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Calculate $lim_{ntoinfty}int_0^nfrac{mathrm{d}x}{n+n^2sinfrac{x}{n^2}}$

Solving $int_0^{infty} sin(ax^2)sin(b/x^2),mathrm dx$Is $int_0^infty left (int_0^infty f(k) k sin kr , mathrm dk right) mathrm dr = int_0^infty f(k) , mathrm dk$ correct?calculate $lim_{ntoinfty}int_{[0,infty)} exp(-x)sin(nx),mathrm{d}mathcal{L}^1(x)$Evaluating $int_0^{infty} frac{xi x^{alpha}}{ e^{x}-xi} :mathrm{d}x$Computing $lim_{epsilon rightarrow 0} int_0^infty frac{sin x}{x} arctan{frac{x}{epsilon}}dx$Evaluating $lim_{nrightarrow infty}int_0^infty frac{e^{sin(ln(x))}}{1+sqrt nx^{42}}dx $Finding $lim_{ntoinfty}int_0^{infty}frac{x^{n-2}}{1+x^n}|sin(nx)|$.Putting limit inside an integral to calculate $int_0^infty frac{sin x}{x}dx = frac{pi}{2}$Computing $limlimits_{n to infty} int_0^{frac{pi}{2}}{frac{(sin(x))^{n}}{1-sin{(x)}},mathrm{d}x} $$lim_{arightarrow 0} int_0^1 frac{1}{1+asin(x)} , mathrm{d}x$

1

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$begingroup$

Calculate $$lim_{ntoinfty}int_0^nfrac{mathrm{d}x}{n+n^2sinfrac{x}{n^2}}$$, well first of all the function isn't monotonic, so I cant use the Lebesgue therem to go with the limit under the integral. What I could do is to express $sinx=x+O(x^3)$ then $$lim_{ntoinfty}int_0^nfrac{mathrm{d}x}{n+n^2sinfrac{x}{n^2}}=lim_{ntoinfty}int_0^nfrac{mathrm{d}x}{n+x}=lim_{ntoinfty} ln2n-ln n=ln2$$, but i am not sure if this is correct and also would like to know why this is correct, as of course I ate the $O(x^3)$ when approximating sin function.

real-analysis calculus integration analysis improper-integrals

share|cite|improve this question

edited Mar 20 at 17:58

Rebellos

15.7k31250

asked Mar 20 at 17:54

ryszard egginkryszard eggink

415110

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Double limit always needs extra care. Although your approach is totally justifiable, it would be easier to first substitute $x=nu$ and the pass limit to the inside of the integral.
  $endgroup$
  – Sangchul Lee
  Mar 20 at 18:00
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  nice suggestion!
  $endgroup$
  – ryszard eggink
  Mar 20 at 18:02
  

  
  
  
  

add a comment | 

1

1

$begingroup$

Calculate $$lim_{ntoinfty}int_0^nfrac{mathrm{d}x}{n+n^2sinfrac{x}{n^2}}$$, well first of all the function isn't monotonic, so I cant use the Lebesgue therem to go with the limit under the integral. What I could do is to express $sinx=x+O(x^3)$ then $$lim_{ntoinfty}int_0^nfrac{mathrm{d}x}{n+n^2sinfrac{x}{n^2}}=lim_{ntoinfty}int_0^nfrac{mathrm{d}x}{n+x}=lim_{ntoinfty} ln2n-ln n=ln2$$, but i am not sure if this is correct and also would like to know why this is correct, as of course I ate the $O(x^3)$ when approximating sin function.

real-analysis calculus integration analysis improper-integrals

share|cite|improve this question

edited Mar 20 at 17:58

Rebellos

15.7k31250

asked Mar 20 at 17:54

ryszard egginkryszard eggink

415110

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Double limit always needs extra care. Although your approach is totally justifiable, it would be easier to first substitute $x=nu$ and the pass limit to the inside of the integral.
  $endgroup$
  – Sangchul Lee
  Mar 20 at 18:00
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  nice suggestion!
  $endgroup$
  – ryszard eggink
  Mar 20 at 18:02
  

  
  
  
  

add a comment | 

$begingroup$

Calculate $$lim_{ntoinfty}int_0^nfrac{mathrm{d}x}{n+n^2sinfrac{x}{n^2}}$$, well first of all the function isn't monotonic, so I cant use the Lebesgue therem to go with the limit under the integral. What I could do is to express $sinx=x+O(x^3)$ then $$lim_{ntoinfty}int_0^nfrac{mathrm{d}x}{n+n^2sinfrac{x}{n^2}}=lim_{ntoinfty}int_0^nfrac{mathrm{d}x}{n+x}=lim_{ntoinfty} ln2n-ln n=ln2$$, but i am not sure if this is correct and also would like to know why this is correct, as of course I ate the $O(x^3)$ when approximating sin function.

real-analysis calculus integration analysis improper-integrals

share|cite|improve this question

edited Mar 20 at 17:58

Rebellos

15.7k31250

asked Mar 20 at 17:54

ryszard egginkryszard eggink

415110

$endgroup$

share|cite|improve this question

edited Mar 20 at 17:58

Rebellos

15.7k31250

asked Mar 20 at 17:54

ryszard egginkryszard eggink

415110

share|cite|improve this question

edited Mar 20 at 17:58

Rebellos

15.7k31250

asked Mar 20 at 17:54

ryszard egginkryszard eggink

415110

edited Mar 20 at 17:58

Rebellos

15.7k31250

edited Mar 20 at 17:58

Rebellos

15.7k31250

asked Mar 20 at 17:54

ryszard egginkryszard eggink

415110

asked Mar 20 at 17:54

ryszard egginkryszard eggink

415110

1 Answer
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3

$begingroup$

Hint

Since $u-{u^3over 6}le sin ule u$ for $0le u<<1$ (small enough $u$) you can write $${1over n+x}lefrac{1}{n+n^2sinfrac{x}{n^2}}le frac{1}{n+x-{x^3over n^4}}le{1over n+x-{1over n}}$$and integrate the sides.

share|cite|improve this answer

answered Mar 20 at 18:13

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

add a comment | 

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3

$begingroup$

Hint

Since $u-{u^3over 6}le sin ule u$ for $0le u<<1$ (small enough $u$) you can write $${1over n+x}lefrac{1}{n+n^2sinfrac{x}{n^2}}le frac{1}{n+x-{x^3over n^4}}le{1over n+x-{1over n}}$$and integrate the sides.

share|cite|improve this answer

answered Mar 20 at 18:13

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

add a comment | 

3

$begingroup$

Hint

Since $u-{u^3over 6}le sin ule u$ for $0le u<<1$ (small enough $u$) you can write $${1over n+x}lefrac{1}{n+n^2sinfrac{x}{n^2}}le frac{1}{n+x-{x^3over n^4}}le{1over n+x-{1over n}}$$and integrate the sides.

share|cite|improve this answer

answered Mar 20 at 18:13

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

add a comment | 

$begingroup$

Hint

Since $u-{u^3over 6}le sin ule u$ for $0le u<<1$ (small enough $u$) you can write $${1over n+x}lefrac{1}{n+n^2sinfrac{x}{n^2}}le frac{1}{n+x-{x^3over n^4}}le{1over n+x-{1over n}}$$and integrate the sides.

share|cite|improve this answer

answered Mar 20 at 18:13

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

share|cite|improve this answer

answered Mar 20 at 18:13

Mostafa AyazMostafa Ayaz

18.1k31040

answered Mar 20 at 18:13

Mostafa AyazMostafa Ayaz

18.1k31040

answered Mar 20 at 18:13

Mostafa AyazMostafa Ayaz

18.1k31040