) right|_-1^1+frac1lambda^2 int_-1^1 fracddx frac2x(x^2+1)^3 e^i lambda (x^3/3+x)dx=O(lambda^-2).$$
answered Mar 9 at 17:10
Delta-uDelta-u
5,6302720
$endgroup$
$begingroup$
Thanks for the explanation. I had previously put it in this form as I was looking for the second term in the expansion and didn't think to do it again to get the bound of O(lambda^-3). Very helpful.
$endgroup$
– D.Dog
Mar 9 at 17:28
add a comment |
$begingroup$
A trick is to integrate by parts once more
$$R(lambda)=frac1i lambda int_-1^1 frac2x(x^2+1)^2 e^i lambda (x^3/3+x)dx=frac1i lambda int_-1^1 frac1i lambdafrac2x(x^2+1)^3 fracddxe^i lambda (x^3/3+x)dx$$
so
$$R(lambda)= -frac1lambda^2 left.frac2x(x^2+1)^3 e^i lambda (x^3/3+x) right|_-1^1+frac1lambda^2 int_-1^1 fracddx frac2x(x^2+1)^3 e^i lambda (x^3/3+x)dx=O(lambda^-2).$$
answered Mar 9 at 17:10
Delta-uDelta-u
5,6302720
$endgroup$
A trick is to integrate by parts once more
$$R(lambda)=frac1i lambda int_-1^1 frac2x(x^2+1)^2 e^i lambda (x^3/3+x)dx=frac1i lambda int_-1^1 frac1i lambdafrac2x(x^2+1)^3 fracddxe^i lambda (x^3/3+x)dx$$
so
$$R(lambda)= -frac1lambda^2 left.frac2x(x^2+1)^3 e^i lambda (x^3/3+x) right|_-1^1+frac1lambda^2 int_-1^1 fracddx frac2x(x^2+1)^3 e^i lambda (x^3/3+x)dx=O(lambda^-2).$$
answered Mar 9 at 17:10
Delta-uDelta-u
5,6302720
answered Mar 9 at 17:10
Delta-uDelta-u
5,6302720
answered Mar 9 at 17:10
Delta-uDelta-u
5,6302720
answered Mar 9 at 17:10
Delta-uDelta-u
5,6302720
5,6302720
$begingroup$
Thanks for the explanation. I had previously put it in this form as I was looking for the second term in the expansion and didn't think to do it again to get the bound of O(lambda^-3). Very helpful.
$endgroup$
– D.Dog
Mar 9 at 17:28
add a comment |
$begingroup$
Thanks for the explanation. I had previously put it in this form as I was looking for the second term in the expansion and didn't think to do it again to get the bound of O(lambda^-3). Very helpful.
$endgroup$
– D.Dog
Mar 9 at 17:28
$begingroup$
Thanks for the explanation. I had previously put it in this form as I was looking for the second term in the expansion and didn't think to do it again to get the bound of O(lambda^-3). Very helpful.
$endgroup$
– D.Dog
Mar 9 at 17:28
$begingroup$
Thanks for the explanation. I had previously put it in this form as I was looking for the second term in the expansion and didn't think to do it again to get the bound of O(lambda^-3). Very helpful.
$endgroup$
– D.Dog
Mar 9 at 17:28
add a comment |
$begingroup$
Hint: This integral is equal to
$$int_-1^1 cos(frackx^33+kx) dx$$
answered Mar 9 at 16:54
Peter ForemanPeter Foreman
5,1941216
$endgroup$
add a comment |
$begingroup$
Hint: This integral is equal to
$$int_-1^1 cos(frackx^33+kx) dx$$
answered Mar 9 at 16:54
Peter ForemanPeter Foreman
5,1941216
$endgroup$
add a comment |
$begingroup$
Hint: This integral is equal to
$$int_-1^1 cos(frackx^33+kx) dx$$
answered Mar 9 at 16:54
Peter ForemanPeter Foreman
5,1941216
$endgroup$
Hint: This integral is equal to
$$int_-1^1 cos(frackx^33+kx) dx$$
answered Mar 9 at 16:54
Peter ForemanPeter Foreman
5,1941216
answered Mar 9 at 16:54
Peter ForemanPeter Foreman
5,1941216
answered Mar 9 at 16:54
Peter ForemanPeter Foreman
5,1941216
answered Mar 9 at 16:54
Peter ForemanPeter Foreman
5,1941216
5,1941216
add a comment |
add a comment |
$begingroup$
rewrite your integral to $int_-1^1[coslambdaint(x^2+1)dx+isinlambdaint(x^2+1)dx]dxsimsinlambdaint(x^2+1)dxvert_-1^1-isinlambdaint(x^2+1)dxvert_-1^1$
therefore, the error's order is $2lambda$, then use the condition of asymptotic exapnasion to assume $O(f(x)/g(x))=1/x^2sim1$. now you can multiple $O(2lambda)$ with $O(lambda^-2)$ to get $|R(lambda)|leq frac2lambda $. thank you!
$endgroup$
add a comment |
$begingroup$
rewrite your integral to $int_-1^1[coslambdaint(x^2+1)dx+isinlambdaint(x^2+1)dx]dxsimsinlambdaint(x^2+1)dxvert_-1^1-isinlambdaint(x^2+1)dxvert_-1^1$
therefore, the error's order is $2lambda$, then use the condition of asymptotic exapnasion to assume $O(f(x)/g(x))=1/x^2sim1$. now you can multiple $O(2lambda)$ with $O(lambda^-2)$ to get $|R(lambda)|leq frac2lambda $. thank you!
$endgroup$
add a comment |
$begingroup$
rewrite your integral to $int_-1^1[coslambdaint(x^2+1)dx+isinlambdaint(x^2+1)dx]dxsimsinlambdaint(x^2+1)dxvert_-1^1-isinlambdaint(x^2+1)dxvert_-1^1$
therefore, the error's order is $2lambda$, then use the condition of asymptotic exapnasion to assume $O(f(x)/g(x))=1/x^2sim1$. now you can multiple $O(2lambda)$ with $O(lambda^-2)$ to get $|R(lambda)|leq frac2lambda $. thank you!
$endgroup$
rewrite your integral to $int_-1^1[coslambdaint(x^2+1)dx+isinlambdaint(x^2+1)dx]dxsimsinlambdaint(x^2+1)dxvert_-1^1-isinlambdaint(x^2+1)dxvert_-1^1$
therefore, the error's order is $2lambda$, then use the condition of asymptotic exapnasion to assume $O(f(x)/g(x))=1/x^2sim1$. now you can multiple $O(2lambda)$ with $O(lambda^-2)$ to get $|R(lambda)|leq frac2lambda $. thank you!
answered Mar 28 at 2:41
user653679
add a comment |
add a comment |
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