Dtjytyk

Processor speed limited at 0.4 Ghz

Notepad++ delete until colon for every line with replace all

How can a day be of 24 hours?

How to prevent "they're falling in love" trope

How does a dynamic QR code work?

Finitely generated matrix groups whose eigenvalues are all algebraic

What Exploit Are These User Agents Trying to Use?

Where would I need my direct neural interface to be implanted?

how do we prove that a sum of two periods is still a period?

How seriously should I take size and weight limits of hand luggage?

Why do I get negative height?

Is it a bad idea to plug the other end of ESD strap to wall ground?

Knowledge-based authentication using Domain-driven Design in C#

Rotate ASCII Art by 45 Degrees

ssTTsSTtRrriinInnnnNNNIiinngg

Machine learning testing data

Fair gambler's ruin problem intuition

Was the old ablative pronoun "med" or "mēd"?

Placement of More Information/Help Icon button for Radio Buttons

What are the G forces leaving Earth orbit?

What exactly is ineptocracy?

Forgetting the musical notes while performing in concert

What historical events would have to change in order to make 19th century "steampunk" technology possible?

What is an equivalently powerful replacement spell for the Yuan-Ti's Suggestion spell?

Find the values of a>0 for which the improper integral $int_{0}^{infty}frac{sin x}{x^{a}} $ converges . [duplicate]

Convergence $I=int_0^infty frac{sin x}{x^s}dx$Improper Integral:$int_{0}^{+infty}frac{sin x}{x+sin x}dx$Improper integral $sin(x)/x $ converges absolutely, conditionaly or diverges?For which values is this improper integral convergent?Does the improper integral $int_{0}^{infty}sin(x^2);mathrm dx$ converge?Improper integral: $int_0^infty frac{sin^4x}{x^2}dx$Determining for which positive values of $p$ the improper integral $int_1^{infty}frac{dx}{e^{px} ln{x}}$ convergesImproper integral - checking convergence of $int_{1}^{infty} x^2 sin(x^4) dx$Can't solve Improper Integral $int_{0}^{infty} frac{sqrt{x}sin(x)}{1+x^2} dx$Find all the values of $alpha$ and $beta$ for which the following integral convergesFor what values of ''$a$'' , does this Improper Integral converges?

-4

$begingroup$

This question already has an answer here:

• 
  Convergence $I=int_0^infty frac{sin x}{x^s}dx$ [duplicate]
  
  5 answers
  
  

Find the values of a>0 for which the improper integral $int_{0}^{infty}frac{sin x}{x^{a}} $ converges .
Do I have to expand integrand using series expansion??

improper-integrals

share|cite|improve this question

edited Mar 18 at 13:54

Bernard

124k741118

asked Mar 18 at 13:47

sejysejy

1589

$endgroup$

marked as duplicate by mrtaurho, uniquesolution, John Omielan, StubbornAtom, Crostul Mar 18 at 20:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not necessarily.
  $endgroup$
  – uniquesolution
  Mar 18 at 13:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Tell me another way !!
  $endgroup$
  – sejy
  Mar 18 at 13:55
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @sejy you shouldn't use this tone against people that you want help from for free
  $endgroup$
  – Henry Lee
  Mar 18 at 14:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/793595/…
  $endgroup$
  – StubbornAtom
  Mar 18 at 14:29
  

  
  
  
  

add a comment | 

-4

$begingroup$

This question already has an answer here:

• 
  Convergence $I=int_0^infty frac{sin x}{x^s}dx$ [duplicate]
  
  5 answers
  
  

Find the values of a>0 for which the improper integral $int_{0}^{infty}frac{sin x}{x^{a}} $ converges .
Do I have to expand integrand using series expansion??

improper-integrals

share|cite|improve this question

edited Mar 18 at 13:54

Bernard

124k741118

asked Mar 18 at 13:47

sejysejy

1589

$endgroup$

marked as duplicate by mrtaurho, uniquesolution, John Omielan, StubbornAtom, Crostul Mar 18 at 20:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not necessarily.
  $endgroup$
  – uniquesolution
  Mar 18 at 13:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Tell me another way !!
  $endgroup$
  – sejy
  Mar 18 at 13:55
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @sejy you shouldn't use this tone against people that you want help from for free
  $endgroup$
  – Henry Lee
  Mar 18 at 14:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/793595/…
  $endgroup$
  – StubbornAtom
  Mar 18 at 14:29
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Convergence $I=int_0^infty frac{sin x}{x^s}dx$ [duplicate]
  
  5 answers
  
  

Find the values of a>0 for which the improper integral $int_{0}^{infty}frac{sin x}{x^{a}} $ converges .
Do I have to expand integrand using series expansion??

improper-integrals

share|cite|improve this question

edited Mar 18 at 13:54

Bernard

124k741118

asked Mar 18 at 13:47

sejysejy

1589

$endgroup$

share|cite|improve this question

edited Mar 18 at 13:54

Bernard

124k741118

asked Mar 18 at 13:47

sejysejy

1589

share|cite|improve this question

edited Mar 18 at 13:54

Bernard

124k741118

asked Mar 18 at 13:47

sejysejy

1589

edited Mar 18 at 13:54

Bernard

124k741118

edited Mar 18 at 13:54

Bernard

124k741118

asked Mar 18 at 13:47

sejysejy

1589

asked Mar 18 at 13:47

sejysejy

1589

2 Answers
2

active

oldest

votes

2

$begingroup$

For $a ge 2$ the singularity at $0$ is non-integrable. For $1 < a < 2$ it's absolutely integrable both at $0$ and at $infty$. For $0 < a le 1$ the series $sum_{n=1}^infty int_{npi}^{(n+1)pi} frac{sin(x)}{x^a}; dx$ is an alternating series.

share|cite|improve this answer

answered Mar 18 at 14:13

Robert IsraelRobert Israel

330k23219473

$endgroup$

add a comment | 

0

$begingroup$

We know that $-1lesin(x)le 1$ and so we can approximate the integral with:
$$int_0^inftyfrac{sin(x)}{x^a}dxleint_0^inftyfrac{1}{x^a}$$
and this integral is convergent for $a>1$ so this narrows our region down to somewhere in $0<a<1$

share|cite|improve this answer

answered Mar 18 at 14:11

Henry LeeHenry Lee

2,189319

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  It converges also for $alpha=1$
  $endgroup$
  – HAMIDINE SOUMARE
  Mar 18 at 14:16
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ Henry Lee You should better consider absolute values in your inequalitiy: $| int_0^infty frac{sin(x)}{x^alpha},dx | le int_0^infty |frac{sin(x)}{x^alpha}|,dxle int_0^infty frac{1}{x^alpha},dx$
  $endgroup$
  – Dr. Wolfgang Hintze
  Mar 18 at 14:50
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For general $alpha$ the integral defines the function $cos left(frac{pi alpha}{2}right) Gamma (1-alpha)$ by analytic continuation.
  $endgroup$
  – Dr. Wolfgang Hintze
  Mar 18 at 14:53
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

For $a ge 2$ the singularity at $0$ is non-integrable. For $1 < a < 2$ it's absolutely integrable both at $0$ and at $infty$. For $0 < a le 1$ the series $sum_{n=1}^infty int_{npi}^{(n+1)pi} frac{sin(x)}{x^a}; dx$ is an alternating series.

share|cite|improve this answer

answered Mar 18 at 14:13

Robert IsraelRobert Israel

330k23219473

$endgroup$

add a comment | 

2

$begingroup$

For $a ge 2$ the singularity at $0$ is non-integrable. For $1 < a < 2$ it's absolutely integrable both at $0$ and at $infty$. For $0 < a le 1$ the series $sum_{n=1}^infty int_{npi}^{(n+1)pi} frac{sin(x)}{x^a}; dx$ is an alternating series.

share|cite|improve this answer

answered Mar 18 at 14:13

Robert IsraelRobert Israel

330k23219473

$endgroup$

add a comment | 

$begingroup$

For $a ge 2$ the singularity at $0$ is non-integrable. For $1 < a < 2$ it's absolutely integrable both at $0$ and at $infty$. For $0 < a le 1$ the series $sum_{n=1}^infty int_{npi}^{(n+1)pi} frac{sin(x)}{x^a}; dx$ is an alternating series.

share|cite|improve this answer

answered Mar 18 at 14:13

Robert IsraelRobert Israel

330k23219473

$endgroup$

share|cite|improve this answer

answered Mar 18 at 14:13

Robert IsraelRobert Israel

330k23219473

answered Mar 18 at 14:13

Robert IsraelRobert Israel

330k23219473

answered Mar 18 at 14:13

Robert IsraelRobert Israel

330k23219473

0

$begingroup$

We know that $-1lesin(x)le 1$ and so we can approximate the integral with:
$$int_0^inftyfrac{sin(x)}{x^a}dxleint_0^inftyfrac{1}{x^a}$$
and this integral is convergent for $a>1$ so this narrows our region down to somewhere in $0<a<1$

share|cite|improve this answer

answered Mar 18 at 14:11

Henry LeeHenry Lee

2,189319

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  It converges also for $alpha=1$
  $endgroup$
  – HAMIDINE SOUMARE
  Mar 18 at 14:16
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ Henry Lee You should better consider absolute values in your inequalitiy: $| int_0^infty frac{sin(x)}{x^alpha},dx | le int_0^infty |frac{sin(x)}{x^alpha}|,dxle int_0^infty frac{1}{x^alpha},dx$
  $endgroup$
  – Dr. Wolfgang Hintze
  Mar 18 at 14:50
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For general $alpha$ the integral defines the function $cos left(frac{pi alpha}{2}right) Gamma (1-alpha)$ by analytic continuation.
  $endgroup$
  – Dr. Wolfgang Hintze
  Mar 18 at 14:53
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

We know that $-1lesin(x)le 1$ and so we can approximate the integral with:
$$int_0^inftyfrac{sin(x)}{x^a}dxleint_0^inftyfrac{1}{x^a}$$
and this integral is convergent for $a>1$ so this narrows our region down to somewhere in $0<a<1$

share|cite|improve this answer

answered Mar 18 at 14:11

Henry LeeHenry Lee

2,189319

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  It converges also for $alpha=1$
  $endgroup$
  – HAMIDINE SOUMARE
  Mar 18 at 14:16
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @ Henry Lee You should better consider absolute values in your inequalitiy: $| int_0^infty frac{sin(x)}{x^alpha},dx | le int_0^infty |frac{sin(x)}{x^alpha}|,dxle int_0^infty frac{1}{x^alpha},dx$
  $endgroup$
  – Dr. Wolfgang Hintze
  Mar 18 at 14:50
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For general $alpha$ the integral defines the function $cos left(frac{pi alpha}{2}right) Gamma (1-alpha)$ by analytic continuation.
  $endgroup$
  – Dr. Wolfgang Hintze
  Mar 18 at 14:53
  
  
  

  
  
  
  

add a comment | 

$begingroup$

We know that $-1lesin(x)le 1$ and so we can approximate the integral with:
$$int_0^inftyfrac{sin(x)}{x^a}dxleint_0^inftyfrac{1}{x^a}$$
and this integral is convergent for $a>1$ so this narrows our region down to somewhere in $0<a<1$

share|cite|improve this answer

answered Mar 18 at 14:11

Henry LeeHenry Lee

2,189319

$endgroup$

share|cite|improve this answer

answered Mar 18 at 14:11

Henry LeeHenry Lee

2,189319

answered Mar 18 at 14:11

Henry LeeHenry Lee

2,189319

answered Mar 18 at 14:11

Henry LeeHenry Lee

2,189319