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Deriving asymptotic expansion of $int_-1^1 e^ilambda(fracx^33+x) dx$

1

$begingroup$

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

share|cite|improve this question

edited Mar 9 at 17:12

Delta-u

5,6302720

asked Mar 9 at 16:26

D.DogD.Dog

229

$endgroup$

add a comment | 

1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

edited Mar 9 at 17:12

Delta-u

5,6302720

asked Mar 9 at 16:26

D.DogD.Dog

229

$endgroup$

share|cite|improve this question

edited Mar 9 at 17:12

Delta-u

5,6302720

asked Mar 9 at 16:26

D.DogD.Dog

229

share|cite|improve this question

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

asked Mar 9 at 16:26

D.DogD.Dog

229

asked Mar 9 at 16:26

D.DogD.Dog

229

3 Answers
3

active

oldest

votes

1

$begingroup$

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
$$R(lambda)= -frac1lambda^2 left.frac2x(x^2+1)^3 e^i lambda (x^3/3+x) right|_-1^1+frac1lambda^2 int_-1^1 fracddx frac2x(x^2+1)^3 e^i lambda (x^3/3+x)dx=O(lambda^-2).$$

share|cite|improve this answer

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 | 

0

$begingroup$

Hint: This integral is equal to
$$int_-1^1 cos(frackx^33+kx) dx$$

share|cite|improve this answer

answered Mar 9 at 16:54

Peter ForemanPeter Foreman

5,1941216

$endgroup$

add a comment | 

0

$begingroup$

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!

share|cite|improve this answer

answered Mar 28 at 2:41

user653679

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add a comment | 

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1

$begingroup$

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
$$R(lambda)= -frac1lambda^2 left.frac2x(x^2+1)^3 e^i lambda (x^3/3+x) right|_-1^1+frac1lambda^2 int_-1^1 fracddx frac2x(x^2+1)^3 e^i lambda (x^3/3+x)dx=O(lambda^-2).$$

share|cite|improve this answer

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 | 

1

$begingroup$

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
$$R(lambda)= -frac1lambda^2 left.frac2x(x^2+1)^3 e^i lambda (x^3/3+x) right|_-1^1+frac1lambda^2 int_-1^1 fracddx frac2x(x^2+1)^3 e^i lambda (x^3/3+x)dx=O(lambda^-2).$$

share|cite|improve this answer

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
$$R(lambda)= -frac1lambda^2 left.frac2x(x^2+1)^3 e^i lambda (x^3/3+x) right|_-1^1+frac1lambda^2 int_-1^1 fracddx frac2x(x^2+1)^3 e^i lambda (x^3/3+x)dx=O(lambda^-2).$$

share|cite|improve this answer

answered Mar 9 at 17:10

Delta-uDelta-u

5,6302720

$endgroup$

share|cite|improve this answer

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

0

$begingroup$

Hint: This integral is equal to
$$int_-1^1 cos(frackx^33+kx) dx$$

share|cite|improve this answer

answered Mar 9 at 16:54

Peter ForemanPeter Foreman

5,1941216

$endgroup$

add a comment | 

0

$begingroup$

Hint: This integral is equal to
$$int_-1^1 cos(frackx^33+kx) dx$$

share|cite|improve this answer

answered Mar 9 at 16:54

Peter ForemanPeter Foreman

5,1941216

$endgroup$

add a comment | 

$begingroup$

Hint: This integral is equal to
$$int_-1^1 cos(frackx^33+kx) dx$$

share|cite|improve this answer

answered Mar 9 at 16:54

Peter ForemanPeter Foreman

5,1941216

$endgroup$

share|cite|improve this answer

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