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Compute integral: $lim_n to +infty int_1^+ infty fraccos^nxx^2dx$

How can I compute the integral $int_0^infty fracdt1+t^4$?Compute $limlimits_ntoinfty int_0^2pi cos x cos 2xcdots cos nx spacedx$Why does $lim_lambda to infty fraccos(lambda x) - cos(lambda y)lambda = 0$How to prove that $lim_Rrightarrowinftyint_0^pifrace^iaRe^ithetaiRe^ithetadthetab^2+R^2e^2itheta=0$Integral $int_-infty^infty fracsin^2xx^2 e^ixdx$How do I find $int_0^infty fracsin^4 xx^2,dx$?Evaluate $int_-infty^inftyfraccos x+x sin xx^2+cos^2x dx$Solving $int_0^fracpi2 e^a cos(x) dx$Evaluating $lim_nrightarrowinfty int_0^pi frac sin x1+ cos^2 (nx) dx$$lim_nto infty prod_k=1^n left( frac 2k2k-1right) int_-1^infty frac (cos x)^2n2^x dx$

2

$begingroup$

I am pretty sure that the answer to $$lim_n to +infty int_1^+ infty fraccos^nxx^2dx$$
is 0, but I am stuck in how to write a good proof, since I found it difficult to discuss the combination of two infinities.
I hope that you can help me solve the problem!

calculus integration

share|cite|improve this question

edited Mar 14 at 17:13

Bernard

123k741117

asked Mar 14 at 16:39

VicSiriusVicSirius

443

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Did you hear of dominated convergence theorem?
  $endgroup$
  – Shashi
  Mar 14 at 17:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Shashi AH, I haven't heard it before..
  $endgroup$
  – VicSirius
  Mar 15 at 1:09
  

  
  
  
  

add a comment | 

2

$begingroup$

I am pretty sure that the answer to $$lim_n to +infty int_1^+ infty fraccos^nxx^2dx$$
is 0, but I am stuck in how to write a good proof, since I found it difficult to discuss the combination of two infinities.
I hope that you can help me solve the problem!

calculus integration

share|cite|improve this question

edited Mar 14 at 17:13

Bernard

123k741117

asked Mar 14 at 16:39

VicSiriusVicSirius

443

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Did you hear of dominated convergence theorem?
  $endgroup$
  – Shashi
  Mar 14 at 17:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Shashi AH, I haven't heard it before..
  $endgroup$
  – VicSirius
  Mar 15 at 1:09
  

  
  
  
  

add a comment | 

$begingroup$

I am pretty sure that the answer to $$lim_n to +infty int_1^+ infty fraccos^nxx^2dx$$
is 0, but I am stuck in how to write a good proof, since I found it difficult to discuss the combination of two infinities.
I hope that you can help me solve the problem!

calculus integration

share|cite|improve this question

edited Mar 14 at 17:13

Bernard

123k741117

asked Mar 14 at 16:39

VicSiriusVicSirius

443

$endgroup$

share|cite|improve this question

edited Mar 14 at 17:13

Bernard

123k741117

asked Mar 14 at 16:39

VicSiriusVicSirius

443

share|cite|improve this question

edited Mar 14 at 17:13

Bernard

123k741117

asked Mar 14 at 16:39

VicSiriusVicSirius

443

edited Mar 14 at 17:13

Bernard

123k741117

edited Mar 14 at 17:13

Bernard

123k741117

asked Mar 14 at 16:39

VicSiriusVicSirius

443

asked Mar 14 at 16:39

VicSiriusVicSirius

443

1 Answer
1

active

oldest

votes

4

$begingroup$

Dominated convergence gives a one-liner proof. Since the integrand $x mapsto cos^n(x)/x^2$ is bounded by $1/x^2$ uniformly in $n$ and it converges to $0$ almost everywhere, we have

$$ lim_ntoinfty int_1^infty fraccos^n xx^2 , mathrmdx
= int_1^infty lim_ntoinfty fraccos^n xx^2 , mathrmdx
= int_1^infty 0 , mathrmdx
= 0. $$

---

If this fancy technique is not available yet, then still we can prove the claim. Let

$$ I_n = int_0^2pi |cos^n x| , mathrmdx = 4 int_0^pi/2 cos^n x , mathrmdx.$$

It is not hard to show that $I_n to 0$ as $ntoinfty$. This can be done in several ways, but here we discuss the following simple trick: substitute $sin x = fracssqrtn$. Then

$$ I_n = frac4sqrtn int_0^sqrtn left( 1 - fracs^2n right)^fracn-12 , mathrmds,$$

and using the inequality $1-x leq e^-x$ which holds for all $x inmathbbR$,

$$ I_n leq frac4sqrtn int_0^sqrtn e^-fracn-12ns^2 , mathrmds. $$

Now it is clear that the integral $int_0^sqrtn e^-fracn-12ns^2 , mathrmds$ is uniformly bounded in $n$, and so, this proves that $I_n leq textconst. / sqrtn$. So $I_n to 0$.

Next, we bound the original integral using $I_n$ as follows:

beginalign*
left| int_1^infty fraccos^n xx^2 , mathrmdx right|
&leq int_1^infty fracx^2 , mathrmdx \
&= sum_k=1^infty int_2pi(k-1) + 1^2pi k + 1 fracx^2 , mathrmdx \
&leq sum_k=1^infty frac1(2pi(k-1) + 1)^2 int_2pi(k-1) + 1^2pi k + 1 |cos^n x| , mathrmdx \
&= C I_n.
endalign*

Here, $C = sum_k=1^infty frac1(2pi(k-1) + 1)^2$ is a finite constant. Now since $I_n to 0$, it follows that the integral converges to $0$ as $ntoinfty$.

share|cite|improve this answer

edited Mar 17 at 7:30

answered Mar 14 at 17:27

Sangchul LeeSangchul Lee

96.2k12171281

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1 for the non-fancy technique in second part.
  $endgroup$
  – Paramanand Singh
  Mar 17 at 8:07
  

  
  
  
  

add a comment | 

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4

$begingroup$

Dominated convergence gives a one-liner proof. Since the integrand $x mapsto cos^n(x)/x^2$ is bounded by $1/x^2$ uniformly in $n$ and it converges to $0$ almost everywhere, we have

$$ lim_ntoinfty int_1^infty fraccos^n xx^2 , mathrmdx
= int_1^infty lim_ntoinfty fraccos^n xx^2 , mathrmdx
= int_1^infty 0 , mathrmdx
= 0. $$

---

If this fancy technique is not available yet, then still we can prove the claim. Let

$$ I_n = int_0^2pi |cos^n x| , mathrmdx = 4 int_0^pi/2 cos^n x , mathrmdx.$$

It is not hard to show that $I_n to 0$ as $ntoinfty$. This can be done in several ways, but here we discuss the following simple trick: substitute $sin x = fracssqrtn$. Then

$$ I_n = frac4sqrtn int_0^sqrtn left( 1 - fracs^2n right)^fracn-12 , mathrmds,$$

and using the inequality $1-x leq e^-x$ which holds for all $x inmathbbR$,

$$ I_n leq frac4sqrtn int_0^sqrtn e^-fracn-12ns^2 , mathrmds. $$

Now it is clear that the integral $int_0^sqrtn e^-fracn-12ns^2 , mathrmds$ is uniformly bounded in $n$, and so, this proves that $I_n leq textconst. / sqrtn$. So $I_n to 0$.

Next, we bound the original integral using $I_n$ as follows:

beginalign*
left| int_1^infty fraccos^n xx^2 , mathrmdx right|
&leq int_1^infty fracx^2 , mathrmdx \
&= sum_k=1^infty int_2pi(k-1) + 1^2pi k + 1 fracx^2 , mathrmdx \
&leq sum_k=1^infty frac1(2pi(k-1) + 1)^2 int_2pi(k-1) + 1^2pi k + 1 |cos^n x| , mathrmdx \
&= C I_n.
endalign*

Here, $C = sum_k=1^infty frac1(2pi(k-1) + 1)^2$ is a finite constant. Now since $I_n to 0$, it follows that the integral converges to $0$ as $ntoinfty$.

share|cite|improve this answer

edited Mar 17 at 7:30

answered Mar 14 at 17:27

Sangchul LeeSangchul Lee

96.2k12171281

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1 for the non-fancy technique in second part.
  $endgroup$
  – Paramanand Singh
  Mar 17 at 8:07
  

  
  
  
  

add a comment | 

4

$begingroup$

Dominated convergence gives a one-liner proof. Since the integrand $x mapsto cos^n(x)/x^2$ is bounded by $1/x^2$ uniformly in $n$ and it converges to $0$ almost everywhere, we have

$$ lim_ntoinfty int_1^infty fraccos^n xx^2 , mathrmdx
= int_1^infty lim_ntoinfty fraccos^n xx^2 , mathrmdx
= int_1^infty 0 , mathrmdx
= 0. $$

---

If this fancy technique is not available yet, then still we can prove the claim. Let

$$ I_n = int_0^2pi |cos^n x| , mathrmdx = 4 int_0^pi/2 cos^n x , mathrmdx.$$

It is not hard to show that $I_n to 0$ as $ntoinfty$. This can be done in several ways, but here we discuss the following simple trick: substitute $sin x = fracssqrtn$. Then

$$ I_n = frac4sqrtn int_0^sqrtn left( 1 - fracs^2n right)^fracn-12 , mathrmds,$$

and using the inequality $1-x leq e^-x$ which holds for all $x inmathbbR$,

$$ I_n leq frac4sqrtn int_0^sqrtn e^-fracn-12ns^2 , mathrmds. $$

Now it is clear that the integral $int_0^sqrtn e^-fracn-12ns^2 , mathrmds$ is uniformly bounded in $n$, and so, this proves that $I_n leq textconst. / sqrtn$. So $I_n to 0$.

Next, we bound the original integral using $I_n$ as follows:

beginalign*
left| int_1^infty fraccos^n xx^2 , mathrmdx right|
&leq int_1^infty fracx^2 , mathrmdx \
&= sum_k=1^infty int_2pi(k-1) + 1^2pi k + 1 fracx^2 , mathrmdx \
&leq sum_k=1^infty frac1(2pi(k-1) + 1)^2 int_2pi(k-1) + 1^2pi k + 1 |cos^n x| , mathrmdx \
&= C I_n.
endalign*

Here, $C = sum_k=1^infty frac1(2pi(k-1) + 1)^2$ is a finite constant. Now since $I_n to 0$, it follows that the integral converges to $0$ as $ntoinfty$.

share|cite|improve this answer

edited Mar 17 at 7:30

answered Mar 14 at 17:27

Sangchul LeeSangchul Lee

96.2k12171281

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  +1 for the non-fancy technique in second part.
  $endgroup$
  – Paramanand Singh
  Mar 17 at 8:07
  

  
  
  
  

add a comment | 

$begingroup$

Dominated convergence gives a one-liner proof. Since the integrand $x mapsto cos^n(x)/x^2$ is bounded by $1/x^2$ uniformly in $n$ and it converges to $0$ almost everywhere, we have

$$ lim_ntoinfty int_1^infty fraccos^n xx^2 , mathrmdx
= int_1^infty lim_ntoinfty fraccos^n xx^2 , mathrmdx
= int_1^infty 0 , mathrmdx
= 0. $$

---

If this fancy technique is not available yet, then still we can prove the claim. Let

$$ I_n = int_0^2pi |cos^n x| , mathrmdx = 4 int_0^pi/2 cos^n x , mathrmdx.$$

It is not hard to show that $I_n to 0$ as $ntoinfty$. This can be done in several ways, but here we discuss the following simple trick: substitute $sin x = fracssqrtn$. Then

$$ I_n = frac4sqrtn int_0^sqrtn left( 1 - fracs^2n right)^fracn-12 , mathrmds,$$

and using the inequality $1-x leq e^-x$ which holds for all $x inmathbbR$,

$$ I_n leq frac4sqrtn int_0^sqrtn e^-fracn-12ns^2 , mathrmds. $$

Now it is clear that the integral $int_0^sqrtn e^-fracn-12ns^2 , mathrmds$ is uniformly bounded in $n$, and so, this proves that $I_n leq textconst. / sqrtn$. So $I_n to 0$.

Next, we bound the original integral using $I_n$ as follows:

beginalign*
left| int_1^infty fraccos^n xx^2 , mathrmdx right|
&leq int_1^infty fracx^2 , mathrmdx \
&= sum_k=1^infty int_2pi(k-1) + 1^2pi k + 1 fracx^2 , mathrmdx \
&leq sum_k=1^infty frac1(2pi(k-1) + 1)^2 int_2pi(k-1) + 1^2pi k + 1 |cos^n x| , mathrmdx \
&= C I_n.
endalign*

Here, $C = sum_k=1^infty frac1(2pi(k-1) + 1)^2$ is a finite constant. Now since $I_n to 0$, it follows that the integral converges to $0$ as $ntoinfty$.

share|cite|improve this answer

edited Mar 17 at 7:30

answered Mar 14 at 17:27

Sangchul LeeSangchul Lee

96.2k12171281

$endgroup$

share|cite|improve this answer

edited Mar 17 at 7:30

answered Mar 14 at 17:27

Sangchul LeeSangchul Lee

96.2k12171281

answered Mar 14 at 17:27

Sangchul LeeSangchul Lee

96.2k12171281

answered Mar 14 at 17:27

Sangchul LeeSangchul Lee

96.2k12171281