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2

$begingroup$

This is problem 42c from chapter 19 from Spivak-Calculus 3rd edition

I need to prove that $$ lim_{x to pi}int_{0}^{pi}sin(x+frac{1}{2})tleft[dfrac{2}{t}-dfrac{1}{sindfrac{t}{2}}right] dt=0$$
So far i have established that the expression within the brackets is bounded on $[0,pi]$,
since $$lim_{t to 0^+}dfrac{2}{t}-dfrac{1}{sindfrac{t}{2}}=0$$which can be shown with a few applications of l´Hospitals rule.

Now the last part of the problem is what has stumped me, the hint in the book suggest that the Reimann-Lebesgue lemma: If $f$ is a integrable function on $[a,b]$ then $$ lim_{x to infty}int_{a}^{b}sin(xt)f(t)dt=0$$ can be used to prove the rest. But, I do not see how the limits can be made equal i.e how to transform $lim_{x to pi}$ into $lim_{u to infty}$.
Any tips or suggestions?

real-analysis

share|cite|improve this question

edited Mar 9 at 17:19

Rodrigo de Azevedo

13k41960

asked Mar 9 at 16:48

André ArmatowskiAndré Armatowski

111

New contributor

André Armatowski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It seems the integral is not $0$, maybe there is a typo ?
  $endgroup$
  – Delta-u
  Mar 9 at 16:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Seems like it, every other copy i could find had the same limit. Thank you for your answer!
  $endgroup$
  – André Armatowski
  Mar 9 at 17:01
  

  
  
  
  

add a comment | 

2

$begingroup$

This is problem 42c from chapter 19 from Spivak-Calculus 3rd edition

I need to prove that $$ lim_{x to pi}int_{0}^{pi}sin(x+frac{1}{2})tleft[dfrac{2}{t}-dfrac{1}{sindfrac{t}{2}}right] dt=0$$
So far i have established that the expression within the brackets is bounded on $[0,pi]$,
since $$lim_{t to 0^+}dfrac{2}{t}-dfrac{1}{sindfrac{t}{2}}=0$$which can be shown with a few applications of l´Hospitals rule.

Now the last part of the problem is what has stumped me, the hint in the book suggest that the Reimann-Lebesgue lemma: If $f$ is a integrable function on $[a,b]$ then $$ lim_{x to infty}int_{a}^{b}sin(xt)f(t)dt=0$$ can be used to prove the rest. But, I do not see how the limits can be made equal i.e how to transform $lim_{x to pi}$ into $lim_{u to infty}$.
Any tips or suggestions?

real-analysis

share|cite|improve this question

edited Mar 9 at 17:19

Rodrigo de Azevedo

13k41960

asked Mar 9 at 16:48

André ArmatowskiAndré Armatowski

111

New contributor

André Armatowski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It seems the integral is not $0$, maybe there is a typo ?
  $endgroup$
  – Delta-u
  Mar 9 at 16:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Seems like it, every other copy i could find had the same limit. Thank you for your answer!
  $endgroup$
  – André Armatowski
  Mar 9 at 17:01
  

  
  
  
  

add a comment | 

$begingroup$

This is problem 42c from chapter 19 from Spivak-Calculus 3rd edition

I need to prove that $$ lim_{x to pi}int_{0}^{pi}sin(x+frac{1}{2})tleft[dfrac{2}{t}-dfrac{1}{sindfrac{t}{2}}right] dt=0$$
So far i have established that the expression within the brackets is bounded on $[0,pi]$,
since $$lim_{t to 0^+}dfrac{2}{t}-dfrac{1}{sindfrac{t}{2}}=0$$which can be shown with a few applications of l´Hospitals rule.

Now the last part of the problem is what has stumped me, the hint in the book suggest that the Reimann-Lebesgue lemma: If $f$ is a integrable function on $[a,b]$ then $$ lim_{x to infty}int_{a}^{b}sin(xt)f(t)dt=0$$ can be used to prove the rest. But, I do not see how the limits can be made equal i.e how to transform $lim_{x to pi}$ into $lim_{u to infty}$.
Any tips or suggestions?

real-analysis

share|cite|improve this question

edited Mar 9 at 17:19

Rodrigo de Azevedo

13k41960

asked Mar 9 at 16:48

André ArmatowskiAndré Armatowski

111

New contributor

André Armatowski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 9 at 17:19

Rodrigo de Azevedo

13k41960

asked Mar 9 at 16:48

André ArmatowskiAndré Armatowski

111

New contributor

André Armatowski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 9 at 17:19

Rodrigo de Azevedo

13k41960

asked Mar 9 at 16:48

André ArmatowskiAndré Armatowski

111

New contributor

André Armatowski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 9 at 17:19

Rodrigo de Azevedo

13k41960

edited Mar 9 at 17:19

Rodrigo de Azevedo

13k41960

asked Mar 9 at 16:48

André ArmatowskiAndré Armatowski

111

New contributor

André Armatowski is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 9 at 16:48

André ArmatowskiAndré Armatowski

111