Deriving asymptotic expansion of $int_{-1}^{1} e^{ilambda(frac{x^3}{3}+x)} dx$Computing the sum $sum...
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Deriving asymptotic expansion of $int_{-1}^{1} e^{ilambda(frac{x^3}{3}+x)} dx$
Computing the sum $sum frac{1}{n (2n-1)}$The Asymptotic Expansion of The Exponential IntegralEvaluate $int _0^{infty}dlambda left(lambda ^2 + 2blambda + cright)^{-frac{epsilon}{2}}$Asymptotic Expansion for a Function involving a Weird IntegralAsymptotic expansion of integrals and solving using integration by parts.Asymptotic expansion of $int_2^x frac{t}{log t}dt$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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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:
begin{align*}int frac{1}{(x^2+1) cdot ilambda} cdot frac{d}{dx}e^{ilambda(frac{x^3}{3}+x)} dx &= frac{e^{ilambda(frac{x^3}{3}+x)}}{(x^2+1) cdot ilambda}biggr{|}_{-1}^{1} +frac{1}{ilambda} int frac{2x}{(x^2+1)^2} cdot e^{ilambda(frac{x^3}{3}+x)} dx\
&=frac{1}{lambda}sinleft(frac{4lambda}{3}right)+R(lambda)end{align*}
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 $|frac{2x}{(x^2+1)^2}|<1$ on the domain, this gives the less useful
$$|R(lambda)|leq frac{2}{lambda} $$
integration analysis numerical-methods asymptotics
edited Mar 9 at 17:12
Delta-u
5,6152720
asked Mar 9 at 16:26
D.DogD.Dog
207
$endgroup$
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:
begin{align*}int frac{1}{(x^2+1) cdot ilambda} cdot frac{d}{dx}e^{ilambda(frac{x^3}{3}+x)} dx &= frac{e^{ilambda(frac{x^3}{3}+x)}}{(x^2+1) cdot ilambda}biggr{|}_{-1}^{1} +frac{1}{ilambda} int frac{2x}{(x^2+1)^2} cdot e^{ilambda(frac{x^3}{3}+x)} dx\
&=frac{1}{lambda}sinleft(frac{4lambda}{3}right)+R(lambda)end{align*}
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 $|frac{2x}{(x^2+1)^2}|<1$ on the domain, this gives the less useful
$$|R(lambda)|leq frac{2}{lambda} $$
integration analysis numerical-methods asymptotics
edited Mar 9 at 17:12
Delta-u
5,6152720
asked Mar 9 at 16:26
D.DogD.Dog
207
$endgroup$
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:
begin{align*}int frac{1}{(x^2+1) cdot ilambda} cdot frac{d}{dx}e^{ilambda(frac{x^3}{3}+x)} dx &= frac{e^{ilambda(frac{x^3}{3}+x)}}{(x^2+1) cdot ilambda}biggr{|}_{-1}^{1} +frac{1}{ilambda} int frac{2x}{(x^2+1)^2} cdot e^{ilambda(frac{x^3}{3}+x)} dx\
&=frac{1}{lambda}sinleft(frac{4lambda}{3}right)+R(lambda)end{align*}
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 $|frac{2x}{(x^2+1)^2}|<1$ on the domain, this gives the less useful
$$|R(lambda)|leq frac{2}{lambda} $$
integration analysis numerical-methods asymptotics
edited Mar 9 at 17:12
Delta-u
5,6152720
asked Mar 9 at 16:26
D.DogD.Dog
207
$endgroup$
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:
begin{align*}int frac{1}{(x^2+1) cdot ilambda} cdot frac{d}{dx}e^{ilambda(frac{x^3}{3}+x)} dx &= frac{e^{ilambda(frac{x^3}{3}+x)}}{(x^2+1) cdot ilambda}biggr{|}_{-1}^{1} +frac{1}{ilambda} int frac{2x}{(x^2+1)^2} cdot e^{ilambda(frac{x^3}{3}+x)} dx\
&=frac{1}{lambda}sinleft(frac{4lambda}{3}right)+R(lambda)end{align*}
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 $|frac{2x}{(x^2+1)^2}|<1$ on the domain, this gives the less useful
$$|R(lambda)|leq frac{2}{lambda} $$
integration analysis numerical-methods asymptotics
integration analysis numerical-methods asymptotics
edited Mar 9 at 17:12
Delta-u
5,6152720
asked Mar 9 at 16:26
D.DogD.Dog
207
edited Mar 9 at 17:12
Delta-u
5,6152720
asked Mar 9 at 16:26
D.DogD.Dog
207
edited Mar 9 at 17:12
Delta-u
5,6152720
edited Mar 9 at 17:12
Delta-u
5,6152720
edited Mar 9 at 17:12
Delta-u
5,6152720
5,6152720
asked Mar 9 at 16:26
D.DogD.Dog
207
asked Mar 9 at 16:26
D.DogD.Dog
207
asked Mar 9 at 16:26
D.DogD.Dog
207
207
add a comment |
add a comment |
2 Answers
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A trick is to integrate by parts once more
$$R(lambda)=frac{1}{i lambda} int_{-1}^1 frac{2x}{(x^2+1)^2} e^{i lambda (x^3/3+x)}dx=frac{1}{i lambda} int_{-1}^1 frac{1}{i lambda}frac{2x}{(x^2+1)^3} frac{d}{dx}e^{i lambda (x^3/3+x)}dx$$
so
$$R(lambda)= -frac{1}{lambda^2} left.frac{2x}{(x^2+1)^3} e^{i lambda (x^3/3+x)} right|_{-1}^1+frac{1}{lambda}^2 int_{-1}^1 frac{d}{dx} frac{2x}{(x^2+1)^3} e^{i lambda (x^3/3+x)}dx=O(lambda^{-2}).$$
answered Mar 9 at 17:10
Delta-uDelta-u
5,6152720
$endgroup$
$begingroup$
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
$$int_{-1}^{1} cos{(frac{kx^3}{3}+kx)} dx$$
answered Mar 9 at 16:54
Peter ForemanPeter Foreman
3,7861216
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
A trick is to integrate by parts once more
$$R(lambda)=frac{1}{i lambda} int_{-1}^1 frac{2x}{(x^2+1)^2} e^{i lambda (x^3/3+x)}dx=frac{1}{i lambda} int_{-1}^1 frac{1}{i lambda}frac{2x}{(x^2+1)^3} frac{d}{dx}e^{i lambda (x^3/3+x)}dx$$
so
$$R(lambda)= -frac{1}{lambda^2} left.frac{2x}{(x^2+1)^3} e^{i lambda (x^3/3+x)} right|_{-1}^1+frac{1}{lambda}^2 int_{-1}^1 frac{d}{dx} frac{2x}{(x^2+1)^3} e^{i lambda (x^3/3+x)}dx=O(lambda^{-2}).$$
answered Mar 9 at 17:10
Delta-uDelta-u
5,6152720
$endgroup$
$begingroup$
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
$$R(lambda)=frac{1}{i lambda} int_{-1}^1 frac{2x}{(x^2+1)^2} e^{i lambda (x^3/3+x)}dx=frac{1}{i lambda} int_{-1}^1 frac{1}{i lambda}frac{2x}{(x^2+1)^3} frac{d}{dx}e^{i lambda (x^3/3+x)}dx$$
so
$$R(lambda)= -frac{1}{lambda^2} left.frac{2x}{(x^2+1)^3} e^{i lambda (x^3/3+x)} right|_{-1}^1+frac{1}{lambda}^2 int_{-1}^1 frac{d}{dx} frac{2x}{(x^2+1)^3} e^{i lambda (x^3/3+x)}dx=O(lambda^{-2}).$$
answered Mar 9 at 17:10
Delta-uDelta-u
5,6152720
$endgroup$
$begingroup$
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
$$R(lambda)=frac{1}{i lambda} int_{-1}^1 frac{2x}{(x^2+1)^2} e^{i lambda (x^3/3+x)}dx=frac{1}{i lambda} int_{-1}^1 frac{1}{i lambda}frac{2x}{(x^2+1)^3} frac{d}{dx}e^{i lambda (x^3/3+x)}dx$$
so
$$R(lambda)= -frac{1}{lambda^2} left.frac{2x}{(x^2+1)^3} e^{i lambda (x^3/3+x)} right|_{-1}^1+frac{1}{lambda}^2 int_{-1}^1 frac{d}{dx} frac{2x}{(x^2+1)^3} e^{i lambda (x^3/3+x)}dx=O(lambda^{-2}).$$
answered Mar 9 at 17:10
Delta-uDelta-u
5,6152720
$endgroup$
A trick is to integrate by parts once more
$$R(lambda)=frac{1}{i lambda} int_{-1}^1 frac{2x}{(x^2+1)^2} e^{i lambda (x^3/3+x)}dx=frac{1}{i lambda} int_{-1}^1 frac{1}{i lambda}frac{2x}{(x^2+1)^3} frac{d}{dx}e^{i lambda (x^3/3+x)}dx$$
so
$$R(lambda)= -frac{1}{lambda^2} left.frac{2x}{(x^2+1)^3} e^{i lambda (x^3/3+x)} right|_{-1}^1+frac{1}{lambda}^2 int_{-1}^1 frac{d}{dx} frac{2x}{(x^2+1)^3} e^{i lambda (x^3/3+x)}dx=O(lambda^{-2}).$$
answered Mar 9 at 17:10
Delta-uDelta-u
5,6152720
answered Mar 9 at 17:10
Delta-uDelta-u
5,6152720
answered Mar 9 at 17:10
Delta-uDelta-u
5,6152720
answered Mar 9 at 17:10
Delta-uDelta-u
5,6152720
5,6152720
$begingroup$
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
$$int_{-1}^{1} cos{(frac{kx^3}{3}+kx)} dx$$
answered Mar 9 at 16:54
Peter ForemanPeter Foreman
3,7861216
$endgroup$
add a comment |
$begingroup$
Hint: This integral is equal to
$$int_{-1}^{1} cos{(frac{kx^3}{3}+kx)} dx$$
answered Mar 9 at 16:54
Peter ForemanPeter Foreman
3,7861216
$endgroup$
add a comment |
$begingroup$
Hint: This integral is equal to
$$int_{-1}^{1} cos{(frac{kx^3}{3}+kx)} dx$$
answered Mar 9 at 16:54
Peter ForemanPeter Foreman
3,7861216
$endgroup$
Hint: This integral is equal to
$$int_{-1}^{1} cos{(frac{kx^3}{3}+kx)} dx$$
answered Mar 9 at 16:54
Peter ForemanPeter Foreman
3,7861216
answered Mar 9 at 16:54
Peter ForemanPeter Foreman
3,7861216
answered Mar 9 at 16:54
Peter ForemanPeter Foreman
3,7861216
answered Mar 9 at 16:54
Peter ForemanPeter Foreman
3,7861216
3,7861216
add a comment |
add a comment |
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