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Calculating limit of two integrals

Compute $lim_sto 0 left(int_0^1 (Gamma (x))^sspacemathrmdxright)^1/s$Uniform Convergence of $sum_n=1^infty -x^2n ln x$Limit involving complicated integralShowing a limit does not exist using Cauchy's $epsilon, delta $ limit definitionCalculating $lim_xrightarrow +infty x sin(x) + cos(x) - x^2 $Limit of a function with exponentiationLimit of a LimitCalculating a Squeeze Theorem limitOn the way finding limit.Calculating limit for sequence that tends to infinity

3

1

$begingroup$

What's $displaystylelim_r to infty fracint_0^pi/2left(fracxpi/2right)^r-1cos x,dxint_0^pi/2left(fracxpi/2right)^rcosx,dx $?

I got the limit as part of Putnam 2011 A3. I know I'm supposed to chop $[0, fracpi2]$ in $[0, fracpi2- epsilon], [ fracpi2- epsilon, fracpi2]$ and the contribution from the first part is very small but I don't know in which direction you approach $(epsilon, r) mapsto (0, infty)$ to make it work properly.

real-analysis calculus

share|cite|improve this question

edited Mar 15 at 8:24

egreg

184k1486206

asked Mar 15 at 4:16

alxchenalxchen

639421

$endgroup$

add a comment | 

3

1

$begingroup$

What's $displaystylelim_r to infty fracint_0^pi/2left(fracxpi/2right)^r-1cos x,dxint_0^pi/2left(fracxpi/2right)^rcosx,dx $?

I got the limit as part of Putnam 2011 A3. I know I'm supposed to chop $[0, fracpi2]$ in $[0, fracpi2- epsilon], [ fracpi2- epsilon, fracpi2]$ and the contribution from the first part is very small but I don't know in which direction you approach $(epsilon, r) mapsto (0, infty)$ to make it work properly.

real-analysis calculus

share|cite|improve this question

edited Mar 15 at 8:24

egreg

184k1486206

asked Mar 15 at 4:16

alxchenalxchen

639421

$endgroup$

add a comment | 

$begingroup$

What's $displaystylelim_r to infty fracint_0^pi/2left(fracxpi/2right)^r-1cos x,dxint_0^pi/2left(fracxpi/2right)^rcosx,dx $?

I got the limit as part of Putnam 2011 A3. I know I'm supposed to chop $[0, fracpi2]$ in $[0, fracpi2- epsilon], [ fracpi2- epsilon, fracpi2]$ and the contribution from the first part is very small but I don't know in which direction you approach $(epsilon, r) mapsto (0, infty)$ to make it work properly.

real-analysis calculus

share|cite|improve this question

edited Mar 15 at 8:24

egreg

184k1486206

asked Mar 15 at 4:16

alxchenalxchen

639421

$endgroup$

share|cite|improve this question

edited Mar 15 at 8:24

egreg

184k1486206

asked Mar 15 at 4:16

alxchenalxchen

639421

share|cite|improve this question

edited Mar 15 at 8:24

egreg

184k1486206

asked Mar 15 at 4:16

alxchenalxchen

639421

edited Mar 15 at 8:24

egreg

184k1486206

edited Mar 15 at 8:24

egreg

184k1486206

asked Mar 15 at 4:16

alxchenalxchen

639421

asked Mar 15 at 4:16

alxchenalxchen

639421

1 Answer
1

active

oldest

votes

1

$begingroup$

Let $x=pi y/2$ to see the integral downstairs equals

$$tag 1 (pi/2) int_0^1 y^rcos (pi y/2),dy.$$

Integrate by parts twice to see $(1)$ equals

$$frac(pi/2)^2(r+1)(r+2) -, (pi/2)^3cdotint_0^1 fracy^r+2(r+1)(r+2)cos (pi y/2), dy.$$

Since $cos (pi y/2)le 1,$ the last integral is majorized by

$$int_0^1 fracy^r+2(r+1)(r+2), dy=O(1/r^3).$$

Thus $(1)$ equals $(pi/2)^2/[(r+1)(r+2)] +O(1/r^3).$ Similarly the upstairs integral equals $(pi/2)^2/[r(r+1)] +O(1/r^3).$ Dividing top by bottom and letting $rto infty$ then gives a limit of $1.$

share|cite|improve this answer

answered Mar 15 at 16:39

zhw.zhw.

74.6k43175

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Good answer. I deleted the hint -- did not see the obvious $cos fracpi2 = 0$
  $endgroup$
  – RRL
  Mar 15 at 17:02
  

  
  
  
  

add a comment | 

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1

$begingroup$

Let $x=pi y/2$ to see the integral downstairs equals

$$tag 1 (pi/2) int_0^1 y^rcos (pi y/2),dy.$$

Integrate by parts twice to see $(1)$ equals

$$frac(pi/2)^2(r+1)(r+2) -, (pi/2)^3cdotint_0^1 fracy^r+2(r+1)(r+2)cos (pi y/2), dy.$$

Since $cos (pi y/2)le 1,$ the last integral is majorized by

$$int_0^1 fracy^r+2(r+1)(r+2), dy=O(1/r^3).$$

Thus $(1)$ equals $(pi/2)^2/[(r+1)(r+2)] +O(1/r^3).$ Similarly the upstairs integral equals $(pi/2)^2/[r(r+1)] +O(1/r^3).$ Dividing top by bottom and letting $rto infty$ then gives a limit of $1.$

share|cite|improve this answer

answered Mar 15 at 16:39

zhw.zhw.

74.6k43175

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Good answer. I deleted the hint -- did not see the obvious $cos fracpi2 = 0$
  $endgroup$
  – RRL
  Mar 15 at 17:02
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $x=pi y/2$ to see the integral downstairs equals

$$tag 1 (pi/2) int_0^1 y^rcos (pi y/2),dy.$$

Integrate by parts twice to see $(1)$ equals

$$frac(pi/2)^2(r+1)(r+2) -, (pi/2)^3cdotint_0^1 fracy^r+2(r+1)(r+2)cos (pi y/2), dy.$$

Since $cos (pi y/2)le 1,$ the last integral is majorized by

$$int_0^1 fracy^r+2(r+1)(r+2), dy=O(1/r^3).$$

Thus $(1)$ equals $(pi/2)^2/[(r+1)(r+2)] +O(1/r^3).$ Similarly the upstairs integral equals $(pi/2)^2/[r(r+1)] +O(1/r^3).$ Dividing top by bottom and letting $rto infty$ then gives a limit of $1.$

share|cite|improve this answer

answered Mar 15 at 16:39

zhw.zhw.

74.6k43175

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Good answer. I deleted the hint -- did not see the obvious $cos fracpi2 = 0$
  $endgroup$
  – RRL
  Mar 15 at 17:02
  

  
  
  
  

add a comment | 

$begingroup$

Let $x=pi y/2$ to see the integral downstairs equals

$$tag 1 (pi/2) int_0^1 y^rcos (pi y/2),dy.$$

Integrate by parts twice to see $(1)$ equals

$$frac(pi/2)^2(r+1)(r+2) -, (pi/2)^3cdotint_0^1 fracy^r+2(r+1)(r+2)cos (pi y/2), dy.$$

Since $cos (pi y/2)le 1,$ the last integral is majorized by

$$int_0^1 fracy^r+2(r+1)(r+2), dy=O(1/r^3).$$

Thus $(1)$ equals $(pi/2)^2/[(r+1)(r+2)] +O(1/r^3).$ Similarly the upstairs integral equals $(pi/2)^2/[r(r+1)] +O(1/r^3).$ Dividing top by bottom and letting $rto infty$ then gives a limit of $1.$

share|cite|improve this answer

answered Mar 15 at 16:39

zhw.zhw.

74.6k43175

$endgroup$

share|cite|improve this answer

answered Mar 15 at 16:39

zhw.zhw.

74.6k43175

answered Mar 15 at 16:39

zhw.zhw.

74.6k43175

answered Mar 15 at 16:39

zhw.zhw.

74.6k43175